Final answer:
The work done in lifting a 10 ft cable weighing 65 lb is found by integrating the force due to the weight over the distance lifted, resulting in 325 ft-lb of work.
Step-by-step explanation:
The student is asking about the work done in lifting the bottom end of a 10 ft flexible cable that weighs 65 lb up to the branch from which it hangs, essentially folding it in half. In this Physics problem, work is defined as the force times the distance over which the force is applied. We would need to account for the fact that as we lift the cable, the distance over which the force is applied changes since the weight of the cable being lifted decreases as we get closer to the branch.
To solve this problem, we consider a small segment of the cable at a distance x from the bottom. The weight of this segment is dW which is a differential amount of force due to gravity acting on that segment. The work done in lifting this differential length dx to the height of the branch is dW dx. Since the cable is uniform, the weight per unit length w is constant, so dW = w dx. Integrating this from 0 to the length L of the cable gives us the total work done, which is W = (1/2) w L^2. Plugging in the values (with L = 10 ft and w = 65 lb/10 ft), we get the work done as 325 ft-lb.