Work EX 1

The diagram below shows the skier sliding downhill and all the forces acting on him. The diagram also shows the vectors of displacement and each of the forces drawn tail-to-tail.

The skier slides 200 m. The friction is given to be 20 N. The gravitational force is 600 N.

The work done by the normal force is zero because this force is perpendicular to the displacement.

(a) The work done by gravity is equal to $W_{F_{G}}=F_{G}\Delta r\cos {\theta }=(600)(200)\cos(65^{\circ })=50714J$.

(b) The work done by friction is $W_{F_{f}}=F_{f}\Delta r\cos {\theta }=(20)(200)\cos(180^{\circ })=-4000J$. Note that the work done by friction will always be negative in problems involving work. We will be always dealing with kinetic friction (when the displacement is zero the work is zero) which is always opposite to the direction of motion.