Question

A 535 kg roller coaster car began at rest at the top of a 93.0 m hill. Now it is at the top of the first loop-de-loop.

An illustration of a roller coaster track with the first hill labeled 93.0 m high and it goes down to a vertical loop with the car at the top. The loop is labeled as 62.0 m high.
This roller coaster’s track is nearly frictionless, so resistance can be ignored. Using g = 9.8 m/s2, what best describes the roller coaster car when it is at the top of the loop-de-loop?

The car has only potential energy, so it is moving at 0 m/s.
The car has both potential and kinetic energy, and it is moving at 24.6 m/s.
The car has both potential and kinetic energy, and it is moving at 34.9 m/s.
The car’s potential energy has all been converted to kinetic energy, so it moves at 42.7 m/s.

Answer 1

Use conservation of energy: mgh_initial = mgh_top + 1/2 m v^2, so v = sqrt(2g(h_initial − h_top)).

v = sqrt(2·9.8·(93.0 − 62.0)) = sqrt(607.6) ≈ 24.6 m/s.

Thus the car has both potential and kinetic energy and is moving at about 24.6 m/s.