A roller coaster of mass 2000 kg is rolling down a track with an instantaneous speed of 10 m/s. The roller coaster has a force of unknown magnitude act on it, so now it is travelling at only 6 m/s. What was the work done on it?

J

We already know from a previous problem that the initial kinetic energy is KE = 1/2 mv^{2} = 1/2 (2000) (10) ^{2} = 100,000J.

The final kinetic energy is, when we plug in m = 2000, and v = 6, KE = 1/2 mv^{2} = 1/2 (2000) (6) ^{2} = 36000J.

We now subtract the initial and final kinetic energies to get the work done on the roller coaster. 100,000-36,000 = 64,000J.