Heat and power.

Step 1 of 7

The total

When you choose On under Show the energy, four columns appear: potential energy, kinetic energy, heat, and the sum of the three. Predict what the total column will do during the ride, then watch.

Step 2 of 7

How much heat

Rolling resistance is friction, and like all friction it does negative work on the car. That work does not vanish: it warms the wheels and the rails. The energy that becomes heat equals the size of the friction’s work, the force times the distance:

$Q = f \cdot d$⁠

Here the force is a fixed fraction of the weight, $f = \mu m g$⁠ with $\mu = 0.015$⁠, and the distance is the length of track the car has covered, not the height it has dropped.

Step 3 of 7

What is left for motion

Conservation of energy still holds, with one more term: what was potential at the start is now shared between kinetic energy and heat.

$m g h = \tfrac12 m v^2 + Q$$m g h =$$\tfrac12 m v^2 + Q$⁠

The farther the car travels, the more energy has become heat, and the less is left for motion.

Step 4 of 7

A higher start

Without friction, 12.5 m is enough to make the loop. Here the Just enough preset of Start height is 13.5 m.

Try it: lower the Start height to 13 m and to 12.5 m, and watch what happens at the top of the loop.

Back to the experiment ↑

Step 5 of 7

Power

Before every ride a chain hauls the car to the top of the hill. The chain’s motor does work, and it also matters how fast it does it. Power is the work done each second:

$P = \frac{W}{t}$⁠

The unit of power is the joule per second, called the watt: 1 W = 1 J/s. A 10 W bulb turns 10 J of electrical energy into light and heat every second, and a car engine delivers tens of thousands of watts, tens of kilowatts.

Step 6 of 7

Faster, the same work

Sometimes it is handy to work out power from a force and a speed. A force $F$⁠ pushing a body at speed $v$⁠ in its own direction does, each second, $F$⁠ times one second’s distance of work, so

$P = F v$⁠

Step 7 of 7

In summary

• Friction turns energy into heat, $Q = f d$⁠, the size of its negative work. The total energy is conserved: $m g h = \tfrac12 m v^2 + Q$$m g h =$$\tfrac12 m v^2 + Q$⁠.
• Because of the heat, the car has to start higher to make the loop.
• Power is the rate of doing work, $P = \dfrac{W}{t} = F v$$P =$$\dfrac{W}{t} =$$F v$⁠, in watts: 1 W = 1 J/s. The same work in half the time is twice the power.