Unit 3 - Work, Energy, and Power

1. Work, Energy, and Power

Title

AP Physics C: Mechanics — Unit 3

The dot product, the work-energy theorem, potential energy diagrams, and conservation as a shortcut past the journey

2. What this deck buys you

Objectives

Newton's laws are complete but expensive. They tell you the acceleration at every instant, so answering a question about the end of a motion means tracking every instant in between.

Energy methods skip the journey. They relate the starting state directly to the finishing state, and a great many problems that need a differential equation with forces need one line with energy.

• Compute work as a dot product, and as an integral when the force varies
• Use the work-energy theorem to get speeds without ever finding the acceleration
• Tell conservative from non-conservative forces and know why only the first kind gets a potential energy
• Apply conservation of mechanical energy, and extend it when friction is present
• Read a potential energy diagram for turning points, equilibria and stability
• Relate force to potential through a derivative, and compute power two different ways

One warning to carry through the deck. Energy is a scalar, which is what makes it so much easier than force — no components, no diagrams full of arrows. The price is that energy methods tell you nothing about direction, and there are questions they simply cannot answer.

Newton's laws are complete. Anything in this course could in principle be solved with them, and nothing in this deck contradicts them. So it is fair to ask what energy is actually for.

The answer is cost. Force methods give you the acceleration at every instant, which means answering a question about the end of a motion requires you to track every instant in between. When the force varies — a spring, a curved track, drag — that means a differential equation.

Energy methods skip the journey entirely. They relate the starting state directly to the finishing state, and a great many problems that need integration with forces need one line with energy. The loop-the-loop and the curved-ramp problems later in this deck are unanswerable with forces alone at this level, and take three lines with energy.

The price is worth knowing up front. Energy is a scalar, which is why there are no components and no vector diagrams — and also why energy can never tell you a direction, a time, or an acceleration. There are questions it simply cannot answer, and recognising them is part of the skill.

3. Work: Force Along a Displacement

Section

Section 1

4. Only the part of the force along the motion does work

Concept

\[ W = Fd\cos\theta \]

The cosine is not decoration. It selects the component of the force that lies along the displacement, and that is the only component that transfers energy.

A force perpendicular to the motion does exactly zero work, no matter how large it is. The normal force on a sliding block, the tension in a conical pendulum's string, and gravity on a horizontally moving object all do no work at all.

That is worth remembering as a labour-saving device. In an energy analysis, any force perpendicular to the motion can be ignored entirely — which is why circular motion problems often collapse to nothing when you look at them with energy.

The cosine in this formula is not a fudge factor. It is doing a specific job: selecting the component of the force that lies along the displacement, because that is the only component that transfers any energy.

The consequence to internalise is that a force perpendicular to the motion does exactly zero work, no matter how large it is. The normal force on a sliding block, the tension in a pendulum string, gravity on something moving horizontally — all of them do nothing at all, energetically speaking.

That is a labour-saving device rather than a technicality. In an energy analysis you can ignore every force perpendicular to the motion outright, which is why pendulum and circular-motion problems collapse so dramatically when you look at them with energy instead of forces.

It is also worth noticing what the picture shows about the perpendicular component. It is not doing nothing physically — it changes the normal force, and therefore the friction. It just does no work, which is a narrower claim than it sounds.

5. Reading the work formula

Notation

\[ W = F_x d_x + F_y d_y \]

The component form is often faster when the force is already given in unit-vector notation, and it is exactly equivalent. Multiply matching components and add — no angle needed.

Units: a newton-metre is a joule. One joule is roughly the work needed to lift an apple one metre, which is a useful benchmark for checking whether an answer is plausible.

Four details, and the second is the one that costs marks most often.

The angle is between the force and the displacement. Not between the force and the horizontal, not between the force and whichever axis you happened to draw. On an incline those are different angles, and using the wrong one is the single most common work error.

The dot means this is a dot product: two vectors go in and a scalar comes out. That is the formal reason energy problems have no components to track — the direction information is consumed by the dot product and does not survive into the answer.

The sign comes entirely from the cosine, and it is worth reading physically rather than arithmetically. Below ninety degrees means the force has a component along the motion, so energy is going into the object. Above ninety means energy is coming out.

And d is the displacement, not the distance. A constant force does zero net work on something that goes out and comes back, because the displacement is zero. Friction is different only because it is not constant in direction — it reverses when the motion does, which is exactly why it is non-conservative.

6. How big is a joule? Benchmarks worth carrying

Concept

Energy answers are easy to get wrong by a factor of a thousand and hard to notice. A few reference points make bad answers visible immediately.

Two habits follow. First, estimate before you compute: a block sliding across a table should have tens of joules, not thousands. Second, watch the units — a kilojoule is a thousand joules, and mixing them is the most common factor-of-1000 error.

The car entry is worth dwelling on. Seven hundred kilojoules is the energy released by dropping that car from a fifty-metre building, and a crash delivers it in about a tenth of a second. That is the whole physics of vehicle safety in one comparison.

Energy answers go wrong by factors of a thousand more often than by small amounts, and a factor of a thousand is invisible unless you have something to compare against.

The first two rows are the ones to memorise. Lifting an apple a metre is about a joule. Dropping a one kilogram book a metre releases about ten. Everything else in mechanics can be scaled from those.

The car entry deserves a moment because it explains something people find surprising. Seven hundred kilojoules is what you would get by dropping that car off a fifty-metre building, and a crash delivers all of it in about a tenth of a second. That is the entire physics of vehicle safety: the energy is fixed by the speed, and all a crumple zone can do is spread the delivery over more time and distance.

The bread entry is there for perspective. Chemical energy densities dwarf mechanical ones, which is why a small tank of petrol moves a tonne of metal for hundreds of kilometres, and why a battery that stores as much as a slice of bread is considered impressive.

The practical habit: estimate the order of magnitude before computing. A block sliding across a table should come out in the tens of joules. If your answer is in the millions, something is wrong and you will find it faster now than after three more lines of algebra.

7. Positive, negative, or zero work?

Sorting

In each case, decide what work the named force does on the named object.

Item g is the one worth remembering. A satellite in a circular orbit has a large force acting on it continuously, and that force does no work whatsoever — which is precisely why its speed never changes.

Sorting these is quick; the value is in noticing why the zero bucket has so many members.

Every item in it is a force perpendicular to the motion. That is the only way to get exactly zero work from a non-zero force, and it happens constantly — normal forces, string tensions in circular motion, gravity on horizontal motion.

Item g is the one to remember for the rest of the course. A satellite in a circular orbit has a very large force acting on it continuously, and that force does no work whatsoever. Which is precisely why the satellite's speed never changes: no work means no change in kinetic energy.

Read the positive and negative buckets as statements about energy flow rather than about direction. Positive work means energy is entering the object. Negative work means energy is leaving it. Friction always does negative work on the sliding object, which is another way of saying it always removes energy.

One subtlety worth flagging: the sign depends on which object you are asking about. Friction does negative work on a sliding box, but the box does positive work on the floor. Always name the object.

8. Holding something heavy is hard work, so you are doing work on it

Trap

9. When the force varies, work becomes an integral

Concept

The formula W = Fd assumes a constant force. When the force changes as the object moves — a spring, a rubber band, a rocket's thrust — you must add up the work in small pieces.

\[ W = \int_{x_i}^{x_f} F(x)\,dx \]

Read the picture rather than the symbol. Each strip is a tiny displacement over which the force is effectively constant, so its area is a tiny amount of work. The integral adds them all up.

This makes the area under a force-position graph the work done, with area below the axis counting as negative — the same signed-area idea as the velocity graph in the kinematics deck.

The formula W equals F d assumes the force is the same throughout the displacement. A spring, a rubber band, a rocket's thrust — none of them are, so that formula is simply unavailable.

Read the picture rather than the integral sign. Slice the displacement into pieces small enough that the force is effectively constant over each. Each slice contributes F times dx, which is a thin rectangle of area. Add them all up and you have the total work.

So the work is the area under the force-position graph, and area below the axis counts as negative. This is the same signed-area idea as the velocity graph in the kinematics deck, and the two are worth holding together: area under v against t is displacement, area under F against x is work.

The practical consequence is that a great many work problems require no integration at all, because the graph is made of triangles and rectangles. Only when the curve is genuinely curved do you need calculus, and even then the integral is usually elementary.

