2.
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 g/4.
1) g/8 m 2)g/16 m 3) g/32 m 4) g/64 m
Answers
answer : option (4) g/64
it is given that a lift of mass m is supported by a cable that can with stand a force of 3mg.
we have to find the shortest distance in which the lift can be stopped when it is descending with a speed of g/4.
so, T = mg + ma
⇒3mg = mg + ma
⇒a = 2g , it means acceleration is directed upward.
now using formula, v² = u² + 2as
let shortest distance is s in which lift can be stopped.
here final velocity, v = 0, u = g/4 , a = 2g
so, 0 = (g/4)² + 2(-2g)s
⇒g²/16 = 4gs
⇒s = g/64
therefore, shortest distance in which lift can be stopped when it is descending with speed of g/4 is g/64.
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