Q7 - This problem has four parts. I will list the questions for each of the four parts, and any other additional info in the "Description" below.

Question

Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 82.6 kg, down a 60.0° slope at a constant speed. The coefficient of kinetic friction between the sled and the snow is 0.100.

Part a) How much work, in joules, is done by friction as the sled moves 32.9 m along the hill?

Part b) How much work, in joules, is done by the rope on the sled over this distance?

Part c) What is the work, in joules, done by the gravitational force on the sled?

Part d) What is the net work, in joules, done on the sled?

Answer 1

Conservation of Energy is the method here.

1) Draw a Free Body diagram of the forces acting on the sled.

There are four: Gravity (mg), the Normal Force, F_N = mg cos θ, friction F_f = μ F_N and the force of the rope lowering the sled F_T .

There is an additional force, the component of gravity parallel to the plane, F_II = mg sin θ, which you will use when summing the forces.

a) W_friction = F_f ⋅d. If you calculated the forces listed above, which you can from the given data, you can calculate this quantity. Notice that the angle of the force relative to the motion is 90°, so it has a negative value as it is removing energy from the system.

b) Since the sled is moving at a constant speed, the Force of the rope, F_T + the Force of friction, F_f have to equal the F_II = mg sin θ. Solve for F_T and calculate the work: W = F_T ⋅ d. This work is also negative as it is in the direction opposite to the motion.

c) You can use W = ΔPE = mgh, where h = d sin θ, or use F_II ⋅ d = mg sin θ d. The answer is the same.

d) Since the sled is not changing its speed, ΔKE = 0, the only net work done is by gravity. You get the answer in c)