Question

With what minimum acceleration can a fireman slide down a rope whose breaking strength is 2/5 th of his weight

Answer 1

ANSWER
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Using Newton's Second law of motion,

As from the figure it is obviously clear that

T - mg = ma

2/5 mg - mg = ma as breaking strength is 2/5th of his weight

=> -3/5 × g = a

=> Acceleration Maximum with which he can slide down the rope is = 3/5 of g

Minimum acceleration with which he can slide down can be zero if he performs a Quasi-static process.


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HOPE IT HELPS ^_^