A 290- kg crate hangs from the end of a 14.0 m long rope. You pull horizontally with a varying force to move it a distance d = 4.9 m to the right.

Question

1.What is the magnitude of the applied force F when the crate is at rest in its final position?
2.What is the work done by the weight of the crate?
3.What is the work you do on the crate?

I am very much stuck, and I would very much appreciate help on all three parts. Thanks.

drwls · Answer

In its new equilibrium position, the rope is inclined to vertical by an angle
arcsin theta = 4.9/14 = 20.49 degrees
In the process, the crate is raised a distance
delta Y = 14 m (1 - cos theta) = 0.886 m
1. Rope tension T is given by
T cos theta = Mg
Solve for T
The applied horizontal force F needed to hold it in equilibrium is
F = T sin theta
3. The work done on the crate is the potential energy change due to the higher elevation, M g delta Y. (See the calculation of delta Y above)

2. The work done by the crate is the negative of the previous answer.

hshs · Answer

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