10. Work done in stretching a spring

Worked example

A spring obeys Hooke's law, F = -kx, where x is the displacement from the natural length. How much work must you do to stretch it from 0 to x?

\[ W = \int_0^{x} kx'\,dx' = \tfrac{1}{2}kx^2 \]

\[ W_{\text{spring}} = -\tfrac{1}{2}kx^2 \]

The factor of one half is worth understanding rather than memorising. It is there because the force starts at zero and builds — you are, in effect, applying the average force kx/2 over the distance x.

Two things to be careful about in this derivation, and one result worth understanding rather than memorising.

First, be clear about whose force you are integrating. The spring pulls back with minus kx; you push with plus kx. The work you do is positive and the work the spring does is negative, and problems will ask for either.

Second, the constant-force formula is unusable here because the force grows continuously from zero. This is the simplest possible example of a variable force, which is why it is the standard one.

Now the factor of one half. It is not a convention and it is not memorisation — it comes from integrating something linear. Physically, since the force rises steadily from zero to kx, you are effectively applying the average force kx over two across the distance x. That is exactly the area of the triangle on the graph.

Notice that same one half appears in the kinetic energy, for exactly the same reason: that came from integrating m v dv, which is also linear. Whenever you see a one half and a square together in this course, an integral of a linear function is behind it.

11. Work from a force-position graph

Worked example

A block is pushed along the x-axis by a force that varies as shown: constant at 8 N from 0 to 3 m, falling linearly to zero between 3 and 5 m, then constant at -4 N from 5 to 8 m. Find the total work.

\[ W_1 = (8)(3) = 24\,\text{J} \]

\[ W_2 = \tfrac{1}{2}(2)(8) = 8\,\text{J} \]

\[ W_3 = (-4)(3) = -12\,\text{J} \]

\[ W_{\text{total}} = 20\,\text{J} \]

The point of this example is that no integration is needed when the graph is piecewise linear, which is how most exam graphs are drawn.

Break the shape into rectangles and triangles, compute each area, and attach the sign according to which side of the axis it sits on. That is the whole method.

The third region is where marks are lost. The force is negative there, meaning it opposes the motion, so its work is negative and must be subtracted. It is easy to compute the area of that rectangle and add it, because areas feel like they should be positive.

The final total is positive, which tells you something physical without any further work: the block ends up faster than it started. If the total had been negative it would have ended slower, and if zero it would have returned to its original speed.

A useful cross-check on any graph problem: does the sign of the total work match what the picture suggests should happen to the object?

12. Two ramps to the same height

Prediction

This is the whole idea of a machine: a ramp, a lever or a pulley system lets you trade force against distance. It never reduces the work required — only the force.

The answer surprises people and it is the foundation of everything in Section 3, so it is worth being sure of the reasoning rather than just the result.

The gentle ramp requires a smaller force because less of the weight acts along the slope. But the distance along it is correspondingly longer. Those two effects do not merely roughly cancel — they cancel exactly, and the work comes out as mgh in both cases.

That exactness is not a coincidence of ramps; it is the defining property of a conservative force. The work depends only on the height gained, never on the route taken to gain it. Everything about potential energy follows from that.

The engineering statement of the same fact is worth knowing: this is what a machine does. A ramp, a lever, a pulley system or a gearbox lets you trade force against distance, and none of them ever reduces the work required. That is why there is no such thing as a machine that gets something for nothing.

One caveat the problem was careful about: both ramps are frictionless. With friction the longer ramp genuinely does cost more, because friction is path-dependent — which is the contrast the next section is built on.

13. Path independence, checked with numbers

Worked example

A 2.0 kg ball is thrown from the ground at 12 m/s and lands on a ledge 3.0 m up. Compute the work done by gravity two ways: along the actual parabola, and along a straight vertical-then-horizontal path.

\[ W_g = \vec{F}\cdot\vec{d} = (-mg\,\hat{\jmath})\cdot(d_x\hat{\imath} + 3.0\,\hat{\jmath}) = -mg(3.0) \]

\[ W_g = -(2.0)(9.8)(3.0) = -58.8\,\text{J} \]

\[ W_g = -mg(3.0) + 0 = -58.8\,\text{J} \]

\[ \tfrac{1}{2}mv_f^2 = \tfrac{1}{2}mv_i^2 - mgh \;\Rightarrow\; v_f = \sqrt{12^2 - 2(9.8)(3.0)} = 9.2\,\text{m/s} \]

This is the payoff of path-independence made concrete. The landing speed depends only on the height gained, so a ball thrown at any angle with the same speed lands on that ledge at the same speed — steeply or shallowly, it makes no difference.

Doing this both ways once makes path-independence concrete rather than a claim you take on trust.

Route one uses the actual displacement. Gravity has no horizontal component, so the horizontal part of the displacement contributes nothing to the dot product, and only the three metres of height survive.

Route two takes a deliberately silly path — straight up, then sideways. The vertical leg gives minus mgh; the horizontal leg gives nothing, because gravity is perpendicular to it. Same answer.

Notice what never entered either calculation: the launch angle, the launch speed, the flight time, the horizontal distance. Gravity's work simply does not depend on any of them.

The payoff is the last step. Once you know gravity's work, the work-energy theorem gives the landing speed in one line with no projectile kinematics at all. And the result says something worth remembering: a ball thrown at any angle with the same speed lands on that ledge at the same speed. Steep or shallow, it makes no difference to how fast it arrives.

14. Kinetic Energy and the Work-Energy Theorem

Section

Section 2

15. Kinetic energy, and where the one half comes from

Concept

\[ K = \tfrac{1}{2}mv^2 \]

Kinetic energy is a scalar and is never negative. It does not care which way the object is moving — reversing the velocity leaves K completely unchanged, because the velocity is squared.

\[ W = \int F\,dx = \int ma\,dx = \int m\,v\frac{dv}{dx}\,dx = \int_{v_i}^{v_f} mv\,dv \]

\[ W = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 \]

So the one half is not a convention. It falls out of integrating v dv, and it is the reason kinetic energy goes as the square of the speed rather than linearly.

The practical consequence of that square is large. Doubling your speed quadruples your kinetic energy, and therefore quadruples the stopping distance for a given braking force. It is the same quadratic you met in the kinematics deck, now with an energy interpretation.

It is worth watching this derivation once, because it explains both the one half and the square.

The move is the v dv/dx identity from the kinematics deck, now with a mass attached. Multiply the second law by dx, convert a dx into v dv, and integrate. What comes out is one half m v squared, evaluated between the endpoints.

So the one half is the signature of integrating something linear — in this case m v dv — exactly as it was for the spring. And the square is there because the integrand was linear in v, not constant.

The physical consequence of that square is large and worth feeling rather than knowing. Doubling your speed quadruples your kinetic energy. Since a given braking force removes energy at a fixed rate per metre, that means quadrupling the stopping distance. It is the same quadratic you met in kinematics, now with an energy explanation attached.

And kinetic energy is a scalar that is never negative, which is the subject of a trap slide a few pages from here.

16. The work-energy theorem

Concept

\[ W_{\text{net}} = \Delta K = \tfrac{1}{2}mv_f^2 - \tfrac{1}{2}mv_i^2 \]

This is the most useful single equation in the deck, and the word that carries it is net. Every force acting must be included, or the theorem is simply false.

What makes it powerful is what it leaves out. There is no time in it, no acceleration, and no reference to the path taken. It connects two speeds and the work between them, and nothing else.

So if a question gives you forces and distances and asks for a speed, this is almost certainly the fastest route — often a single line where a force analysis would need two or three steps.

This is the most useful single equation in the deck, and the word that carries it is net.

Every force acting must be included in that sum, with its own sign. Leaving one out does not give an approximate answer; it gives a wrong one. This is where the free-body-diagram discipline from the previous deck pays off again.

What makes the theorem powerful is what is absent from it. No time. No acceleration. No reference to the path. It relates two speeds and the work done between them and nothing else.

So the recognition rule is: if a problem gives you forces and distances and asks for a speed, this is almost certainly the fastest route. Often one line where a force analysis would take three steps and a kinematic equation.

It is worth noticing that the theorem is exact, not an approximation, and it applies whether or not the forces are conservative. Friction and applied pushes go into the net work exactly like anything else. The distinction between conservative and non-conservative matters for potential energy, not for this theorem.

17. Energy or forces? Choosing your tool

Pattern

The rule of thumb: energy answers questions about start and end states; forces answer questions about instants. If the word 'when' or 'how long' appears, energy alone will not be enough.

