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Work and Kinetic Energy

7.3 The Work-Kinetic Energy Theorem

The Braking Problem

A car is moving at 60 km/h when the driver slams the brakes. The tires lock. The car skids across the asphalt and comes to a stop. How far does it skid?

You could solve this. Set up Newton's second law: the friction force gives a constant deceleration $a = -\mu_k g$. Use kinematics: $v_f^2 = v_i^2 + 2a\Delta x$. Solve for $\Delta x$. Two equations, some algebra, done.

But notice something about that kinematic equation. It connects speed, acceleration, and displacement --- and time never appears. That equation is doing something quietly powerful: it is relating force (through $a$) directly to how far the car travels and how fast it ends up going, without asking when anything happens.

This section makes that quiet power explicit. We will derive a single equation --- the work-kinetic energy theorem --- that connects the net work done on an object to its change in speed. No time variable. No step-by-step tracking of the force history. Just a direct link between force, distance, and motion.

It turns out this equation is not new physics. It is Newton's second law, repackaged. But the repackaging is so useful that it deserves its own name and its own section.

Prediction

Before you read on: A car traveling at 60 km/h brakes to a stop and skids a distance $d$. A second car, identical in every way, is traveling at 120 km/h --- exactly double the speed --- and brakes with the same force on the same road.

How far does the second car skid?

(a) $d$ --- the same distance

(b) $2d$ --- twice as far

(c) $4d$ --- four times as far

(d) $8d$ --- eight times as far

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. The student selects one of the four options. After committing, the student sees: "The answer is (c): four times the distance. Doubling the speed quadruples the stopping distance. This is not a minor correction --- it is a dramatic, life-and-death difference. The reason traces to the fact that kinetic energy depends on $v^2$, not $v$. We will derive this relationship below."]

If you guessed twice the distance, you are in good company --- most people do. The intuition is that double the speed should mean double everything. But that intuition is wrong, and the reason it is wrong reveals something deep about how energy works.

A car at 120 km/h does not have twice the kinetic energy of a car at 60 km/h. It has four times the kinetic energy. And since the braking force is the same, it takes four times as much work to bring it to a stop --- which means four times the distance. This $v^2$ relationship is not a mathematical accident. It is the central structural fact of the work-energy theorem, and it has real consequences every time someone steps on a brake pedal.

The Guiding Question

Why can net work summarize a change in motion without solving the full force-time history?

When you use Newton's second law in the standard way --- draw a free-body diagram, write $\sum F = ma$, integrate to get $v(t)$, integrate again to get $x(t)$ --- you are tracking the complete history of the motion. You know the velocity at every instant, the position at every instant, the acceleration at every instant.

But many problems do not ask for the complete history. They ask: "What is the final speed?" or "How far does it travel before stopping?" For these questions, the full time history is more information than you need. The work-energy theorem extracts exactly the information that matters --- speed and distance --- and discards the rest.

How is this possible? What mathematical operation compresses the entire force-time story into a single equation relating speed and displacement? That is what we will uncover.

Exploration: Work and Kinetic Energy in Action

[Interactive: Work-Energy Explorer. A block of mass $m$ slides on a horizontal surface. Sliders control: initial speed $v_i$ (0 to 10 m/s), kinetic friction coefficient $\mu_k$ (0 to 1.0), and an optional applied horizontal push $F_{\text{push}}$ (0 to 50 N). When the student clicks "Run," the block slides across the surface. Two quantities are displayed prominently in real time as the block moves: net work done $W_{\text{net}}$ (accumulated from the start) and change in kinetic energy $\Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mv_i^2$. A running comparison shows these two values side by side, updating at each moment.]

Prompt 1: Set $v_i = 5$ m/s, $\mu_k = 0.3$, and $F_{\text{push}} = 0$. Run the simulation. The block slides and slows. Watch $W_{\text{net}}$ and $\Delta KE$ as the block moves. What do you notice about their values?

Prompt 2: Now change the initial speed to $v_i = 8$ m/s, keeping everything else the same. Run it again. Are $W_{\text{net}}$ and $\Delta KE$ still equal at every moment?

