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 rope. The tension in the rope is (g = 10 m/s²).

 

For man T–mg = m(a–a¹)
For load T–Mg = Ma¹

$\mathrm{Mg}-\mathrm{mg}=\mathrm{ma}-\left(\mathrm{m}+\mathrm{M}\right){\mathrm{a}}^1$
Solving a¹ = 2.19 m/s²
T–100g = 100(2.19) ; T = 1219 N

Here a = Acceleration of the man relative to the rope and a¹ is the acceleration of the 100 kg mass i.e., acceleration of the rope. So acceleration of the man relative to the ground = (a-a¹).