Question

A 1150 kg car drives up a hill that is 15.7 m high. During the drive, two nonconservative forces do work on the car: (i) the force of friction, and (ii) the force generated by the car's engine. The work done by friction is −2.91×105 J ; the work done by the engine is 6.74×105 J . Find the change in the car's kinetic energy from the bottom of the hill to the top of the hill.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the car = M = 1150 kg

Height of the hill = H = 15.7 m

Work done by friction on the car = W₁ = -2.91 x 10⁵ J

Work done by the engine of the car on it = W₂ = 6.74 x 10⁵ J

Kinetic energy of the car at the bottom of the hill = KE₁

Kinetic energy of the car at the top of the hill = KE₂

Change in kinetic energy of the car = KE = KE₂ - KE₁

By conservation of energy the initial kinetic energy plus the work done by the two non-conservative forces is equal to the potential plus kinetic energy of the car at the top of the hill.

KE₁ + W₁ + W₂ = MgH + KE₂

KE₂ - KE₁ = W₁ + W₂ - MgH

KE = (-2.91x10⁵) + (6.74x10⁵) - (1150)(9.81)(15.7)

KE = 2.06 x 10⁵ J

Change in the car's kinetic energy from the bottom of the hill to the top of the hill = 2.06 x 10⁵ J