Kinetic energy and the work-energy theorem

Energy of motion

A moving hammer drives a nail; a moving car crumples a fence. A moving body can do work, so it has energy: kinetic energy,

It is a scalar, never negative, measured in joules. In terms of momentum, $K = \dfrac{p^2}{2m}$.

The work-energy theorem

Push a body with a steady net force $F$ through a distance $d$. Then $v^2 = u^2 + 2ad$ with $a = \dfrac{F}{m}$. Multiply by $\tfrac12 m$:

This holds for any forces, steady or not:

The net work on a body equals the change in its kinetic energy. Explore it.

You should find that the energy gained, $Fd$, does not depend on the mass at all. A heavier body gains the same energy, but a smaller speed.

Watch out. The theorem uses the work of the net force, the total of all the forces. If you push a box at a steady speed against friction, your work and friction's work cancel, the net work is zero, and the kinetic energy does not change.

A worked example

A $1100\ \text{kg}$ car speeds up from $10\ \text{m/s}$ to $25\ \text{m/s}$ over $150\ \text{m}$ of road.

1. Change in kinetic energy. $\tfrac12 \times 1100\ \text{kg} \times ((25\ \text{m/s})^2 - (10\ \text{m/s})^2) = 550\ \text{kg} \times 525\ \text{m}^2/\text{s}^2 = 288\,750\ \text{J}$.
2. Net work. By the theorem, the net work is the same, about $2.89 \times 10^{5}\ \text{J}$.
3. Average net force. $F = \dfrac{W}{d} = \dfrac{288\,750\ \text{J}}{150\ \text{m}} = 1925\ \text{N}$.

Think first. Square each speed before you subtract: $(25\ \text{m/s})^2 - (10\ \text{m/s})^2$ is $525\ \text{m}^2/\text{s}^2$, but $(25\ \text{m/s} - 10\ \text{m/s})^2$ is only $225\ \text{m}^2/\text{s}^2$.

Practise with new numbers

Try it in Python

The program finds the stopping distance at several speeds, on a dry road and a wet one.

Remember these

• $K = \tfrac12 mv^2 = \dfrac{p^2}{2m}$.
• Work-energy theorem: $W_{\text{net}} = K_{\text{f}} - K_{\text{i}}$.
• With a fixed braking force, stopping distance goes as $v^2$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the kinetic energy of a $70\ \text{kg}$ cyclist moving at $8.0\ \text{m/s}$?

2. Which has more kinetic energy: a $2.0\ \text{kg}$ ball at $3.0\ \text{m/s}$, or a $1.0\ \text{kg}$ ball at $4.0\ \text{m/s}$?

3. A $0.20\ \text{kg}$ ball speeds up from $5.0\ \text{m/s}$ to $15\ \text{m/s}$. How much net work was done on it?

4. Why does doubling a car's speed make its braking distance four times as long?