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A block of mass 0.18 kg is attached to a spring of force-constant 2 N/m. 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 m/s is V = N/10. Then N is 


Solution


12mV2=μmgx+12kx2

12(0.18)V2=(0.1)(0.18)(10)(0.06)+122(36×104)

0.09V2=108×104+36×104

0.09V2=104[144]

0.3V=102(12)

V=0.4 ms1=N10

N=4

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