Question

A chain 75 meters long whose mass is 25 kilograms is hanging over the edge of a tall building and does not touch the ground. How much work is required to lift the top 3 meters of the chain to the top of the building? Use that the acceleration due to gravity is 9.8 meters per second squared.

Answer 1

Answer:

dW = F ds     differential unit of work

F = w (L - x)    force at top and w is the weight / length

W = w (L x - x^2 /2)     work done in moving 3 meters of chain

W = w [3 L - 9 / 2) = w (3 * 75 - 9/2) = 220.5 w

Check:

Initial PE = 0      with w * 75 / 2 = 0

Final PE = w (72 * 1.5 + 3 * 37.5) = w 220.5

Increase in PE due to 72 m of chain on side and 3 meters of chain atop

If w is weight / length then w = g / 3

The final result is 220.5 w = 73.5 g = 720.3 J