A 76-kg crate starts from rest at A and is moving at 8 m/s. Using integration, find the work done between A and B of the variable x component of force F= 756 N. (See attached jpeg for the geometry of the problem)
EDIT: We are doing a variation of the problem shown in the jpeg below, the only difference being the length from A to B (5 m instead of 6 m) and the vertical position of the pulley (5 m instead of 6 m).
U1-2= ∫F cos θ dx
The Attempt at a Solution
Since F is constant and cos θ is variable, I've tried to relate cos θ in terms of the displacement in x. Since cos θ = Δx/Δ(Length of rope), and because the vertical displacement of the pulley remains constant (5 m), I obtained a value of:
cos θ = x/(x2 + 25)-1/2
Whenever I solve this integral I end up with a value equal to the magnitude of the force multiplied by a displacement of 7.5 m in x which, from the geometry of the problem, is impossible. Obviously, I'm just not looking at this correctly. Any hints would be very helpful, and thank you in advance.