A 70 kg trunk is pushed 7.6 m at constant speed up a 30° incline by a constant horizontal force. The coefficient of kinetic friction between the trunk and the incline is 0.21.

Question

(1) From now suppose the 70 kg trunk is pushed 7.6 m at constant speed up a 30° incline by a force along the plane. The coefficient of kinetic friction between the trunk and the incline is 0.21.

Calculate the work done by the applied force.

(2) Calculate the work done by the weight of the trunk.

(3) How much energy was dissipated by the frictional force acting on the trunk?

Answer 1

First draw the picture of the problem and the forces acting on the system i.e the applied force, the frictional force, the weight and normal force.

Second, find the x- and y-components of the applied force. i.e

F_x = Fcos(30°) and

F_y = Fsin(30°)

Note that only the x-component of the applied force does work on the trunk.

How do you find the magnitude of the force? We know that the trunk is moving with constant speed. So, the x-component of the applied force is equal to the x-component of the gravitational force plus the force of friction.

==> 0 = Fcos(30°) - mgsin(30°) - mgcos(30°)μ_k [μ_k is coefficient of kinetic friction]

==> 0.86F = 70(9.8)(0.5) + 70(9.8)(0.86)(0.21) ==> F = 542.9N

Now,

F_x = 542.9cos(30°) = 470.2N and

F_y = 542.9sin(30°) = 271.5N

Therefore:

1) W_Fx = F_xd = (470.9N)(7.6m) = 3578.8J

2) The work done by the weight of the trunk can be calculated similarly. Only the x-component of the weight does work on the trunk.

W_g = -mgsin(30°)d = -2644.0J

Notice the direction of the weight force is opposite of the direction of the motion, so this force does negative work on the trunk.

The energy dissipated by the frictional force can be calculated using:

W_f = -mgcos(30°)μ_kd = -70(9.8)(0.86)(0.21)(7.6) = -941.6J

We expected the sum of work done by the friction and weight to be approximately equal in magnitude to the work done by the applied force. This shows the consistency of our calculations.