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a loadw is to b raised by a rope from rest to rest through a height h. the greatest tension the rope can safely bear is nw. th eleast time in which the ascent cabn b made will be
2 forces will act on the load: 1) tension upwards 2) mg downwards. its to b pulled upwards in least time so there must b an acceleration upwards. let max tension be nw. so, (nw)-mg=mamax amax = (nw)-mg/m---------(1) using equation s=ut+1/2at2 we get h= 0 + 1/2 {(nw) - mg/m}t2 (from 1) t2 = 2mh/(nw)-mg t= √2mh/nw-mg
2 forces will act on the load:
1) tension upwards
2) mg downwards.
its to b pulled upwards in least time
so there must b an acceleration upwards.
let max tension be nw.
so, (nw)-mg=mamax
amax = (nw)-mg/m---------(1)
using equation s=ut+1/2at2
we get h= 0 + 1/2 {(nw) - mg/m}t2 (from 1)
t2 = 2mh/(nw)-mg
t= √2mh/nw-mg
This is not the correct answer. Since the block will first accelerate and then decelerates. In this solution you have considered only one aspect. It should be in 3 steps , acceleration, constant speed amd then deceleration. Think again
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