Inclined plane with massless, frictionless winch

[r34racer01]

A 100 kg box is pulled up an inclined plane by a massless, frictionless winch to which it is attached by a massless rope. The plane makes an angle of 45° with respect to the horizontal. The coefficient of friction between the box and the plane is μ = 0.25 (with μs = μk). The winch has a radius of 0.25 m and is turned at a constant angular velocity of ω = 3.14 radians/sec.

How much work is done on the 100 kg box by the rope when the winch makes 10 complete revolutions?

(a) 4380 J
(b) 5235 J
(c) 9262 J
(d) 12531 J
(e) 13620 J


So I did d = [2(.25)(3.34)]10 since we need 10 rev. And then...
W = mgh + uNd
W = mg(d sin 45) + u(mg cos 45)d
but I'm not getting the right answer, I think I'm suppose to get 13620J, but how do I get that?

 

[Rake-MC]

I'm not so sure that W = mgh + uNd
Isn't mgh the gravitational potential that is gained by doing the work?

I would sum up the forces that are opposing the pull of the rope (parallel to the plane) and multiply it by the distance it moves

 

[thomate1]

To calculate the work done on the box by the rope, we need to consider the forces acting on the box as it is pulled up the inclined plane by the winch.

First, we need to calculate the force of gravity acting on the box, which is equal to its mass (100 kg) multiplied by the acceleration due to gravity (9.8 m/s^2), giving us a force of 980 N.

Next, we need to calculate the force of friction between the box and the inclined plane. The coefficient of friction (μ) is given as 0.25, so we can calculate the maximum frictional force using the equation Ff = μN, where N is the normal force acting on the box. Since the plane makes an angle of 45° with the horizontal, the normal force is equal to the component of the force of gravity acting perpendicular to the plane, which is mg cos 45. Plugging in the values, we get a maximum frictional force of 245 N.

Now, we can calculate the net force acting on the box by subtracting the frictional force from the force of gravity. This gives us a net force of 735 N.

Since the winch is turning at a constant angular velocity, we can use the equation W = Fd, where F is the force and d is the distance traveled. In this case, the force is the net force acting on the box and the distance traveled is the circumference of the winch (2πr) multiplied by the number of revolutions (10). Plugging in the values, we get a work done by the rope of 13620 J, which matches the answer given.

Therefore, the correct answer is (e) 13620 J.