Prompt 3: Set $\mu_k = 0$ (frictionless surface) and $F_{\text{push}} = 10$ N. Now the block speeds up instead of slowing down. Are $W_{\text{net}}$ and $\Delta KE$ still equal? Is the work positive or negative this time?

Prompt 4: Try $\mu_k = 0.3$ and $F_{\text{push}} = 20$ N. Now two forces do work: friction (negative work, opposing motion) and the push (positive work, along the motion). The net work is the sum of the two. Is $W_{\text{net}} = \Delta KE$ still holding?

Prompt 5: Can you find any combination of settings where $W_{\text{net}} \neq \Delta KE$?

If you worked through those prompts carefully, you discovered the central fact: the net work done on the block always equals its change in kinetic energy. Always. It does not matter whether the block speeds up or slows down, whether one force acts or five, whether the surface is rough or smooth. The net work and the change in kinetic energy march in lockstep.

This is not a coincidence. It is a theorem --- provable directly from Newton's second law. Let us prove it.

Deriving the Work-Kinetic Energy Theorem

Start with Newton's second law for a particle of mass $m$ moving in one dimension under a net force $F_{\text{net}}$:

$$F_{\text{net}} = ma$$

The acceleration is $a = dv/dt$. But we want a relationship involving displacement, not time. Here is the key move: use the chain rule to rewrite the acceleration.

$$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v\frac{dv}{dx}$$

This substitution replaces the time derivative with a spatial derivative. It is exactly the maneuver you practiced with the chain rule in Chapter 2. Substituting into Newton's second law:

$$F_{\text{net}} = mv\frac{dv}{dx}$$

Now multiply both sides by $dx$:

$$F_{\text{net}}\,dx = mv\,dv$$

The left side is the infinitesimal net work done over a displacement $dx$. The right side is $mv\,dv$. Integrate both sides --- the left from position $x_i$ to $x_f$, the right from speed $v_i$ to $v_f$:

$$\int_{x_i}^{x_f} F_{\text{net}}\,dx = \int_{v_i}^{v_f} mv\,dv$$

The left side is the total net work:

$$W_{\text{net}} = \int_{x_i}^{x_f} F_{\text{net}}\,dx$$

The right side integrates cleanly:

$$\int_{v_i}^{v_f} mv\,dv = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$

And there it is:

$$\boxed{W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \Delta KE}$$

This is the work-kinetic energy theorem. The net work done on an object equals the change in its kinetic energy. The quantity $\frac{1}{2}mv^2$ is the kinetic energy --- the energy an object possesses by virtue of its motion.

What This Theorem Says --- and Where It Comes From

Let us be precise about what just happened.

We started with $F = ma$ --- Newton's second law, the foundational equation of mechanics. We did not add any new physics. We did not invoke any new principle. We performed a mathematical operation: we multiplied both sides by $dx$ and integrated. The calculus from Chapters 2 through 4 did the heavy lifting.

The result is a repackaging of Newton's second law. Everything the work-energy theorem tells you, Newton's second law already knew. But the repackaging is powerful because it collects force and displacement on one side (work) and speed on the other (kinetic energy). Time has been integrated out of the picture.

Think of it this way. Newton's second law is a differential equation --- it tells you what happens at each instant. The work-energy theorem is the integral of that equation over a displacement interval. Integration compresses the instantaneous story into a before-and-after summary. You lose the detailed time history (when was the speed 3 m/s? at what time did the block reach position $x = 2$ m?) but you gain a direct connection between net force accumulated over distance and the resulting speed change.

This tradeoff --- less information, but faster answers --- is exactly what makes energy methods valuable.

Why $v^2$ and Not $v$?

Go back to the derivation. The right-hand side was $\int mv\,dv$, which integrates to $\frac{1}{2}mv^2$. The $v^2$ is not an arbitrary choice. It comes from the chain-rule substitution $a = v\,dv/dx$. When you write $F\,dx = mv\,dv$, the velocity $v$ appears as a factor on the right side. Integrating $v\,dv$ gives $\frac{1}{2}v^2$.

This is why kinetic energy is proportional to the square of speed, not speed itself. And this is why doubling the speed quadruples the kinetic energy and quadruples the stopping distance. The $v^2$ is baked into the mathematics of integrating Newton's second law over displacement.

