Energy with friction and in curved paths

Where does the lost energy go?

A child slides down a real slide and arrives slower than $\sqrt{2gh}$. Mechanical energy has gone missing. It has not vanished: friction has turned it into heat in the slide and the child's clothes. Total energy is always conserved:

The heat is the work done against friction, $f \times$ the path length.

Energy on curved paths

On a curved track the forces change all the time, but energy does not care about the shape. Use energy for the speed, then Newton's second law toward the centre for the forces.

At the top of a vertical loop of radius $R$, the weight and the track's push both point down, to the centre:

The cart stays on only while $N \ge 0$, so $v_t^2 \ge gR$. With energy from a start at height $h$, $v_t^2 = 2g(h - 2R)$, which gives

Try it.

You should find the push at the top drops to zero at $h = 2.5R$, for every loop.

Watch out. Starting at exactly $2R$ is not enough. The cart would need zero speed at the top, but then nothing holds it on the track, and it falls away before it gets there.

A worked example

A $60\ \text{kg}$ skier starts from rest at the top of a slope $25\ \text{m}$ high and reaches the bottom at $18\ \text{m/s}$. The path down the slope is $80\ \text{m}$ long.

1. Energy at the top. $mgh = 60\ \text{kg} \times 9.8\ \text{m/s}^2 \times 25\ \text{m} = 14\,700\ \text{J}$.
2. Energy at the bottom. $\tfrac12 \times 60\ \text{kg} \times (18\ \text{m/s})^2 = 9720\ \text{J}$.
3. Lost to friction and drag. $14\,700\ \text{J} - 9720\ \text{J} = 4980\ \text{J}$, so the average resisting force is $\dfrac{4980\ \text{J}}{80\ \text{m}} = 62.3\ \text{N}$.

Think first. When the speed at the end is known, energy tells you the heat. When the friction is known, it tells you the speed.

Practise with new numbers

Try it in Python

The program tries several starting heights for a loop of radius $5.0\ \text{m}$.

Remember these

• With friction: $K_1 + U_1 = K_2 + U_2 + E_{\text{heat}}$.
• On curves: energy for the speed, $F = \dfrac{mv^2}{r}$ for the forces.
• A loop of radius $R$ needs a start at least $2.5R$ high; a string in a vertical circle needs $\sqrt{5gL}$ at the bottom.

Check yourself

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

1. What least height does a cart need to loop a smooth loop of radius $4.0\ \text{m}$?

2. A $2.0\ \text{kg}$ ball is dropped from $10\ \text{m}$ and lands at $13\ \text{m/s}$. How much energy did air resistance take? ($g = 9.8\ \text{m/s}^2$.)

3. Rub your hands together hard. Where does the work you do go?

4. A car goes over a hump-backed bridge of radius $20\ \text{m}$. Above what speed do its wheels leave the road at the top? ($g = 9.8\ \text{m/s}^2$.)