A great many exam problems are deliberately built so that the force route is possible but painful and the energy route is three lines. Recognising which you are looking at is worth real marks.

And the two are never in conflict — they are the same physics. Anything you can get one way you can get the other way, eventually.

Most of the time lost on mechanics problems is lost by starting with the wrong tool and getting three-quarters of the way through before noticing.

The rule of thumb is that energy answers questions about start and end states, and forces answer questions about instants. If the question contains how fast and gives you heights or distances, try energy. If it contains when, how long, or which way, you need forces.

Look particularly at the last row of the table. A force given as a function of position is a signal, not a difficulty. Integrating it for the work is nearly always cheaper than setting up and solving the differential equation that the force approach would require.

And the two approaches are never in conflict. They are the same physics — the work-energy theorem was derived from the second law a slide ago. Anything you can get one way you can get the other way, eventually. The question is only ever which is faster.

Many AP free-response questions are deliberately built so the force route is possible but painful and the energy route is short. Spotting that is worth real marks under time pressure.

18. Which tool for this question?

Check

A 2.0 kg block slides down a curved, frictionless ramp of height 1.8 m, starting from rest. The ramp's shape is not given.

The phrase to notice in the question is that the ramp's shape is not given. That is a deliberate signal.

Because gravity is conservative, the work it does depends only on the height dropped and not at all on the shape of the path. So the speed at the bottom is available without knowing anything about the ramp — one line, and the mass cancels too.

The acceleration and the time both genuinely require the shape. The acceleration depends on the local slope, which changes along a curved ramp, so there is not even a single acceleration to find. The time depends on the whole path.

That contrast is the whole lesson. The same missing information makes one question unanswerable and leaves another completely unaffected, and which is which depends on whether the quantity you want is path-dependent.

When an exam question conspicuously withholds a detail, it is usually telling you which method to use.

19. How far does friction stop a block?

Worked example

A 3.0 kg block slides onto a rough horizontal surface at 8.0 m/s. The coefficient of kinetic friction is 0.35. How far does it travel before stopping?

\[ f = \mu_k mg = (0.35)(3.0)(9.8) = 10.3\,\text{N} \qquad W_f = -fd \]

\[ -fd = 0 - \tfrac{1}{2}mv_i^2 \;\Rightarrow\; d = \frac{mv_i^2}{2f} \]

\[ d = \frac{(3.0)(8.0)^2}{2(10.3)} = 9.3\,\text{m} \]

\[ d = \frac{v_i^2}{2\mu_k g} \]

The mass cancellation is worth noticing. More mass means more kinetic energy to dissipate, but also proportionally more friction to dissipate it with — the same structure as the angle of repose.

The setup step is where the saving happens. Gravity and the normal force are both perpendicular to the motion, so both do zero work and can be ignored entirely. Friction is the only force in the energy equation.

Friction opposes the motion, so the angle is a hundred and eighty degrees and its work is negative. Getting that minus sign in is the whole of the physics; the rest is arithmetic.

The symbolic form at the end is more valuable than the number. Substituting the friction force in terms of mu and mg makes the mass cancel, leaving the stopping distance as v squared over two mu g.

Two things follow from that expression. Heavy and light blocks slide the same distance, because more kinetic energy is exactly matched by more friction to dissipate it — the same structure as the angle of repose in the forces deck. And the distance goes as the square of the speed, which is the stopping-distance result again.

Getting into the habit of simplifying symbolically before substituting is what makes these structural facts visible. Substitute early and you get a number that tells you nothing about why.

20. Pulling a sled at an angle, with friction

Worked example

A 20 kg sled is pulled 15 m along level ground by a 90 N force at 25 degrees above the horizontal. The coefficient of kinetic friction is 0.15. Find its final speed if it started from rest.

\[ N = mg - F\sin 25^\circ = 196 - 38.0 = 158\,\text{N} \]

\[ f = \mu_k N = (0.15)(158) = 23.7\,\text{N} \]

\[ W_F = (90)(15)\cos 25^\circ = 1224\,\text{J} \qquad W_f = -(23.7)(15) = -356\,\text{J} \]

\[ \tfrac{1}{2}(20)v^2 = 1224 - 356 = 868 \;\Rightarrow\; v = 9.3\,\text{m/s} \]

This problem combines the two ideas that make energy questions non-trivial: a force at an angle, and a normal force that is not mg.

The order matters. Find the normal force first, because friction depends on it. The rope pulls partly upward, so it relieves the ground of some of the weight and the normal force is well below mg.

Then compute each force's work separately, remembering that gravity and the normal force contribute nothing. The applied force's work uses the cosine of the angle; friction's work is negative over the full distance.

The angle is a genuine trade-off and worth thinking about. Raising the rope reduces the normal force and therefore the friction, which helps, but it also reduces the useful horizontal component of the pull, which hurts. There is an optimal angle, and finding it is a nice calculus exercise if you want one.

Contrast this with the pushed crate in the forces deck, where the force pointed downward and both effects worked against you. Pulling upward is genuinely better than pushing downward, and this is the calculation that shows why.

21. An object moving in the negative direction has negative kinetic energy

Trap

22. Why is there no time in the work-energy theorem?

Socratic

This is a good question to sit with, because the answer explains the whole character of energy methods.

The second law is full of time — acceleration is a rate. Yet the work-energy theorem is derived directly from it and contains no time at all. Something must have removed it.

You can watch it happen. The derivation multiplies through by dx and integrates over position. Writing dx as v dt introduces a dt that cancels the dt in the denominator of the acceleration. Time enters and leaves within one step, and what survives is a relation between speeds and positions.

This is the same manoeuvre as v dv/dx from the kinematics deck, which is why that identity kept being described as secretly the work-energy theorem. It is exactly that.

The price is symmetrical with the benefit. You get a relation between start and end that does not care how long the journey took — and for the same reason, the theorem can never tell you a time.

There is a companion result waiting in the next deck. Integrate the second law over time instead of position and you get impulse and momentum, which relates forces to durations and knows nothing about distances. Two integrals of one law, two complementary tools.

23. Find the mistake in this energy calculation

Error analysis

A student computes the speed of a 2.0 kg block at the bottom of a rough incline, having slid 4.0 m down a 30-degree slope from rest with mu_k = 0.20.

\[ \tfrac{1}{2}mv^2 = mgd\sin\theta - \mu_k mgd\cos\theta \]

\[ v^2 = 2gd(\sin\theta - \mu_k\cos\theta) = 2(9.8)(4.0)(0.5 - 0.173) = 25.6 \]

\[ v = 5.06\,\text{m/s} \]

Both errors are geometric rather than conceptual, and both come from the same source: reusing a formula from the flat-ground case on a slope. On an incline, h and d are different quantities and N is not mg.

The structure of this student's equation is completely correct: kinetic energy gained equals gravitational energy released minus energy lost to friction. Both errors are geometric.

The first is using the distance along the slope where the vertical drop is needed. Gravitational potential energy depends on height, and on a thirty-degree slope the height is only half the slope distance.

The second is using mg for the normal force. On an incline the surface supports only mg cos theta, so the friction is smaller than the flat-ground formula suggests.

Both mistakes come from the same source: reusing a formula memorised for flat ground on a slope. On an incline, h and d are different quantities and N is not mg, and those two facts are what the geometry of the problem is for.

The reliable defence is to write the substitutions explicitly — h equals d sin theta, N equals mg cos theta — as separate lines rather than doing them in your head. The sine goes with the height and the cosine with the normal force, and mixing them is the most common error in rough-incline problems.

24. Potential Energy and Conservative Forces

Section

Section 3

25. Conservative forces do path-independent work

Concept

A force is conservative if the work it does between two points is the same for every path — equivalently, if the work it does around any closed loop is zero.

Gravity is conservative: lift a book by two metres along any route you like and the work against gravity is mgh every time. Spring forces are conservative too.

Friction is not. Drag a box in a circle back to where it started and friction has done a large negative amount of work, not zero — because friction always opposes the motion, so it takes energy on every leg of the trip.

Only conservative forces can have a potential energy, and the reason is now visible: potential energy is a function of position, so it can only exist if the energy stored genuinely depends on position alone and not on how you got there.

This is the definition that makes potential energy possible, so it is worth being precise about.

A force is conservative if the work it does between two points is the same along every path. Equivalently — and this version is often easier to test — the work it does around any closed loop is zero.

Gravity passes: lift a book two metres by any route and the work against gravity is mgh every time. Spring forces pass too.

