A sled is dragged along a horizontal path at a constant speed of 1.5 m/s by a rope that is inclined at an angl

Question

A sled is dragged along a horizontal path at a constant speed of 1.5 m/s by a rope that is inclined at an angle of 30.0° with respect to the horizontal (the figure below). The total weight of the sled is 470 N. The tension in the rope is 230 N. How much work is done by the rope on the sled in a time interval of 20.0 s?

Answer 1

well if the tension in the rope is T the force that the sled is experiencing in the horizontal direction is simply:

Sum of horizontal forces = T(cos 30) - Fk = 0 (Fk = kinetic friction force) The sum of the forces equals zero because the sled has constant velocity (i.e. no acceleration).

T = 230N in the problem.
So Fk = 199.2N

Work = force x distance

distance = (1.5 m/s)(20s) = 30m

So the work done by the rope on the sled is
199.2N(30m) = 5976 Nm = 5976 Joules