What is the speed of this crate?

Question

A crate of mass 13.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.95 m. The change in kinetic energy of the crate is 50.26 J.

What is the speed of the crate after being pulled 5.95 m?

Answer 1

We can get this directly from the change in KE: initial KE is 13.0(1.50)^2 / 2 = 14.625 J, so the final KE is 64.88J and the velocity is therefore √(2KE/m) = 3.16 m/s.

I haven't checked to see if this is consistent with the other figures given in the problem. It's probably a worthwhile exercise... [Edit: I said earlier that it didn't turn out to be consistent. This was an error on my part; I miscalculated the frictional force. Correct calculations are below.]

Force of gravity applies a 13.0(9.81)sin20° = 43.6N force down the incline, and a normal force of 13.0(9.81)cos20° = 119.8N, so the friction force is 0.400(119.8) = 47.9N. So the net force is 100N - 43.6 - 47.9 = 8.44N, yielding an acceleration of 0.650 m/s^2. From v^2 = u^2 + 2as we get v^2 = 1.50^2 + 2(0.650)(5.95) = 9.98 and hence v = √9.98 = 3.16 m/s.
We can also check the change of KE = work done on the crate = mas = 13.0(0.650)(5.95) = 50.26 J.