Question

Kathy is changing the tire of her car on a steep hill 20m high. She trips and drops the 10kg spare tire which rolls down the hill. What is the speed of the tire at the top of the next hill if the height of the hill is 5m high?

Answer 1

Potential Energy at top of hill:
PE = mgh = (10)(9.8)(20) = 1960 J

Kinetic Energy at bottom of hill is equal to PE at top of hill, since all energy is converted:
KE = PE = 1960 J

Potential Energy at top of next hill:
PE = mgh = (10)(9.8)(5) = 490 J

Since energy is conserved, we can say that the kinetic energy at the top of the next hill summed with the potential energy of being at the top of the next hill must be equal to the energy that was initially in the system (1960 J).

Kinetic Energy at top of next hill:
KE = KE(after first hill) - PE(next hill) = PE(first hill) - PE(next hill) = 1960-490 = 1470 J

Using a representation of Kinetic Energy:
KE = mv^2 / 2
v^2 = 2KE / m
v = sqrt(2KE / m)

v = sqrt[2(1470) / (10)] = sqrt(294) = 17.15 m/s

The tire is traveling at a speed of 17.15 m/s at the top of the next hill.