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- A block of mass m lies on a wedge of mass M. The wedge in turn lies on a smooth horizontal surface. Friction is absent everywhere. The wedge-block system is released from rest. All situations given in Column I are to be estimated in duration the block undergoes a vertical displacement h starting from rest. Match the statements in Column I with the results in Column II. ( g is acceleration due to gravity.) <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/28b489ee0ee028ded1fb6228db9d596d-.png /> Column I Column II i Work done by normal reaction acting on the block is a positive ii Work done by normal reaction (exerted by block) acting on the wedge is b negative iii The sum of work done by normal reaction on the block and work done by normal on wedge c zero iv Net work done by all forces on the block is d less than mgh in magnitude
Q.
A block of mass $m$ lies on a wedge of mass $M$. The wedge in turn lies on a smooth horizontal surface. Friction is absent everywhere. The wedge-block system is released from rest. All situations given in Column I are to be estimated in duration the block undergoes a vertical displacement $h$ starting from rest. Match the statements in Column I with the results in Column II. ( $g$ is acceleration due to gravity.)
Column I
Column II
i
Work done by normal reaction acting on the block is
a
positive
ii
Work done by normal reaction (exerted by block) acting on the wedge is
b
negative
iii
The sum of work done by normal reaction on the block and work done by normal on wedge
c
zero
iv
Net work done by all forces on the block is
d
less than mgh in magnitude
Column I | Column II | ||
---|---|---|---|
i | Work done by normal reaction acting on the block is | a | positive |
ii | Work done by normal reaction (exerted by block) acting on the wedge is | b | negative |
iii | The sum of work done by normal reaction on the block and work done by normal on wedge | c | zero |
iv | Net work done by all forces on the block is | d | less than mgh in magnitude |
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Solution:
(i) Angle between velocity of the block and normal reaction on the block is obtuse. Hence, work done by normal reaction on the block is negative.
As the block falls by vertical distance $h$, from work-energy theorem,
Work done by $m g$ + work done by $N= KE$ of the block
$\because \mid$ work done by $N \mid=\frac{1}{2} m v^{2}-m g h<0$
$\because \frac{1}{2} m v^{2}< m g h$
(ii) Work done by normal reaction on wedge is positive.
Since loss in PE of the block = $K E$ of wedge $+K E$ of block, work done by normal reaction on the wedge $=K E$ of the wedge.
Therefore, work done by $N$ on the wedge is less than $m g h$.
(iii) Net work done by normal reaction on the block and wedge is zero.
(iv) The net work done by all forces on the block is positive, because its kinetic energy has increased.
Also $KE$ of the block is less than $m g h$.
Hence, net work done on the block $=$ final $KE$ of the block $
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