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Question

A 1.0kg block collides with a horizontal weightless spring of force constant 2.75Nm1 as shown in figure. The block compresses the spring 4.0m from the rest position. If the coefficient of kinetic friction between the block and horizontal surface is 0.25 , the speed of the block at the instant of collision is :
1034027_b3a91d956bf24929b45bbefb36f05f78.png

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Solution

Let the velocity of the block at the time of collision is v.

The kinetic energy of the block is given as,

12mv2=12kx2+μkmgx

12×1×v2=12×2.75×(4)2+(0.25)×1×9.8×4

v2=63.6

v=7.975m/s

Thus, the speed of block at the instant of collision is 7.975m/s.


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