help?? application integral [Work]

Question

Answer 1

(3)

Step 1:

A cable weighs 2 kg/meter.

Length of the shaft is 75 meter.

Intially weight of the load at bottom is 50kg.

First find force at any point.

Let the position of the load at any point is x.

At the bottom of the shaft  and at the top of the shaft .

At any point in the shaft it is .

The force is any point x is then nothing more than the weight of the cable and load at that point.

Force is .

Step 2:

Work done is integral of force function.

Here the limits are 0 to 40. (Since the load travelling from the bottom to 40 m)

Sum/difference property of Integral :.

Therefore the amount of workdone is 24000 N-m.

Solution :

The amount of workdone is 24000 N-m.