How fast is the car moving at the bottom of the hill?

Question

A 1200-kg car, initially at rest, rolls 80 m down a hill inclined at a 6.0⁰ angle. A 400-N force of friction opposes it motion. How fast is the car moving at the bottom of the hill?

Answer 1

Potential energy at the top of the hill is all the energy the car has that can be converted into kinetic energy 80 m down the hill. PE = mgh = (1200)(9.8)(80) = 940800 J.

Friction performs negative work on the car as it goes 80 m down the hill. W = Fd = (400)(80) = 32000 J.

This negative work reduces the kinetic energy at the 80 m mark: KE = 940800 - 32000 = 908800 J.

Set this equal to the KE equation: (0.5)(1200)v² = 908800; solve for v = Sqrt(908800 / 600) = 38.9 m/s