Pause and think: If kinetic energy were proportional to $v$ instead of $v^2$, what would happen to our braking prediction? A car at double the speed would have double the kinetic energy, requiring double the distance to stop. The actual $v^2$ dependence means the situation is far more dangerous than linear intuition suggests. This is why highway speed limits matter so much --- the energy cost of going faster grows much more quickly than the speed itself.

Returning to the Braking Problem

Now we can solve the opening problem with a single equation.

A car of mass $m$ travels at initial speed $v_i$ and brakes to a stop ($v_f = 0$). The braking force is kinetic friction: $f_k = \mu_k mg$, directed opposite to the displacement. The work done by friction over a skidding distance $d$ is:

$$W_{\text{net}} = -\mu_k mg \cdot d$$

The change in kinetic energy is:

$$\Delta KE = 0 - \frac{1}{2}mv_i^2 = -\frac{1}{2}mv_i^2$$

Setting $W_{\text{net}} = \Delta KE$:

$$-\mu_k mg \cdot d = -\frac{1}{2}mv_i^2$$

The mass cancels:

$$d = \frac{v_i^2}{2\mu_k g}$$

Notice: the stopping distance is proportional to $v_i^2$. Double the speed, quadruple the distance. The prediction is confirmed, and we did not need to find the acceleration, write a kinematic equation, or track time at all.

For the numbers: at 60 km/h ($v_i \approx 16.7$ m/s) with $\mu_k = 0.7$:

$$d = \frac{(16.7)^2}{2(0.7)(9.8)} = \frac{279}{13.7} \approx 20 \text{ m}$$

At 120 km/h ($v_i \approx 33.3$ m/s):

$$d = \frac{(33.3)^2}{2(0.7)(9.8)} = \frac{1109}{13.7} \approx 81 \text{ m}$$

That is $81/20 \approx 4$ times the distance, exactly as predicted by the $v^2$ relationship.

Connection: Newton's Law vs. Energy

[Video: Side-by-side solution comparison. On the left, the braking problem is solved using Newton's second law and kinematics: draw FBD, write $\sum F = ma$, find $a = -\mu_k g$, apply $v_f^2 = v_i^2 + 2a\Delta x$, solve for $\Delta x$. On the right, the same problem is solved using the work-energy theorem: compute $W_{\text{net}} = -f_k d$, set equal to $\Delta KE = -\frac{1}{2}mv_i^2$, solve for $d$. Both paths arrive at the same answer. The energy path is shorter. A voiceover notes: "Both methods use the same physics --- Newton's second law. The energy approach just gets to the answer faster by skipping the intermediate step of finding acceleration."]

The two approaches are not competing theories. They are different routes through the same physics. The force-based route passes through acceleration as an intermediate quantity. The energy route bypasses acceleration entirely and goes straight from force-times-distance to speed change.

When should you use which?

  • Use Newton's second law when you need the acceleration, the time history, or the direction of motion.
  • Use the work-energy theorem when you need the final speed (or distance) and do not care about time or direction.

The expert physicist does not commit to one method. The expert reads the problem, identifies what is asked, and picks the tool that gets there fastest. You are beginning to build that judgment now.

The Net-Work Requirement

There is a word in the theorem that deserves emphasis: net. The theorem says

$$W_{\text{net}} = \Delta KE$$

It does not say that the work done by any single force equals the change in kinetic energy. It says the total work done by all forces equals the change in kinetic energy.

If a block slides across a rough surface while you push it, three forces act: your push (forward), friction (backward), and gravity and normal force (both perpendicular to the motion, doing zero work). The net work is:

$$W_{\text{net}} = W_{\text{push}} + W_{\text{friction}} + \underbrace{W_{\text{gravity}} + W_{\text{normal}}}_{= 0}$$

Only the sum of all the individual work contributions equals $\Delta KE$. Using just $W_{\text{push}}$ or just $W_{\text{friction}}$ alone would give the wrong answer --- unless there happens to be only one force doing work.

This is a common source of error. Keep it in mind for the practice problems.

Practice

Layer 1: Concrete

Problem 1. A 1500 kg car is moving at 20 m/s. The driver applies the brakes, and a net braking force of 7500 N acts on the car.

(a) What is the car's kinetic energy before braking?

