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Question

A mass of 1 kg is dropped from a height of 2m on a horizontal spring board. The vertical spring supporting the board has a spring constant of 87.5 N/m .The maximum distance by which the mass compress the spring is close to..?

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Solution

the above answer is correct ,only mistake in calculations
So will explain only calculating step from this onward ,rest are same
x=[(2*1*10 + √4*10^2*1 + 4*87.5(2*10*2*1)] /87.5*2
=[20+√(400+14000)]/175
=20+√(14400)]/175
=20+120/175=140/175
=0.8m
The method is same as above

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