Kinetic and potential energy.

Step 1 of 7

Energy stored in height

To lift a car of mass $m$⁠ to a height $h$⁠ at a steady speed, you pull it up with a force equal to its weight, $mg$⁠, through a vertical distance $h$⁠. That work is $mgh$⁠, and it is not lost: it is stored in the car as gravitational potential energy,

$E_p = m g h$⁠

where $h$⁠ is the height above a level you choose as zero, here the bottom of the track. The car waits at rest at the start height, so at the moment of release this is all of its energy.

Step 2 of 7

Height becomes speed

As the car goes down, its weight does positive work on it, and the potential energy turns into kinetic energy, $E_k = \tfrac12 m v^2$$E_k =$$\tfrac12 m v^2$⁠. Without friction their sum does not change for the whole ride. This is conservation of mechanical energy:

$m g h_1 + \tfrac12 m v_1^2 = m g h_2 + \tfrac12 m v_2^2$$m g h_1 + \tfrac12 m v_1^2 =$$m g h_2 + \tfrac12 m v_2^2$⁠

Look inside is at Energy, and there you can see it: the orange potential column and the light-blue kinetic column trade places while the total column stays level. At the bottom of the loop the height is zero, and all the energy is kinetic.

Step 3 of 7

Does the mass matter?

Every term on both sides of the equation has an $m$⁠ in it. Predict what that says about the speed of a heavier car, then watch.

Step 4 of 7

At the top of the loop

The top of the loop is 10 m up, twice its radius. The car is still high there, so only part of its energy is kinetic. What sets the speed is how far the car has dropped from where it started, not how high it is.

Step 5 of 7

Is the loop’s height enough?

It seems that a car starting at the height of the top of the loop would just reach it. Predict what will happen, then watch.

Step 6 of 7

The lowest start

At the top of the loop the car is upside down, and only gravity and the rail bend its path. For the rail to stay pressed against it, it has to be moving fast enough up there. So it must start higher than the top of the loop, and arrive there with some speed left. How much higher? Find out for yourself.

Try it: once you have found it, change the Car mass at that height. Does the car still make the loop?

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Step 7 of 7

In summary

• A body at height $h$⁠ stores gravitational potential energy $E_p = m g h$⁠, and a moving body carries kinetic energy $E_k = \tfrac12 m v^2$$E_k =$$\tfrac12 m v^2$⁠.
• Without friction their sum is conserved, so the speed depends only on how far the body has dropped: $v = \sqrt{2 g \Delta h}$$v =$$\sqrt{2 g \Delta h}$⁠. The mass cancels.
• In a loop of radius $R$⁠, without friction, the car must start at least $2.5R$⁠ up. In the next lesson you will add friction and see where part of the energy goes.