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Q.
A man weighing $100\, kg$ slides down a light rope with an acceleration of $1.8\, ms ^{-2}$. If $g=9.8\, ms ^{-2}$, the tension of the rope is
ManipalManipal 2017
Solution:
The resultant of tension and weight given acceleration to the man, hence
$m g-T=m a$
$\Rightarrow T=m g-m a=m(g-a)$
Given, $m=100\, kg$
$g=9.8\, m / s ^{2}, a=1.8\, ms ^{-2}$
$\therefore T=100(9.8-1.8)=800\, N$