Friction fails, and the closed-loop test shows exactly why. Drag a box in a circle back to its starting point and friction has done a large negative amount of work, not zero. The reason is that friction always opposes the motion, so it reverses direction whenever the motion does, and takes energy on every single leg of the journey.

The link to potential energy is now visible. A potential energy is a function of position — one number for each place. That can only exist if the energy stored genuinely depends on position alone. For a path-dependent force there is no such function to write down, which is why friction has no potential energy and never will.

26. Which forces are conservative?

Discrimination

A quick test that usually works: could this force ever give the energy back? A stretched spring will. A compressed one will. Friction never does — the energy has become heat and it is not coming back as motion.

The quick test that usually works: could this force ever give the energy back?

A stretched spring will. A compressed spring will. A raised mass will. Friction never does — the energy has become heat and it is not coming back as organised motion.

The two gravitational entries are the same force in two regimes, and both are conservative. The near-Earth version gives U equal to mgy; the general version gives minus GMm over r. The gravitation deck derives the second from the first.

The normal force is the interesting entry. It is listed as non-conservative, which surprises people, because it usually does zero work and therefore seems harmless. The reason it is not conservative is that its value depends on the situation rather than on position alone — the same point in space can have very different normal forces depending on what else is happening. In practice it rarely matters, because zero work is zero work.

A hand's push is non-conservative for the same kind of reason, and this one does matter: an applied force must go into the work-energy theorem explicitly, because there is no potential energy to hide it in.

27. Potential energy is defined by the work a force can do

Concept

\[ \Delta U = -W_{\text{conservative}} \]

Read the minus sign carefully. When a conservative force does positive work, the stored energy decreases — the store is being spent. When you do work against the force, the store grows.

Notice that the first two have a choice of zero point and the third does not. That is not an inconsistency — the general form's zero is fixed at infinity by the mathematics, and near the Earth's surface the choice is yours because only differences matter.

Which brings us to the rule that saves the most confusion: only changes in potential energy are physical. You may put the zero of gravitational potential energy at the floor, at the table, or at the ceiling, and every physical answer will come out the same.

The minus sign in this definition is where the confusion usually starts, so read it physically.

When a conservative force does positive work, the stored energy decreases. A falling ball has gravity doing positive work on it, and the gravitational potential energy is being spent. When you do work against the force — lifting the ball — the store grows.

The table has one column worth attention: the zero point. The first two forms let you choose it, and the third does not. That is not an inconsistency. The general gravitational form has its zero fixed at infinity by the mathematics, while near the Earth's surface only differences ever appear in any physical answer, so the choice is free.

That is the rule which saves the most confusion in this unit: only changes in potential energy are physical. You may set the zero at the floor, the table, or the ceiling, and every answer you compute will be identical. Choose whichever makes the most terms vanish.

One practical note: whatever you choose, write it on your diagram. Changing the reference height halfway through a problem is the potential-energy equivalent of changing your axis halfway through a force problem, and it produces the same kind of unfindable error.

28. Potential energy belongs to the object

Trap

29. Force is minus the slope of the potential

Concept

\[ F_x = -\frac{dU}{dx} \qquad \text{and conversely} \qquad U(x) = -\int F_x\,dx \]

Everything about a conservative force is contained in its potential function. Differentiate U and you get the force; integrate the force and you get U back, up to the arbitrary constant.

The minus sign encodes something intuitive: the force always pushes toward lower potential energy. A ball rolls downhill; a stretched spring pulls back toward its natural length. Both are the system moving toward a smaller U.

Check it on the two familiar cases. Differentiating mgy gives mg, so the force is -mg — downward, correct. Differentiating one half k x squared gives kx, so the force is -kx — Hooke's law, correct.

This relation says something quite strong: everything about a conservative force is contained in its potential function. Differentiate U and you get the force. Integrate the force and you recover U, up to the arbitrary constant you were always allowed.

The minus sign encodes something you already believe. The force always pushes toward lower potential energy — a ball rolls downhill, a stretched spring pulls back toward its natural length. Both are systems moving toward smaller U.

Check it on the two cases you know. Differentiating mgy gives mg, so the force is minus mg: downward, correct. Differentiating one half k x squared gives kx, so the force is minus kx: Hooke's law, correct. The relation reproduces both without being told anything new.

The figure is the picture to keep. Where the potential curve slopes downward to the right, the force points right. At the bottom of the valley the slope is zero and so is the force — which is the definition of equilibrium, and the subject of the next few slides.

This is also your first meeting with a gradient in physics. In three dimensions the force is minus the gradient of U, and the same statement holds: force points downhill on the potential landscape.

30. Getting the force from a potential function

Worked example

A particle moves along the x-axis in a potential U(x) = 2x^3 - 9x^2 + 12x, with U in joules and x in metres. Find the force, and locate the equilibrium positions.

\[ F_x = -\frac{dU}{dx} = -(6x^2 - 18x + 12) = -6(x-1)(x-2) \]

\[ x = 1\,\text{m} \quad \text{and} \quad x = 2\,\text{m} \]

\[ \frac{d^2U}{dx^2} = 12x - 18 \]

The test with the second derivative is fast; the test with the force is convincing. Use the first to get the answer and the second when you want to be sure.

This is a standard AP question type, and the procedure is short enough to be worth having automatic.

Differentiate and negate for the force. Set the force to zero for the equilibria. Use the second derivative to classify them: positive means a minimum and therefore stable, negative means a maximum and therefore unstable.

The verification step at the end is worth doing at least until the second-derivative test feels reliable. Test the force just to one side of each equilibrium. If it points back toward the equilibrium, the equilibrium is stable; if away, unstable.

That test is slower but far more convincing, and it is the physical meaning of the calculus. A stable equilibrium is one where being displaced produces a restoring force.

Notice that the factored form of the force made the equilibria visible immediately. Factoring before solving is worth the extra line here, as it was in the kinematics deck when finding when a particle was momentarily at rest.

31. Match each energy form to its expression

Matching

Two structural observations worth more than the memorisation. The one half always signals an integral of something linear — m v dv or k x dx. And work and power differ only by swapping d for v, because power is work per unit time and v is displacement per unit time.

Two structural observations here are worth more than memorising six formulas.

The first is that the one half always signals an integral of something linear. Kinetic energy is the integral of m v dv; spring energy is the integral of k x dx. Both integrands are linear, so both integrals produce a one half and a square. If you ever forget one of these expressions, you can rebuild it from the force in one line.

The second is that work and power differ only by swapping d for v. That is not a coincidence either: power is work per unit time, and v is displacement per unit time, so dividing the work formula by dt turns d into v and leaves everything else alone.

Noticing patterns like these is what turns a formula sheet from a list into a structure. There are far fewer independent ideas in this deck than there are equations.

32. Shape the potential well

Tweak it

A particle sits in a potential well. Change the well's depth and watch what becomes possible.

This picture — a horizontal energy line drawn across a potential curve — is the single most useful diagram in the deck, and the next section is built on reading it.

This picture — a horizontal energy line drawn across a potential curve — is the single most useful diagram in the deck, and Section 6 is entirely built on it.

The particle's total energy does not change, so it is a horizontal line. The potential curve is fixed by the physics. Everything about the motion follows from where the line sits relative to the curve.

When the line is below the rim, the particle is trapped. It oscillates between the two points where the line meets the curve — the turning points, where all its energy is potential and its speed is momentarily zero.

When the line is above the rim, the particle escapes over the barrier and never returns. It arrives at the top with kinetic energy to spare.

The borderline case is the interesting one: with the line exactly at the rim height, the particle approaches the top asymptotically and never quite arrives, slowing forever. That is the same mathematical situation as escape velocity in the gravitation deck, and it is worth recognising when you meet it there.

33. Conservation of Mechanical Energy

Section

Section 4

34. When no friction acts, the total is fixed

Concept

\[ E = K + U = \text{constant} \qquad \Longrightarrow \qquad K_i + U_i = K_f + U_f \]

If only conservative forces do work, the sum of kinetic and potential energy never changes. Energy sloshes between the two forms, and the total is a number you can write down at the start and use at the end.

The gain in usefulness is enormous. You do not need to know the shape of the track, the forces at any instant, or how long anything took. Two states, one equation.

The cost is what got left behind: no time, no acceleration, no direction. Conservation of energy will tell you how fast the ball is going at the bottom and will never tell you when it gets there.

