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Question

The arrangement of pulley-block system is released from rest at t=0 as shown in figure. The velocity of block B is vB=45 m/s upwards at some instant. Then, find the magnitude of velocity of block A in m/s. Consider the string to be massless and inextensible, all pulleys are light and smooth.


A
9 m/s
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B
22.5 m/s
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C
1.5 m/s
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D
3 m/s
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Solution

The correct option is D 3 m/s
Since tension will be an internal force for the system of (blocks+pulleys+strings), the sum of work done by all internal forces on the system will be zero.
Wint=0...(i)

Let the displacement of block B be xB upwards and displacement of block A be xA downwards at an instant. Distributing tension in the strings as shown below.


For block A, work done by net tension force 15T will be ve due to velocity and displacement being in opposite directions.
Simillarly for block B, work done by net tension force T will be +ve
Wint=0
(15T×xA)+(T×xB)=0
xB=15 xA ..(ii)

Differentiating Eq. (ii) w.r.t time, we get the relation between velocities of the blocks.
vB=15 vA
vA=vB15=4515=3 m/s (downward)

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