A block is equilibrium at rest in a lift as shown in figure. Now lift starts acclerating. The minimum magnitude of acceleration of lift for which block will touch the floor of lift, is
A
`(W_(1))/(M)`
B
`(2kh)/(M)`
C
`(kh)/(2M)`
D
`(3)/(2)(kh)/(M)`
Text Solution
Verified by Experts
The correct Answer is:
C
From work energy theorem `W_(S) + W_("pseudo") + W_("spring") = DeltaKE` `rArr Mgh + Mah + (1)/(2)kx^(2) - 1/2k(h+x)^(2) = 0` where `Mg = kx` `rArr Mgh + Mah - 1/2kh^(2) - khx = 0 :. a = (kh)/(2M)`
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