Strategy: let energy do the work

Two ways to solve a problem

A stone falls from a height of $20\ \text{m}$. How fast is it moving when it lands?

The force way. Find the acceleration, then the time, then the speed: three steps, and each can go wrong.

The energy way. Compare the energy at the start and at the end: one step. This lesson is about the energy way.

The method in four steps

1. Pick two moments: the start and the end.
2. Write the energy at each: $mgh$ for height, $\tfrac{1}{2}mv^2$ for motion.
3. Say what happens in between: with no friction or air, nothing is lost; with friction, some becomes heat.
4. Make them equal, and solve.

The speed after a fall

With no air and no friction, $mgh = \tfrac{1}{2}mv^2$. The mass cancels:

The speed does not depend on the mass, and it does not depend on the path. A block sliding down a smooth slope from a height $h$ reaches the bottom as fast as a stone dropped from $h$. Only the height counts.

When something is thrown

If the stone starts with a speed $u$, it has kinetic energy at the start too: $\tfrac{1}{2}mu^2 + mgh = \tfrac{1}{2}mv^2$. Thrown straight up, it stops at the top, so the greatest height is $h = \dfrac{u^2}{2g}$.

When there is friction

The heat is the work done against friction: friction force times distance. Braking is the same problem: a car that stops has given all its kinetic energy to the brakes, $\tfrac{1}{2}mv^2 = F \times d$.

You should find that without friction every slope of the same height gives the same speed. With friction, the energy turned to heat grows with the length, so long gentle slopes lose the most.

Watch out. Use the vertical height for $mgh$, and the length along the slope for the friction.

A worked example

A $3.2\ \text{kg}$ block slides from rest down a slope $2.5\ \text{m}$ high and $8\ \text{m}$ long, with a friction force of $4\ \text{N}$ along it. Find its speed at the bottom.

1. Start. $mgh = 3.2\ \text{kg} \times 9.8\ \text{m/s}^2 \times 2.5\ \text{m} = 78.4\ \text{J}$.
2. Heat. $4\ \text{N} \times 8\ \text{m} = 32\ \text{J}$.
3. End. $\tfrac{1}{2} \times 3.2\ \text{kg} \times v^2 = 78.4\ \text{J} - 32\ \text{J} = 46.4\ \text{J}$, so $v^2 = 29\ \text{m}^2/\text{s}^2$ and $v = 5.39\ \text{m/s}$.
4. Check. Without friction it would be $\sqrt{2 \times 9.8\ \text{m/s}^2 \times 2.5\ \text{m}} = 7.0\ \text{m/s}$: friction made it slower, as it should.

Practise with new numbers

Try it in Python

The program finds the speed after falls from four heights.

Remember these

• Pick two moments, write the energy at each, say what happens in between, make them equal.
• After a fall of $h$: $v = \sqrt{2gh}$. The mass and the path do not matter.
• Thrown up at $u$: it rises $h = \dfrac{u^2}{2g}$.
• With friction: heat $=$ friction force $\times$ distance.

Check yourself

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

1. A $1\ \text{kg}$ stone and a $5\ \text{kg}$ stone are dropped together from the same height. Which lands faster?

2. A ball is thrown up at $11\ \text{m/s}$. How high does it rise?

3. A car of $1200\ \text{kg}$ at $15\ \text{m/s}$ brakes with a steady force of $5400\ \text{N}$. How far does it go before it stops?

4. Two smooth slopes have the same height; one is twice as long. Which gives the faster finish?