Work Power and Energy 1 Question 1
1. A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $\frac{1}{n}$ 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
(2019 Main, 9 April I)
(a) $\frac{2 M g L}{n^{2}}$
(b) $n M g L$
(c) $\frac{M g L}{n^{2}}$
(d) $\frac{M g L}{2 n^{2}}$
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Answer:
Correct Answer: 1. (d)
Solution:
Given, mass of the cable is $M$.
So, mass of $\frac{1}{n}$ th part of the cable, i.e. hanged part of the cable is
$$ =M / n $$
Now, centre of mass of the hanged part will be its middle point. So, its distance from the top of the table will be $L / 2 n$.
$\therefore$ Initial potential energy of the hanged part of cable,
$$ \begin{aligned} U _i & =\frac{M}{n}(-g) \frac{L}{2 n} \\ \Rightarrow \quad U _i & =-\frac{M g L}{2 n^{2}} \end{aligned} $$
When whole cable is on the table,
its potential energy will be zero.
$$ \therefore \quad U _f=0 $$
Now, using work-energy theorem,
$$ \begin{aligned} & W _{\text {net }}=\Delta U=U _f-U _i \\ & \Rightarrow \quad W _{\text {net }}=0–\frac{M g L}{2 n^{2}} \\ & \Rightarrow \quad W _{\text {net }}=\frac{M g L}{2 n^{2}} \end{aligned} $$
[using Eqs. (ii) and (iii)]
2 Normal reaction force on the block is
$$ N=m a _{\text {net }} $$
where, $a _{\text {net }}=$ net acceleration of block.
$$ \begin{aligned} & =g+a=g+\frac{g}{2}=\frac{3 g}{2} \\ \Rightarrow \quad N & =m g+\frac{g}{2}=\frac{3 m g}{2} \end{aligned} $$
Now, in time ’ $t$ ’ block moves by a displacement $s$ given by
$$ s=0+\frac{1}{2} a t^{2}=\frac{1}{2} \frac{g}{2} \quad t^{2} \quad(\because u=0) $$
Here, $\quad a=\frac{g}{2}$ (given)
$\therefore$ Work done $=$ Force $\times$ Displacement
$$ \Rightarrow \quad W=\frac{3 m g}{2} \times \frac{g t^{2}}{4}=\frac{3 m g^{2} t^{2}}{8} . $$
