A car with a mass of 10kg is pushed 100m parallel with the horizontal surface with a force of 500N. The surface has a coefficient of friction of 0.2. How much work does each of the following forces do on the box: (a) 500N push, (b) the friction force, (c) What is the net work done on the car?

work done by 500 n force = 500 N * 100 m = 50,000 Joules input

work done by friction = -0.2 * 10 * 9.81 * 100 m = -1962 Joules
net work = increase in kinetic energy = 48,038 Joules

Did you know?

Did you know that work is a measure of the energy transferred when a force is applied to an object and it moves in the direction of the force applied? In the given scenario, a force of 500N is applied to a car with a mass of 10kg, pushing it 100m parallel to the horizontal surface. The friction force on the surface, determined by the coefficient of friction (0.2), opposes the motion of the car. (a) The force of 500N does work on the car by transferring energy to it as it is pushed along the surface. (b) The friction force also does work on the car, but in the opposite direction due to its opposing nature. It reduces the kinetic energy of the car. (c) The net work done on the car is the sum of the work done by the force pushing it and the work done by the friction force.