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Question

A 20 - kg block attached to a spring of spring constant 5 N m−1 is released from rest at A. The spring at this instant is having an elongation of 1 m. The block is allowed to move in smooth horizontal slot with the help of a constant force 50 N in the rope as shown. The velocity of the block as it reaches B is (assume the rope to be light)


A

4 ms1

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B

2 ms1

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C

1 ms1

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D

3 ms1

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Solution

The correct option is B

2 ms1


Consider the rope and the blocks as a system.From work-energy theorem, k=Wnet

Here only two forces do non-zero work on the system: one is the spring force and the other is the constant force of 50 N acting on the cable.Let v be the speed of the block when it reaches B, then

mv220=[5×5225×122]+50×2

= 20×v22=40

v=2ms1


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