(b) Using the work-energy theorem, how far does the car travel before stopping?

(c) What is the car's speed after it has traveled half that distance?

Check your answer **(a)** Kinetic energy: $$KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(1500)(20)^2 = 300{,}000 \text{ J} = 300 \text{ kJ}$$ **(b)** The work done by the braking force over distance $d$ is $W_{\text{net}} = -7500 \cdot d$ (negative because the force opposes the displacement). Setting $W_{\text{net}} = \Delta KE$: $$-7500d = 0 - 300{,}000$$ $$d = 40 \text{ m}$$ **(c)** At half the distance ($d = 20$ m): $$W_{\text{net}} = -7500 \times 20 = -150{,}000 \text{ J}$$ $$\Delta KE = \frac{1}{2}mv^2 - 300{,}000 = -150{,}000$$ $$\frac{1}{2}(1500)v^2 = 150{,}000$$ $$v = \sqrt{\frac{150{,}000}{750}} = \sqrt{200} \approx 14.1 \text{ m/s}$$ Note: at half the stopping distance, the car is not at half the initial speed. It is at $v_i / \sqrt{2} \approx 0.707 \, v_i$. This is the $v^2$ relationship showing up again --- half the distance removes half the kinetic energy, but kinetic energy is proportional to $v^2$, so the speed drops by $1/\sqrt{2}$, not $1/2$.

Problem 2. A 0.5 kg hockey puck is sliding at 8 m/s on ice. A player strikes it with a stick, applying a constant 40 N force in the direction of motion over a distance of 0.3 m.

(a) What is the work done by the stick on the puck?

(b) What is the puck's speed after the hit?

Check your answer **(a)** The force is in the direction of motion, so: $$W_{\text{stick}} = F \cdot d = 40 \times 0.3 = 12 \text{ J}$$ **(b)** The ice is nearly frictionless, so $W_{\text{net}} \approx W_{\text{stick}} = 12$ J. Apply the work-energy theorem: $$W_{\text{net}} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$$ $$12 = \frac{1}{2}(0.5)v_f^2 - \frac{1}{2}(0.5)(8)^2$$ $$12 = 0.25\,v_f^2 - 16$$ $$0.25\,v_f^2 = 28$$ $$v_f = \sqrt{112} \approx 10.6 \text{ m/s}$$ The puck sped up from 8 m/s to about 10.6 m/s. Positive net work increases kinetic energy and therefore increases speed.

Layer 2: Pattern

Problem 3. A 2 kg block starts from rest at the top of a frictionless ramp that is 3 m long and inclined at 30 degrees to the horizontal. Find the block's speed at the bottom of the ramp using:

(a) Newton's second law and kinematics.

(b) The work-energy theorem.

(c) Which approach required fewer steps?

Check your answer **(a)** Newton's second law approach: The component of gravity along the ramp is $mg\sin\theta = 2(9.8)\sin 30^\circ = 9.8$ N. The normal force is perpendicular to motion and does no work, but we do not need it for the acceleration along the ramp: $$a = g\sin\theta = 9.8 \times 0.5 = 4.9 \text{ m/s}^2$$ Using $v_f^2 = v_i^2 + 2a\Delta x$: $$v_f^2 = 0 + 2(4.9)(3) = 29.4$$ $$v_f = \sqrt{29.4} \approx 5.42 \text{ m/s}$$ **(b)** Work-energy approach: The only force doing work is the component of gravity along the ramp: $F_{\parallel} = mg\sin\theta = 9.8$ N. $$W_{\text{net}} = F_{\parallel} \cdot L = 9.8 \times 3 = 29.4 \text{ J}$$ $$W_{\text{net}} = \frac{1}{2}mv_f^2 - 0$$ $$29.4 = \frac{1}{2}(2)v_f^2$$ $$v_f = \sqrt{29.4} \approx 5.42 \text{ m/s}$$ **(c)** For this problem, both approaches are about the same length. The work-energy approach did not need to separately identify the acceleration --- it went directly from force and distance to speed. On a frictionless ramp with constant force, the savings are modest. The work-energy method shines more dramatically in problems with variable forces or complex geometry, where finding the acceleration is hard but computing work is straightforward.

