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Q.

A blocks of mass 0.18 kg is attached to a spring of force constant 2Nm1 . The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in  ms1  is  V=N/10.  Then N is

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answer is 4.

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Detailed Solution

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12mu2=μmg×0.06+12kx2
12×0.18u2=0.1×0.18×10×0.06+12×2×(0.06)2 

u=0.4ms1=N10 N=4

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A blocks of mass 0.18 kg is attached to a spring of force constant 2Nm−1 . The coefficient of friction between the block and the floor is 0.1. Initially the block is at rest and the spring is un-stretched. An impulse is given to the block as shown in the figure. The block slides a distance of 0.06 m and comes to rest for the first time. The initial velocity of the block in  ms−1  is  V=N/10.  Then N is