The bar charts are the picture to hold. As the ball descends, the potential bar shrinks and the kinetic bar grows by exactly the same amount, so the combined height never changes.

That combined height is the total mechanical energy, and when only conservative forces act it is a number you can write down at the start and reuse at the end. You never need to know what happened in between.

The gain in power is enormous. No need for the shape of the track, the forces at any instant, or the duration. Two states, one equation, done.

The cost is what got left behind, and it is worth restating because people forget it under exam pressure: no time, no acceleration, no direction. Conservation of energy will tell you how fast the ball is moving at the bottom and will never tell you when it arrives or which way it is going.

Also note the condition carefully. It is not that no non-conservative forces are present — it is that none of them do work. A normal force is non-conservative and is present in nearly every problem, and mechanical energy is conserved anyway, because the normal force does no work.

35. Energy as a bank balance

Picture it

The bookkeeping analogy is worth adopting because it makes the sign conventions stop being arbitrary.

Two accounts, and money moves freely between them. Going up transfers from kinetic to potential; coming down transfers back. The total is unchanged, which is exactly what conservation means.

Friction is a leak. It drains from whichever account is being used into a third one — heat — and the arrow points only one way. That is why a problem with friction needs an extra term and a problem without one does not.

The analogy also explains the arbitrary zero. A bank balance is meaningful because of the transactions, not because of where someone once decided to start counting. Add a million to every entry in a ledger and every transaction is unchanged, which is precisely what happens when you move the reference height.

This analogy is worth adopting because it makes the sign conventions stop feeling arbitrary.

Two accounts, kinetic and potential, and money moves freely between them. Going up is a transfer from kinetic to potential; coming down transfers it back. The total is unchanged, which is exactly what conservation means.

Friction is a leak rather than a transfer. It drains from whichever account is in use into a third account — heat — and the arrow points one way only. That asymmetry is why a problem with friction needs an extra term and one without does not.

The analogy also explains the arbitrary zero rather nicely. A bank statement is meaningful because of the transactions, not because of where someone once decided to start counting. Add a million to every entry and every transaction is unchanged — which is precisely what happens when you move the reference height.

Where the analogy breaks down is worth knowing too: the heat account is not just hard to withdraw from, it is fundamentally restricted by the second law of thermodynamics. That is a real physical asymmetry, not an accounting convention.

36. The energy bookkeeping recipe

Pattern

1. Choose the system and say what is in it. Usually the object plus the Earth plus any springs.
2. Choose the zero of potential energy and mark it on your diagram. Any choice is legal; a convenient one saves arithmetic.
3. Pick two states: where you know the most, and where the question is asked.
4. List K and U at each state. Zero out everything you can — at rest means K is zero, at the reference height means U is zero.
5. Check for non-conservative work. Friction, air resistance, an applied push. If there is any, it goes on the right-hand side as an extra term.
6. Write one equation and solve. Look for masses that cancel before you substitute numbers.

\[ K_i + U_i + W_{\text{non-cons}} = K_f + U_f \]

Step four is where the work is saved. The whole reason to choose your two states carefully is to make as many terms as possible vanish — a well-chosen pair often leaves an equation with two terms in it.

Step five is where marks are lost. Read the problem for the words rough, friction, air resistance or drag, and if any appear, mechanical energy is not conserved and you need the extra term.

Six steps, and the value is concentrated in steps four and five.

Step four is where the work is saved. The whole reason to choose your two states carefully is to make as many terms as possible vanish. At rest means K is zero; at the reference height means U is zero. A well-chosen pair of states often leaves an equation with two terms in it.

The spring launcher later in this deck is a good example: both chosen states have zero kinetic energy, which is exactly why that choice was made.

Step five is where marks are lost. Read the problem for the words rough, friction, air resistance or drag. If any of them appear, mechanical energy is not conserved and you need the extra term. If none appear, the problem is telling you it is conserved.

Step two is worth doing on paper rather than in your head. Mark the zero of potential energy on your diagram. It costs two seconds and it prevents the error of measuring one height from the floor and another from the table.

37. Speed at the bottom of a frictionless track

Worked example

A 0.40 kg cart is released from rest at a height of 2.5 m on a frictionless track of unspecified shape. Find its speed at the bottom, and at a point 1.0 m above the ground.

\[ K_i + U_i = K_f + U_f \;\Rightarrow\; 0 + mgh_i = \tfrac{1}{2}mv^2 + mgh_f \]

\[ v = \sqrt{2g(h_i - h_f)} \]

\[ v = \sqrt{2(9.8)(2.5)} = 7.0\,\text{m/s} \]

\[ v = \sqrt{2(9.8)(1.5)} = 5.4\,\text{m/s} \]

The mass cancelling is not an accident of these numbers. On a frictionless track, every object released from the same height arrives at the same speed, which is the free-fall result generalised to any shape of track.

The striking feature of this problem is what it does not require: the shape of the track is never given and never needed.

That is path-independence doing real work. Gravity's contribution depends only on the height dropped, so a straight ramp, a curved ramp and a corkscrew all deliver the same speed at the bottom.

The mass cancels as well, which means every object released from the same height arrives at the same speed. That is the free-fall result generalised from a vertical drop to any shape of frictionless track.

The second part is worth checking your instincts against. After falling sixty percent of the way, the cart has reached seventy-seven percent of its final speed. Speed goes as the square root of the drop, so most of the speed is gained early and the last part of the descent adds surprisingly little.

That square root is worth carrying. It is why doubling the height of a drop only multiplies the impact speed by about 1.4, and why the height needed to reach a given speed grows quadratically.

38. A pendulum released from an angle

Worked example

A pendulum bob on a 1.5 m string is released from rest at 40 degrees from the vertical. Find its speed at the lowest point.

\[ h = L - L\cos\theta = L(1 - \cos\theta) \]

\[ h = 1.5(1 - \cos 40^\circ) = 1.5(1 - 0.766) = 0.351\,\text{m} \]

\[ v = \sqrt{2gh} = \sqrt{2(9.8)(0.351)} = 2.62\,\text{m/s} \]

That last point is why pendulum problems are so much easier with energy than with forces. The tension varies continuously through the swing and is genuinely hard to track — and energy methods never need to know it.

The only genuinely difficult step here is the geometry, and it is worth drawing rather than recalling.

The bob starts at a height L cos theta below the pivot and ends at L below it. So it drops by L minus L cos theta, which is L times one minus cos theta. Writing that expression down from memory is risky; deriving it from the picture takes ten seconds and is reliable.

The energy step is then trivial, and the reason it is trivial is worth naming. The tension does no work at any point in the swing, because the string is a radius and the motion is along the arc — the two are perpendicular throughout.

That is why pendulum problems are so much easier with energy than with forces. The tension varies continuously through the swing and is genuinely awkward to track, and energy methods never need to know it at all.

It also means the answer is independent of the string's length in one specific sense: the speed depends only on the height dropped, so a short string released from a large angle and a long string released from a small one can give the same speed.

39. A spring launcher

Worked example

A spring with k = 800 N/m is compressed 0.12 m and used to launch a 0.25 kg ball vertically. How high does the ball rise above the launch point?

\[ \tfrac{1}{2}kx^2 = mg(h + x) \]

\[ \tfrac{1}{2}(800)(0.12)^2 = 5.76\,\text{J} \]

\[ h + x = \frac{5.76}{(0.25)(9.8)} = 2.35\,\text{m} \;\Rightarrow\; h = 2.23\,\text{m} \]

Two things make this problem more than routine, and the second is the one exams reward.

The choice of states is the first. Both the compressed state and the highest point have zero kinetic energy, which makes the equation as short as it can be — spring energy in, gravitational energy out, nothing else.

The second is the height bookkeeping. The ball rises h above the launch point, but it also travels the compression distance x on the way up, so the total climb from the compressed position is h plus x. Forgetting that gives an answer about five percent too high here.

Five percent sounds negligible, and on a free-response question it is worth a point. Rubrics reward exactly this kind of careful accounting, and the correction is one extra symbol.

Notice also that the mass does not cancel in this problem. The spring stores a fixed amount of energy regardless of what sits on it, so a heavier ball launches lower. That is the exception discussed on the mass-cancellation slide.