Problem 4. Two identical balls are launched horizontally from a cliff. Ball A is launched at 10 m/s from a cliff that is 20 m high. Ball B is launched at 20 m/s from a cliff that is 5 m high.

Using the work-energy theorem (with gravity as the only force doing work in the vertical direction, and no air resistance), which ball has a greater speed just before hitting the ground?

Hint: Think about the total work done on each ball.

Check your answer For each ball, the only force doing work is gravity (the normal force from the launch mechanism is only momentary, and there is no air resistance). Gravity does work $W = mgh$ where $h$ is the height fallen. Ball A: Total kinetic energy at impact = initial KE + work by gravity $$KE_f = \frac{1}{2}m(10)^2 + mg(20) = 50m + 196m = 246m$$ $$v_A = \sqrt{2 \times 246} = \sqrt{492} \approx 22.2 \text{ m/s}$$ Ball B: $$KE_f = \frac{1}{2}m(20)^2 + mg(5) = 200m + 49m = 249m$$ $$v_B = \sqrt{2 \times 249} = \sqrt{498} \approx 22.3 \text{ m/s}$$ They end up with nearly the same speed. Ball B's extra initial kinetic energy almost exactly compensates for Ball A's extra height. The work-energy theorem makes this comparison clean: you just add up the kinetic energy contributions without needing to decompose the motion into horizontal and vertical components or solve projectile equations.

Layer 3: Structure

Problem 5. A student asks: "The work-energy theorem has $v^2$ in it. Why not $v$? It would be simpler."

Explain, using the derivation, why kinetic energy must involve $v^2$ rather than $v$. Your explanation should trace the $v^2$ to a specific step in the derivation.

Check your answer The $v^2$ enters at the integration step. When we write Newton's second law as $$F\,dx = mv\,dv$$ the right-hand side contains $v\,dv$. The integral of $v\,dv$ is $\frac{1}{2}v^2$, not $v$. This is a direct consequence of the chain rule substitution $a = v\,dv/dx$: the factor of $v$ appears because $dx/dt = v$, and we are converting a time derivative into a spatial derivative. If kinetic energy were $mv$ instead of $\frac{1}{2}mv^2$, then the integral of the right side would need to be $\int m\,dv = mv$. That would require the equation $F\,dx = m\,dv$, which would mean $F = m\,dv/dx$. But Newton's second law says $F = m\,dv/dt$, not $F = m\,dv/dx$. The conversion from $dt$ to $dx$ introduces the extra factor of $v$, which upon integration produces $v^2$. In short: the $v^2$ is the mathematical fingerprint of converting Newton's second law from a time-based equation to a displacement-based equation. It is not a choice or a convention. It is forced by the calculus.

Problem 6. The work-energy theorem says $W_{\text{net}} = \Delta KE$. Newton's second law says $\sum F = ma$.

(a) What information does Newton's second law give you that the work-energy theorem does not?

(b) What information does the work-energy theorem give you that Newton's second law (by itself, without further integration) does not?

Check your answer **(a)** Newton's second law gives you the instantaneous acceleration --- a vector, with both magnitude and direction. From this you can find the velocity as a function of time, the position as a function of time, and the direction of motion at every instant. The work-energy theorem does not provide any of this. It tells you nothing about time, nothing about direction, and nothing about the path between the initial and final states. **(b)** The work-energy theorem directly gives you the final speed from the net work done, without requiring you to first find the acceleration, then integrate to get velocity. If the force varies with position in a complicated way, computing $W = \int F\,dx$ (a single integral) may be far easier than solving the differential equation $F(x) = m\ddot{x}$ for $x(t)$. The two methods contain the same physics. They differ in what they make easy to extract. Newton's second law is the complete story; the work-energy theorem is a useful summary of one chapter of that story.

Layer 4: Debug

Problem 7. A 5 kg block is pushed across a rough horizontal surface by a 30 N horizontal force. The kinetic friction force is 10 N. The block starts from rest and is pushed for 4 m.

A student writes:

$$W = Fd = 30 \times 4 = 120 \text{ J}$$

$$\frac{1}{2}mv_f^2 = 120$$

$$v_f = \sqrt{\frac{2 \times 120}{5}} = \sqrt{48} \approx 6.93 \text{ m/s}$$

The correct answer is $v_f \approx 5.66$ m/s. What did the student do wrong?

