PhysicsLesson 10 of 18About 35 minutes

Kinetic energy and the work-energy theorem

When net work is done on an object, its speed changes. That change is stored as energy of motion, called kinetic energy. This lesson defines kinetic energy and connects it to work through one of the most useful results in mechanics.

By the end of this lesson you can

name several forms of energy;
calculate the kinetic energy of a moving object;
explain why speed matters more than mass for kinetic energy;
solve problems by linking net work to the change in kinetic energy.

Energy

Energy appears in many forms: the energy of moving things, energy stored by height or by a stretched spring, chemical energy in food and fuel, thermal energy in hot objects, light, sound and electrical energy. Energy can change from one form into another, and work is one way of moving energy from one object to another. Like work, energy is a scalar measured in joules.

This lesson is about the first form on the list: the energy an object has because it is moving.

Where net work goes

Think first

On an air track with no friction, you keep pushing a glider with a steady force, so positive net work is done on it. How does its speed change? Where does that work end up?

Compare your answer

The cart speeds up. The work is stored in the cart's motion: the faster it goes, the more energy it carries, and the more work it could do if it ran into something.

You can see the connection with a little algebra. For a constant net force along the motion, $W_{\text{net}} = F d = (ma)d$. From kinematics, $v_f^2 = v_i^2 + 2ad$, so $ad = \tfrac12(v_f^2 - v_i^2)$. Put the two together:

$$W_{\text{net}} = \tfrac12 mv_f^2 - \tfrac12 mv_i^2$$

The quantity $\tfrac12 mv^2$ keeps showing up. It gets a name.

Kinetic energy

$$KE = \tfrac12 mv^2$$

Kinetic energy is the energy an object has because of its motion. With $m$ in kilograms and $v$ in meters per second, $KE$ is in joules.

Three things to notice:

Kinetic energy is a scalar. A ball moving north and an identical ball moving south at the same speed have the same kinetic energy.
It is never negative, because $v^2$ is never negative.
It depends on the square of the speed. Double the speed and the kinetic energy becomes four times larger. Double the mass and it only doubles.

Kinetic energy and stopping distance

Interactive

SpeedMass

Scroll sideways to see the whole demo, or turn your phone sideways.

Try this

Brake at 10 m/s and note the stopping distance. Then brake at 20 m/s. How many times farther does the car go?
Now keep 10 m/s and double the mass. What changes? (The demo uses a braking force that grows with the car's weight, like real tires.)
Explain why speed limits matter so much near schools.

Worked example 1

A 1200 kg car creeps through a parking lot at 10.0 m/s. A 70.0 kg sprinter wants to match the car's kinetic energy. How fast would the sprinter have to run? Is that possible?

Kinetic energy of the car. $$KE = \tfrac12 (1200\ \text{kg})(10.0\ \text{m/s})^2 = 6.00 \times 10^4\ \text{J}$$
Solve $KE = \tfrac12 mv^2$ for $v$. $v = \sqrt{\dfrac{2\,KE}{m}}$, with $m = 70.0\ \text{kg}$.
Substitute. $$v = \sqrt{\frac{2(6.00 \times 10^4\ \text{J})}{70.0\ \text{kg}}} = 41.4\ \text{m/s}$$
About 41 m/s (150 km/h). The fastest humans reach about 12 m/s, so no: even a slow car carries far more energy than a running person.

The work-kinetic energy theorem

Work-kinetic energy theorem

$$W_{\text{net}} = \Delta KE = KE_f - KE_i$$

The net work done on an object equals the change in its kinetic energy.

If the net work is positive the object gains speed; if it is negative the object loses speed; if it is zero the speed does not change. Be careful to use the net work, from all the forces, not the work of one force.

The theorem is often quicker than Newton's laws, because it links forces and distances directly to speeds, without needing the acceleration or the time.

Worked example 2

A 0.170 kg hockey puck leaves a player's stick at 8.00 m/s and glides across level ice. The coefficient of kinetic friction is 0.050. How far does it travel before stopping?

Which forces do work? Gravity and the normal force are at right angles to the motion and do no work; only friction does. $F_k = \mu_k mg = 0.050 \times 0.170 \times 9.81 = 0.0834\ \text{N}$.
Write the theorem. Friction acts against the motion, so $W_{\text{net}} = -F_k d$. The puck ends at rest, so $KE_f = 0$: $$-F_k d = 0 - \tfrac12 mv_i^2$$
Solve for $d$. $$d = \frac{\tfrac12 mv_i^2}{F_k} = \frac{\tfrac12(0.170)(8.00)^2}{0.0834} = \frac{5.44\ \text{J}}{0.0834\ \text{N}} = 65\ \text{m}$$
About 65 m, further than the length of most rinks. The mass cancels: $d = v_i^2/(2\mu_k g)$.

Worked example 3

During a pitch, a baseball (0.145 kg) starts from rest and leaves the pitcher's hand at 38.0 m/s. The hand pushes it over a distance of 2.00 m. What average net force acts on the ball?

Change in kinetic energy. It starts at rest: $\Delta KE = \tfrac12 (0.145)(38.0)^2 - 0 = 105\ \text{J}$.
Apply the theorem. $W_{\text{net}} = F_{\text{net}} d = \Delta KE$.
Solve. $$F_{\text{net}} = \frac{105\ \text{J}}{2.00\ \text{m}} = 52.3\ \text{N}$$
About 52 N, roughly 37 times the ball's weight

Check yourself

Question 1

What is the kinetic energy of a 1200 kg car moving at 25.0 m/s?

Show solution

$KE = \tfrac12 (1200)(25.0)^2 = 3.75 \times 10^5\ \text{J}$

Question 2

A 0.145 kg baseball has 120 J of kinetic energy. How fast is it moving?

Show solution

$v = \sqrt{\dfrac{2(120)}{0.145}} = 40.7\ \text{m/s}$

Question 3

Car A and car B have the same mass. Car B moves twice as fast as car A. How does car B's kinetic energy compare with car A's?

The same
Twice as large
Half as large
Four times as large

Show solution

$KE$ depends on $v^2$: $(2v)^2 = 4v^2$, so four times as large.

Question 4

A car moving at 20.0 m/s has $2.00 \times 10^5\ \text{J}$ of kinetic energy. What is its mass?

Show solution

$m = \dfrac{2\,KE}{v^2} = \dfrac{2(2.00 \times 10^5)}{(20.0)^2} = 1.00 \times 10^3\ \text{kg}$

Question 5

An air-hockey puck of mass 0.500 kg rests on a frictionless table. A paddle pushes it with a steady 2.00 N. Over what distance must the paddle push for the puck to reach 3.00 m/s?

Show solution

$Fd = \tfrac12 mv^2$, so $d = \dfrac{\tfrac12(0.500)(3.00)^2}{2.00} = \dfrac{2.25}{2.00} = 1.13\ \text{m}$

Question 6

A cyclist and bike (70.0 kg in total) start from rest. Pedaling gives a forward force of 120 N, while air and road resistance total 50.0 N. How fast are they going after 35.0 m?

Show solution

Net force: $120 - 50.0 = 70.0\ \text{N}$. $W_{\text{net}} = (70.0)(35.0) = 2450\ \text{J} = \tfrac12 (70.0)v^2$, so $v = \sqrt{70.0} = 8.37\ \text{m/s}$.

Summary

Kinetic energy is the energy of motion: $KE = \tfrac12 mv^2$, a scalar in joules.
Kinetic energy grows with the square of the speed: twice the speed, four times the energy.
The work-kinetic energy theorem: $W_{\text{net}} = \Delta KE$. The sign of the net work tells you whether the object gains or loses speed.

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