40. Minimum speed to complete a vertical loop

Worked example

A cart on a frictionless track must complete a vertical loop of radius R. From what minimum height must it be released?

\[ mg = \frac{mv_{\text{top}}^2}{R} \;\Rightarrow\; v_{\text{top}}^2 = gR \]

\[ mgh = \tfrac{1}{2}mv_{\text{top}}^2 + mg(2R) \]

\[ gh = \tfrac{1}{2}gR + 2gR \;\Rightarrow\; h = 2.5R \]

The structure here is the one to remember: use forces for the condition at a single instant, and energy to connect two positions. Almost every hard mechanics problem is a combination like this, and knowing which tool answers which part is most of the difficulty.

This is the template hybrid problem, and the structure is worth more than the answer.

The condition at the top is a force question, not an energy one. Energy methods cannot tell you when the cart leaves the track, because leaving the track is about the normal force going to zero, and energy knows nothing about forces at an instant. So you get that condition from the second law: at minimum speed, gravity alone supplies the centripetal force.

Then you switch tools. Energy connects the release height to the top of the loop, because it is a relation between two positions and does not care about the shape of the track in between.

The answer, two and a half radii, is worth sanity-checking. The loop itself is only two radii tall, so the cart must start above the top of the loop. It needs that extra half-radius of height to still be moving fast enough at the top. Starting at exactly two radii would put it at the top with zero speed, and it would fall off.

The general lesson: use forces for conditions at a single instant, and energy to connect two positions. Almost every hard mechanics problem is that combination, and knowing which tool answers which part is most of the difficulty.

41. Why does the mass keep cancelling?

Explain it to yourself

The speed at the bottom of a frictionless track, the acceleration on a frictionless incline, the stopping distance under friction, the angle of repose, the safe speed on a curve — in every one of them the mass dropped out.

This has now happened in five or six different problems, and there is a single reason behind all of them.

Mass cancels whenever every force in the problem is itself proportional to mass. Gravity is mg; friction is mu times mg cos theta. Meanwhile kinetic energy and the ma term are also proportional to mass. Same factor on both sides, so it goes.

Physically: a heavier object has more energy to move and gets proportionally more force to move it with. The two effects scale together exactly, which is a genuinely non-obvious fact about gravity and one that Einstein eventually built a theory on.

The exception is the spring launcher. A spring stores one half k x squared and that number knows nothing about what is sitting on top of it. So the energy is fixed and mgh must absorb it, which means a heavier ball goes lower.

The same is true of any externally fixed force — a stated push, a given rope tension, an engine's thrust. A ninety newton push accelerates a light sled more than a heavy one, and the mass survives into the answer.

So the diagnostic is: does every force scale with the mass of the object? If yes, expect cancellation and look for it before substituting numbers. If any force is fixed externally, the mass will stay.

42. Energy methods versus force methods

Comparison

Three of the five rows say energy cannot answer the question at all. That is not a weakness to apologise for — it is the price of a method that ignores the path, and it is exactly why energy is so much less work when it does apply.

A useful reflex: if the question contains the words how fast and gives you distances or heights, try energy first. If it contains when, how long, or which way, you need forces.

Three of the five rows in this table say energy cannot answer the question at all, and that is worth reading as a feature rather than an apology.

Energy ignores the path, which is why it is so much less work when it applies. Ignoring the path necessarily means losing the information that lives in the path — the timing, the instantaneous acceleration, the direction.

The fourth row is where the trade pays off most dramatically. Finding a speed on a curved track with forces would require knowing the local slope everywhere and integrating along the path. With energy it is one line, and the shape is irrelevant.

The reflex worth building: if the question says how fast and gives distances or heights, try energy first. If it says when, how long, or which way, you need forces. If it says both — and hard problems usually do — you will need both, in the order the loop problem demonstrated.

43. When Friction Is Present

Section

Section 5

44. Mechanical energy is not conserved, but energy is

Concept

\[ K_i + U_i + W_{\text{nc}} = K_f + U_f \]

Friction does negative work, so W_nc is a negative number and the mechanical energy at the end is less than at the start. The missing energy has not vanished — it has become thermal energy in the two surfaces.

\[ |W_{\text{friction}}| = f_k d = \Delta E_{\text{thermal}} \]

Note that d here is the path length, not the displacement. Friction takes energy on every leg of the journey, so a box dragged out and back loses twice as much as one dragged out alone — even though its displacement is zero.

So the grand statement is unaffected: energy is always conserved. What friction breaks is the smaller claim that K plus U alone stays constant.

The distinction in this title is the whole point, and the sloppy phrase energy is lost to friction is what obscures it.

Nothing is lost. The mechanical energy — the kinetic plus potential total — decreases, and exactly that much energy appears as thermal energy in the two rubbing surfaces. The grand total is untouched.

The extra term in the equation is negative because friction does negative work. Its magnitude is the friction force times the path length, and the phrase path length matters: friction takes energy on every leg of the journey, so a box dragged out and back loses twice as much as one dragged out alone, even though its displacement is zero.

That is the path-dependence from the conservative-force slide, showing up as a concrete calculational rule. For gravity you use the height difference; for friction you use the total distance travelled.

The bar chart makes the accounting visible. The mechanical bars are shorter at the end, and the difference is exactly the height of the thermal bar. Drawing that chart on a rough-surface problem is a good way to check you have not mislaid a term.

45. Sliding down a rough incline

Worked example

A 2.5 kg block slides 3.0 m down a 35-degree incline with mu_k = 0.25, starting from rest. Find its speed at the bottom, and how much energy became heat.

\[ h = d\sin\theta = 3.0\sin 35^\circ = 1.72\,\text{m} \]

\[ N = mg\cos\theta = 20.1\,\text{N} \qquad f_k = (0.25)(20.1) = 5.02\,\text{N} \]

\[ mgh = (2.5)(9.8)(1.72) = 42.1\,\text{J} \qquad W_f = -(5.02)(3.0) = -15.1\,\text{J} \]

\[ \tfrac{1}{2}mv^2 = 42.1 - 15.1 = 27.0\,\text{J} \;\Rightarrow\; v = \sqrt{\frac{2(27.0)}{2.5}} = 4.65\,\text{m/s} \]

The two geometric factors — sine for the height and cosine for the normal force — are the ones to be careful with. Mixing them up is the most common error in rough-incline problems, and it is what the error-analysis slide earlier in this deck was about.

This problem needs both geometric factors, and using the wrong one for either is the standard error.

The height drop uses the sine: the block travels three metres along the slope but drops only 1.72 metres vertically, and gravitational potential energy depends on the vertical drop.

The friction uses the cosine, through the normal force. On an incline the surface supports only mg cos theta, so the friction is smaller than it would be on level ground with the same coefficient.

And friction acts over the full slope distance, not the vertical drop. So the two lengths in the problem are used for different purposes: the vertical component for gravity, the full slope length for friction.

The final check is worth making a habit. Thirty-six percent of the released energy became heat, which is why the answer is well below the frictionless value of 5.8 metres per second. If your rough-surface answer ever comes out faster than the frictionless one, a sign has gone wrong.

46. A block, a spring, and a rough floor

Worked example

A 1.2 kg block slides at 4.0 m/s across a floor with mu_k = 0.30 toward a spring of k = 500 N/m, 1.5 m away. How far does it compress the spring?

\[ \tfrac{1}{2}mv^2 = \tfrac{1}{2}kx^2 + \mu_k mg(d + x) \]

\[ \tfrac{1}{2}(1.2)(4.0)^2 = 9.6\,\text{J} \qquad \mu_k mg = (0.30)(1.2)(9.8) = 3.53\,\text{N} \]

\[ 9.6 = 250x^2 + 3.53(1.5 + x) \]

\[ 250x^2 + 3.53x - 4.30 = 0 \]

\[ x = 0.124\,\text{m} = 12.4\,\text{cm} \]

Two features make this a typical hard exam problem: the friction path includes the compression, and the result is a quadratic rather than a linear equation. Both are easy to miss and both are worth points.

Two features make this a typical hard exam problem, and both are easy to miss on a first reading.

The first is that friction acts over d plus x, not just d. The block does not stop being on the rough floor when it touches the spring — it keeps sliding while compressing it, so the friction path is longer than the open floor.

The second is that the result is a quadratic rather than a linear equation, because the unknown appears squared in the spring term and linearly in the friction term. That is not a complication to be avoided; it is what the physics produces, and the quadratic formula handles it.

Discard the negative root deliberately rather than silently. It corresponds to the spring being stretched rather than compressed, which is not what is happening.

The verification at the end is the habit worth copying: account for every joule. Spring energy plus friction energy should equal the initial kinetic energy, and here they agree to rounding. That check catches both algebra errors and missed terms.

