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A lift of mass m is supported by a cable that can with stand a force of 3mg. Find the shortest distance in which the lift can be stopped when it is descending with a speed of g4

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a

g8m

b

g16m

c

g32m

d

g64m

answer is D.

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Detailed Solution

Given T=m(g+a) ⇒3mg=m(g+a) ⇒a=2g V2−u2=2as 0−(g4)2=2(2g)×s⇒s=g64m
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