In order to raise a mass of 100,kg,a man of m | vedclass

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(C) Let $T$ be the tension in the rope and $a$ be the acceleration of the $100\,kg$ mass upwards.For the $100\,kg$ mass: $T - 100g = 100a$  --- $(i)$F

Vedclass · Accepted Answer

$1218$

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(C) Let $T$ be the tension in the rope and $a$ be the acceleration of the $100\,kg$ mass upwards.For the $100\,kg$ mass: $T - 100g = 100a$  --- $(i)$For the man of mass $60\,kg$: The man climbs the rope with acceleration $5g/4$ relative to the rope. Since the rope itself is accelerating upwards with acceleration $a$,the absolute acceleration of the man is $(5g/4 - a)$ upwards.Thus,$T - 60g = 60(5g/4 - a)$  --- $(ii)$From $(i)$,$a = (T - 100g) / 100 = T/100 - g$.Substitute $a$ into $(ii)$:$T - 60g = 60(5g/4 - (T/100 - g))$$T - 60g = 60(5g/4 - T/100 + g)$$T - 60g = 75g - 0.6T + 60g$$1.6T = 195g$Given $g = 10\,m/s^2$,$1.6T = 1950$$T = 1950 / 1.6 = 1218.75\,N \approx 1218\,N$.

In order to raise a mass of $100\,kg$,a man of mass $60\,kg$ fastens a rope to it and passes the rope over a smooth pulley. He climbs the rope with an acceleration $5g/4$ relative to the rope. The tension in the rope is ....... $N$. $(g = 10\,m/s^2)$

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