47. Where does the energy actually go?

Anomaly

A block slides to a halt on a rough table. Its kinetic energy was, say, 40 J. That energy is now gone from the mechanical picture entirely.

This is worth being precise about, because 'energy is lost to friction' is a phrase that quietly teaches the wrong thing. Nothing is lost. Mechanical energy is converted, and the conversion happens to run one way.

A small amount does become sound, and a small amount can go into deforming the surfaces. The overwhelming majority becomes heat, which is why brakes get hot and why rubbing your hands warms them.

The one-way nature is the interesting part. You can convert ordered motion into heat with perfect efficiency and no effort at all. Converting heat back into ordered motion requires an engine, and even a perfect one cannot do it completely.

This deserves a precise answer because the usual phrasing teaches something false.

The energy is still entirely present. It is now the random kinetic energy of enormous numbers of molecules in both surfaces, which is what a rise in temperature physically is. Energy is perfectly conserved; nothing was destroyed.

What was lost is order. Before, a huge number of molecules were all moving the same way — that is what a moving block is. After, they are moving in random directions with the same total energy. The energy is the same; the organisation is gone.

And that is a one-way street. Converting ordered motion into heat happens spontaneously, with perfect efficiency, requiring no effort. Converting heat back into ordered motion requires an engine, and even a perfect engine cannot do it completely — that is the content of the second law of thermodynamics.

A small amount does become sound, and some goes into deforming the surfaces. The overwhelming majority becomes heat, which is why brakes glow, why rubbing your hands warms them, and why every machine with moving parts needs cooling.

So the phrase to use is that mechanical energy was converted, not lost. It is a small change of wording and it points at the right physics.

48. Reading a Potential Energy Diagram

Section

Section 6

49. Everything about the motion, in one curve

Concept

Draw the total energy as a horizontal line, because it does not change. The potential curve is fixed by the physics. The vertical gap between them is the kinetic energy, and everything else follows from that one reading.

• Where the gap is large, the particle moves fast
• Where the gap is zero, the particle is momentarily at rest — a turning point
• Where the curve is above the line, the motion is forbidden: kinetic energy cannot be negative
• Where the curve has zero slope, the force is zero — an equilibrium

This one diagram answers questions that would each require a separate calculation: where the particle can go, where it is fastest, where it turns around, and where it could sit at rest.

This diagram answers half a dozen questions that would each otherwise need a separate calculation, so it repays the effort of learning to read fluently.

The total energy is a horizontal line because it does not change. The potential curve is fixed by the physics. The vertical gap between them is the kinetic energy at that position, and everything else follows from that single reading.

Where the gap is large, the particle is moving fast. Where it closes to zero, the particle is momentarily at rest — a turning point, where it reverses. Where the curve rises above the line, the region is forbidden outright, because kinetic energy cannot be negative.

Where the curve has zero slope, the force is zero, which is an equilibrium. Note that this is a statement about the curve's shape, independent of where the energy line happens to be.

Practise reading these before practising calculating with them. A great many exam questions about potential diagrams can be answered by looking, and the ones that need numbers still start with the same reading.

50. Stable, unstable, and neutral equilibrium

Concept

The physical test is more memorable than the calculus one: nudge it, and ask which way the force points. Toward the equilibrium is stable; away from it is unstable.

Stable equilibria matter beyond this deck. A particle near the bottom of any smooth potential well behaves like a mass on a spring — the valley looks parabolic close up — which is why simple harmonic motion appears everywhere in physics. That is the subject of Unit 7.

The calculus test is fast and the physical test is convincing, and it is worth having both.

The calculus: the second derivative of U is positive at a minimum, which is stable, and negative at a maximum, which is unstable. Zero over an interval is neutral.

The physical test: nudge it and ask which way the force points. Back toward the equilibrium means stable. Away from it means unstable. Nowhere in particular means neutral.

Unstable equilibria are real but never observed for long, which is why they feel strange. A pencil balanced perfectly on its point is genuinely in equilibrium — but any disturbance, however small, grows, so it does not last.

The last point on this slide is the one that matters most for the rest of the course. Zoom in close enough to any smooth minimum and it looks like a parabola. A parabolic potential is exactly a spring, so a particle near any stable equilibrium behaves like a mass on a spring. That is why simple harmonic motion turns up everywhere in physics, and it is the whole justification for Unit 7.

51. Reading a potential curve

Worked example

A particle of mass 0.50 kg moves in the potential shown, with a total energy of 12 J. Describe its motion completely.

\[ K_{\text{max}} = E - U_{\text{min}} \qquad v_{\text{max}} = \sqrt{\frac{2(E - U_{\text{min}})}{m}} \]

Work through this systematically rather than by inspection, because the systematic version is what generalises to curves you have not seen.

First, find where motion is allowed: everywhere the energy line is above the curve. Second, find the turning points: where they intersect. Third, find the fastest point: where the gap is widest, which is the bottom of whichever well the particle occupies.

Fourth, and this is the question exams like, decide whether the particle can cross the barrier. It can only if the barrier's peak lies below the energy line. If the peak is higher, the particle is trapped in whichever well it started in and can never reach the other, no matter how long you wait.

That last point is worth pausing on. Two allowed regions can exist simultaneously with the particle permanently confined to one of them. Classical mechanics is absolutely strict about this — which is exactly the rule quantum mechanics famously breaks, through tunnelling.

The check at the end is the useful habit: ask what changes as the energy line rises. Turning points move outward, the maximum speed rises, and at some critical energy the line clears the barrier and the character of the motion changes qualitatively.

52. Reading a turning point

Check

A particle of mass 0.20 kg moves in a one-dimensional potential. At x = 2.0 m the potential is 6.0 J; the particle's total energy is 10.0 J.

The mistake this question is designed to catch is using the total energy as though it were all kinetic.

The kinetic energy is the gap between the total and the potential: ten minus six is four joules. Only then do you convert to a speed.

The habit that prevents the error is to write K equals E minus U as a separate line every time, before touching the square root. It takes two seconds and it makes the subtraction impossible to forget.

Option D is worth understanding rather than just eliminating. A turning point is where the potential equals the total energy, leaving nothing kinetic. Here there are still four joules to spare, so the particle is moving — and moving reasonably quickly, since the mass is small.

Note also how the small mass amplifies the speed. Four joules in a 0.20 kilogram particle gives 6.3 metres per second; the same four joules in a 20 kilogram block would give 0.63. Speed goes as one over the square root of the mass for a fixed energy.

53. Raise the energy line until something changes

Edge cases

The point of this exercise is that the interesting energies are not arbitrary — they are the local extrema of the potential.

Below the lowest point of the potential, nothing is possible: the kinetic energy would have to be negative. At exactly that value, the particle sits at rest at the bottom of the well.

Between the two well bottoms, only the lower well is accessible, and not because of the barrier — because the floor of the other well is above the energy line. That distinction is worth noticing, since it is a different reason for exclusion.

Between the higher well bottom and the barrier peak, something genuinely interesting happens: there are two separate allowed regions, and the particle is confined to whichever it started in. The allowed region is disconnected.

Above the barrier, the two regions merge and the particle roams freely, slowing over the barrier and speeding up in each well.

The general statement: the critical energies are the local maxima and minima of U, because those are exactly the values at which the shape of the allowed region changes. Identifying them is the standard exam question about these diagrams, and it requires no calculation at all — just reading the curve.

54. Power: How Fast the Energy Moves

Section

Section 7

55. Two formulas for power, and when to use each

Concept

\[ P_{\text{avg}} = \frac{W}{\Delta t} \qquad P = \frac{dW}{dt} = \vec{F}\cdot\vec{v} \]

Power is the rate at which work is done. Lifting a crate onto a shelf takes the same work whether you do it in one second or one minute; the power differs by a factor of sixty.

The second form is the more useful of the two in this course, and it is worth deriving in one line: dW = F dx, so dW/dt = F dx/dt = Fv. Same dot product, same cosine, same sign rules.

Units: a joule per second is a watt. A useful benchmark set: a bright lamp is about 10 W, a person working hard sustains roughly 100 W, a kettle is 2000 W, and a family car is around 100,000 W.

Power is the rate at which work is done, and the distinction from work itself is worth a concrete example: lifting a crate onto a shelf takes the same work whether you do it in one second or one minute. The power differs by a factor of sixty.

