A cat of mass m = 1 kg climbs to a rope hung over a leight frictionless pulley. The opposite end of the rope is tied to a weight of mass M = 2m lying on a smooth horizontal plane. What is the tension of the rope when the cat move upwards with in acceleration `a = 2 m//s^(2)` relative to the rope ?

Explanation

Let a be the absolute upward acceleration of the monkey and a’ be the absolute downward acceleration of the rope. a’ is also the rightward acceleration of M. Then b = a - (-a (since relative acceleration is the vector difference between the absolute accelerations.) or b = a + a'
Considering the upward motion of the monkey
T - mg = ma ...(i)
Considering the rightward motion of M.
T = Ma' = M (o - a)
Eliminating a between (i) and (ii), we get T in terms of b
`T = (nM)/( n + M) (g + b) = ((m xx 2m)/(m + 2m)) (10 + 2) = (2m)/(3) xx 12 = 80 N`