Question

Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 55.05 kg, down a theta= 64.84º slope at constant acceleration a=-2.00 m/s², as shown in Figure. The positive direction is downward along the slope. Hence, the negative sign in acceleration, if present, means the direction of the acceleration is pointing along the slope upwards. The coefficient of friction between the sled and the snow is 0.100. How many Joules of work is done by the tension in the rope as the sled moves 3.45 m along the hill? Use g= 10 m/s².

Answer 1

Using Newton's second law

F_net = ma

T + f - mgsin = ma

T + umgcos - mgsin = ma

T = ma - umgcos + mgsin

T = 55.05 * (-2) - 0.1 * 55.05 * 10 * cos 64.84 + 55.05 * 10 * sin 64.84

T = 364.77 N

so,

work done,

W = 364.77 * 3.45

W = 1258.44 J