The second form, force dotted with velocity, is the more useful in this course, and it is one line to derive: dW equals F dx, so dividing by dt turns dx into v. Same dot product, same cosine, same sign rules as work.

Choose between the two by what you are given. Total work and total time gives you an average. Force and current speed gives you the instantaneous value. They are different numbers and questions do distinguish between them.

The benchmarks at the end are worth memorising, because power estimates go wrong easily. A bright lamp is around ten watts; a person working hard sustains about a hundred; a kettle is two thousand; a car is about a hundred thousand.

That last comparison is the one that puts industrialisation in perspective. A modest family car delivers the sustained mechanical power of roughly two hundred and fifty people.

56. Engine power at constant speed

Worked example

A 1400 kg car travels at a steady 30 m/s against a total resistive force of 600 N. What power is the engine delivering? What if it also climbs a 4-degree hill at the same speed?

\[ F_{\text{drive}} = 600\,\text{N} \]

\[ P = Fv = (600)(30) = 18000\,\text{W} = 18\,\text{kW} \]

\[ mg\sin 4^\circ = (1400)(9.8)\sin 4^\circ = 957\,\text{N} \]

\[ P = (600 + 957)(30) = 46700\,\text{W} = 47\,\text{kW} \]

Notice that the answer would be the same for a car going down the same hill with the engine braking, except the sign: the power would be negative, meaning energy is being taken out of the car's motion rather than put in.

The step that carries this problem is the first one, and it is easy to read past: constant speed means zero acceleration, which means zero net force, which means the driving force exactly equals the resistance.

Once you have that, the power is a single multiplication. The interesting part is the second half.

Adding a four-degree hill requires the engine to overcome the down-slope component of the weight as well as the drag. For a 1400 kilogram car that component is nearly a thousand newtons — larger than the entire drag force at 30 metres per second.

So a gentle-looking hill more than doubles the power required, which matches ordinary experience exactly: a car that cruises effortlessly on the flat needs a downshift on a modest incline.

One extension worth thinking about. Going down the same hill with the engine braking gives the same magnitude with a negative sign — energy is being taken out of the car's motion rather than put in. Negative power is a perfectly ordinary thing, and regenerative braking in an electric car is exactly that, recovered.

57. How much power can a person produce?

Estimation

The comparison is worth sitting with. A trained athlete can sustain about 400 W for an hour and can peak at around 1500 W for a few seconds. Human beings are, by the standards of machines, very low-power devices.

That is why the invention of engines mattered so much. A modest family car delivers the sustained power of roughly two hundred and fifty people, continuously, for as long as there is fuel.

This estimate is worth doing because the answer recalibrates most people's sense of what human beings are capable of.

Running up four metres of stairs in five seconds is close to flat out for most people, and it comes to about five hundred and fifty watts — roughly a quarter of a kettle.

And that is a peak, sustainable for seconds. A trained athlete can hold about four hundred watts for an hour and can peak near fifteen hundred for a few seconds. By the standards of machines, humans are very low-power devices.

The comparison also explains why manual labour was replaced so completely and so fast. One car engine delivers the sustained mechanical power of a few hundred people, continuously, without rest, for as long as there is fuel.

For exam purposes, the technique is the thing: work first, using mgh, then divide by the time. Estimation questions like this appear on the AP exam and are usually testing whether you can do that in two lines without a calculator.

58. Why cars have a top speed

Real world

An engine delivering constant power against air resistance that grows with speed reaches a maximum speed. Nothing breaks; the car simply stops accelerating.

This is a nice piece of reasoning because it combines two formulas you already have and produces something genuinely non-obvious.

At top speed there is no acceleration, so the driving force equals the drag, which goes as v squared. And the engine delivers P equals Fv. Combine them and P goes as v cubed.

So the top speed goes as the cube root of the power. Doubling the power multiplies the top speed by about 1.26 — a twenty-six percent gain for twice the engine.

That explains something you can observe. The fastest production cars have enormous engines for very modest gains at the top end, and manufacturers chasing top speed work at least as hard on reducing the drag coefficient, because that appears in the same cube root but is often cheaper to improve.

The same cube law explains everyday fuel consumption. Driving at 130 kilometres per hour instead of 100 requires roughly twice the power against drag, which is why motorway speed limits are also fuel-economy measures.

And notice the technique: two simple relations combined to give a scaling law. Scaling arguments like this are often more useful than exact solutions, because they tell you what matters.

59. Putting It Together

Section

Section 8

60. Deciding what to write down

Pattern

1. Is anything non-conservative doing work? If no, use K + U conserved. If yes, add the W term.
2. Are you asked for a speed or a distance? Energy. A time, an acceleration or a direction? Forces.
3. Does the problem have a condition at one instant — barely completes the loop, just leaves the surface, momentarily at rest? That condition is a force statement; get it separately and feed it into the energy equation.
4. Is a force given as a function of position? Integrate it for the work rather than solving a differential equation.
5. Is a potential given? Differentiate for the force; draw the energy line for the motion.
6. Does anything cancel? Mass usually does when only gravity acts. Check before substituting.

Point three is the structural insight of the whole unit. The hard problems are hybrids: a force condition at a single point, connected to an energy equation across two points. The loop-the-loop problem earlier in this deck is the template.

Once you see that combination, a large fraction of AP free-response mechanics questions become recognisable rather than novel.

Six questions, and the third is the structural insight of the whole unit.

Hard mechanics problems are hybrids. There is a condition at a single instant — barely completes the loop, just leaves the surface, momentarily at rest — and that condition is always a force statement. Then there is a relation between two positions, and that is always an energy statement.

Get the force condition separately, then feed it into the energy equation. The loop-the-loop problem earlier in this deck is the template, and once you recognise the pattern a large fraction of AP free-response mechanics becomes familiar rather than novel.

The first question is the gatekeeper: is anything non-conservative doing work? Read the problem for rough, friction, drag, air resistance, or an applied push. Their presence or absence decides whether you write two terms or three.

And the last question — does anything cancel — is worth asking before you substitute rather than after. Symbolic answers show you structure; numerical ones hide it.

61. Four questions before you close the deck

Exit ticket

Answer these from memory first. All four are about the boundaries of the concepts rather than about calculation, which is where exam questions concentrate.

The first tests the definition of work. Zero, because the force you exert is vertical and the displacement is horizontal. Your tiredness is physiological, not physical.

The second tests why potential energy exists at all. Friction's work depends on the path taken, and a potential energy is a function of position, so no such function can exist for it.

The third tests fluency with the potential diagram. The gap is the kinetic energy — everything else on that diagram follows from reading that gap.

The fourth tests the scaling argument. Cube root of two, about twenty-six percent, because power goes as v cubed against quadratic drag.

Any of these that gave you pause points at a specific slide rather than at the whole deck, which is the more efficient thing to reread.

62. What you can do now

Recap

• Compute work as a dot product, as an integral, and as the signed area under a force-position graph
• Use the work-energy theorem to find speeds without touching the acceleration
• Distinguish conservative from non-conservative forces, and say why only the first kind has a potential
• Apply conservation of mechanical energy, and add the friction term when it is needed
• Move between force and potential with a derivative in one direction and an integral in the other
• Read a potential energy diagram for turning points, forbidden regions, and the stability of equilibria
• Compute power both as work over time and as force dotted with velocity

Next deck: momentum. Energy told you about speeds without needing the path; momentum will tell you about collisions without needing the forces — which is essential, because in a collision the forces are enormous, brief, and completely unknown.

The change from the start of this deck is that you now have a second, independent way of solving mechanics problems — one that ignores the path entirely and often turns a differential equation into a single line.

The organising idea is that work is force integrated over distance, and that integrating the second law over distance produces the work-energy theorem. Everything else in the deck is that one relation specialised: conservative forces let you store the work as a potential energy, non-conservative ones do not, and power is the same thing per unit time.

Three specific things are worth carrying beyond this deck. Potential energy belongs to a system, not an object. Only differences in potential energy are physical. And the force is minus the slope of the potential, which means a potential curve contains everything.

The stable-equilibrium result is the one with the longest reach. Any smooth minimum looks parabolic close up, so any particle near a stable equilibrium behaves like a mass on a spring. That is the entire justification for the oscillations unit later in the course.

Next comes momentum, and the relationship is worth anticipating. Energy came from integrating the second law over position. Momentum comes from integrating it over time. Energy tells you about speeds without needing the path; momentum will tell you about collisions without needing the forces — which is essential, because in a collision the forces are enormous, brief, and entirely unknown.