Question 132: Work, Energy and Power & Circular Motion

Question

A uniform cable of mass 'M' and length 'L' is
placed on a horizontal surface such that its (1/n)th
part is hanging below the edge of the
surface. To lift the hanging part of the cable upto
the surface, the work done should be :

Accepted Answer

To solve this problem, we need to determine the work done to lift the hanging part of the cable up to the surface. The work done lifting a small element of the cable will be the weight of the element times the distance it has to be lifted.
Let's take a small section of the cable at a depth $x$ below the surface. This section has a length of $dx$, so its mass is $(M/L)dx$ where $(M/L)$ is the linear mass density of the cable.
The work $dW$ done to lift this small section up to the surface is the weight of the section times the distance it has to be lifted :
$dW = (M/L)gdx  	imes  x$.
Integrating this expression from 0 to L/n (the length of the hanging part of the cable) gives the total work done :
$$W = \int\limits_{0}^{L/n} (M/L)gxdx$$
$$= (Mg/L) \int\limits_{0}^{L/n} xdx$$
$$= (Mg/L) 	imes [x^2/2]_{0}^{L/n}$$
$$= (Mg/L) 	imes [L^2/(2n^2)]$$
$$= MgL/(2n^2)$$
So the work done to lift the hanging part of the cable up to the surface is $MgL/(2n^2)$.
Therefore, the correct answer is Option C :
$$\frac{MgL}{2n^2}$$

Answer

$${{2MgL} \over {{n^2}}}$$

Answer

nMgL

Answer

$${{MgL} \over {2{n^2}}}$$

Answer

$${{MgL} \over {{n^2}}}$$

Work, Energy and Power & Circular Motion

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface such that its (1/n)th part is hanging below the edge of the surface. To lift the hanging part of the cable upto the surface, the work done should be :

Choose the correct answer: