Answer:
315,000 ft·lb
Explanation:
At 300 ft and 5 lb/ft, the weight of the cable is (300 f)(5 lb/ft) = 1500 lb. The work done to raise it is equivalent to the work done to raise the cable's center of mass. Since the cable is of uniform density, its center of mass is half the cable length below the winch.
total work done = work to raise cable + work to raise load
= (1500 lb)(150 ft) +(300 lb)(300 ft) = 315,000 ft·lb
