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Question

A block of mass m is placed on the floor of lift which is moving with velocity v=4t2, where t is time in second and velocity in m/s. Find the time at which normal force on the block is three times of its weight.

A
(3g8)s
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B
g s
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C
g4s
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D
3g s
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Solution

The correct option is C g4s
Given velocity v=4t2
acceleration a=dvdt=8t
Now as lift is moving up
N=m(g+a)
N=m(g+8t)
Given N=3 mg
3mg=mg+3mt
=2/mg=8/mt
t=g4s

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