Check your answer The student used the work done by the applied force alone ($W_{\text{push}} = 120$ J) instead of the **net work**. Friction also does work on the block: $$W_{\text{friction}} = -10 \times 4 = -40 \text{ J}$$ The net work is: $$W_{\text{net}} = W_{\text{push}} + W_{\text{friction}} = 120 + (-40) = 80 \text{ J}$$ Now apply the theorem correctly: $$80 = \frac{1}{2}(5)v_f^2$$ $$v_f = \sqrt{\frac{160}{5}} = \sqrt{32} \approx 5.66 \text{ m/s}$$ The error: using work by one force instead of net work. The work-energy theorem requires $W_{\text{net}}$ --- the total work done by all forces. When only one force does work (e.g., on a frictionless surface with no other horizontal forces), then $W_{\text{net}}$ equals the work done by that one force, and the shortcut happens to work. But when multiple forces do work, ignoring any of them gives the wrong answer. This is the single most common mistake with the work-energy theorem. Always ask: "Have I accounted for the work done by every force?"

Problem 8. A block slides down a curved, frictionless track from height $h$ to the ground. A student says: "I can't use the work-energy theorem here because the force changes direction along the curved track, and I don't know the shape of the track."

Is the student correct? Can the work-energy theorem be applied here?

Check your answer The student is wrong --- the work-energy theorem works perfectly here, and the shape of the track does not matter. On a frictionless curved track, two forces act on the block: gravity and the normal force. The normal force is always perpendicular to the displacement (it pushes the block away from the surface, while the block moves along the surface), so it does zero work. The only force doing work is gravity. The work done by gravity depends only on the vertical displacement, not on the path: $$W_{\text{gravity}} = mgh$$ So: $$W_{\text{net}} = mgh = \frac{1}{2}mv_f^2 - 0$$ $$v_f = \sqrt{2gh}$$ The student's error is thinking that the work-energy theorem requires knowing the force at every point along the path. For gravity on a frictionless track, the net work depends only on the height change. The curved shape of the track is irrelevant. This is actually a preview of a deeper idea you will meet in Section 7.4: gravity is a *conservative force*, meaning the work it does is path-independent. The work-energy theorem combined with a conservative force is an extremely powerful tool --- it lets you solve problems where the path is complicated or even unknown.

Reflection

Think about what the work-energy theorem gives you and what it takes away.

The theorem tells you how speed changes when forces act over distances. It compresses the full dynamical story --- force at every instant, acceleration at every instant, velocity at every instant --- into a single before-and-after equation. This compression is its power. You do not need the complete time history to find the final speed.

But compression always has a cost. The work-energy theorem is a scalar equation. It involves speed ($v = |\vec{v}|$), not velocity (which has direction). It tells you how fast the object is moving at the end, but not which way. It tells you nothing about how long the process took. It says nothing about the path between start and finish (unless the forces are path-dependent).

So here is the question to sit with: What information does the work-energy theorem give you that Newton's second law doesn't --- and what does it hide?

The answer is that it gives you nothing new in principle. Every result from the work-energy theorem can be obtained from Newton's second law with enough work. But it gives you a faster route to certain answers by eliminating time and direction from the calculation. What it hides is precisely time and direction --- the information it integrated out. Knowing when to accept that tradeoff is part of becoming a skilled problem solver.

Looking Ahead

You now have the work-energy theorem: $W_{\text{net}} = \Delta KE$. It is Newton's second law repackaged as a scalar equation connecting force, distance, and speed. It is powerful, concise, and --- as you saw --- it naturally explains why stopping distance grows as the square of speed.

But there is a limitation lurking. The theorem uses net work --- the total work from all forces. For every problem, you need to compute the work done by each individual force and add them up. For some forces, like gravity, this work turns out to depend only on where you start and where you end, not on the path you took. For other forces, like friction, the work depends on every detail of the path.

This distinction --- forces whose work is path-independent versus forces whose work is path-dependent --- is the subject of the next section. It leads to one of the most powerful ideas in all of physics: potential energy and the principle of energy conservation.