Question

A bucket filled with water weighing $ 20kg $ is raised from a well of depth $ 20m $ . If the linear density of the rope is $ 0.2 $ kilogram per meter. The amount of work done is:
(A) $ 4000J $
(B) $ 4040J $
(C) $ 4400J $
(D) $ 4200J $

Answer 1

Hint :To meet the need of the answer in the options use $ g = 10m{s^{ - 1}} $ . We have to calculate the work done for each meter depth. To reduce the addition and lengthy process try to use integral methods. While raising the bucket weight of bucket and rope will be added.

Complete Step By Step Answer:
Given that the mass of the bucket filled with water is $ 20kg $ and the depth of the well is $ 20m $ .
Linear density of Rope is $ 0.2 $ $ kg{m^{ - 1}} $
 $ \therefore $ weight of rope of length $ l $ $ = 0.2l $
To find out overall work done in raising the bucket we have to calculate mass of the bucket and rope together
i.e. $ m = 20 + 0.2l $
If we say that the work done for raising the bucket from depth $ dl $ $ = dW = mg(dl) $
But we have been given the depth of well as $ 20m $
Therefore let us set the range of integral as $ l = 0m $ to $ l = 20m $ .
 $ \Rightarrow dW = mg(dl) $
Integrating both the sides in the set range of $ l $ , we get
 $ \Rightarrow W = \int\limits_0^{20} {mg(dl)} $
 $ = \int\limits_0^{20} {(20 + 0.2l)(10)dl} $
On solving this integral we get,
 $ W = 4400J $
Thus the work done in raising the the bucket from the well with the rope is $ 4400J $
The correct answer is option C.

Note :
The Work done is the amount of force required to move the object to the particular distance i.e. $ W = Fs $ . But if the work is variable with variable distance we have to add all the different works done until we reach the end, this process of adding is much lengthy so we have used the integral form of method to solve this question. It is easier to find work done at any length.