A bucket with mass 100kg when filled with oil is lifted at a

A bucket with mass 100kg when filled with oil is lifted at a constant rate by a pulley from the bottom to the top of a 60-meter hole. Assuming the chain used to lift the bucket has negligible mass, compute the work needed to lift the bucket from the bottom of the hole to the top. Assuming the chain used to lift the bucket has an evenly-distributed mass of 12kg, set up an integral that will compute the amount of work needed to lift the bucket from the bottom to the top of the hole. Assuming the chain used to lift the bucket has an evenly-distributed mass of 12kg and that oil is leaking out of a hole in the bucket at a constant rate such that the bucket and the remaining oil has a mass of 21kg when it reaches the top, set up an integral that will compute the amount of work to lift the bucket from the bottom of the hole to the top (assuming that the bucket is lifted at a constant rate).

Solution

a.) The bucket has a mass of 100kg and is pulled up by 60 metres. Therefore the work done on it will be equal to the gain in potential energy.

Hence, work = mgH = 100 * 9.81 * 60 = 58860 J

b.) Now, for any distance x which has already been pulled, the weight of the chain left hanging would be: (12 - x/5) Kgs [Since 12 kg of mass is distributed over 60 metres, unit length will have mass 1/5 kgs]

That is the net force being applied on the chain and bucket system would be: (12 - x/5) + 100 kgs

Further, we know that dW/ds = F

that is, dW = F.ds = (112 - 0.2*x) dx

The required integral would be: dW = (112x - 0.2*x^2) dx [Limits of integral being from x -= 0 to x =60 metres]

c.) Now, 73 kg of oil leaks over a distance of 100 metres. Therefore, for a unit distance, the oil leaked would be:

73/60 = 1.217

For any distance x, the leaked oil would be 1.217x

Using this in the above formed integral, we obtain the new integral as:

dW = (112 - 1.217x - 0.2*x) dx