Weight of empty lift WL = 8KN = 8000N
Weight of man Wm = 600N
Distance covered by lift s = 187.5m
Velocity of lift after 187.5m v = 25m/sec
Tension in rope T = ?
Force exerted on the man Fm =?
Apply the relation v2 = u2 + 2as, for finding acceleration.
(25)2 = 0 + 2a(187.5)a = 1.67m/sec2
Cage moves down only when WL + Wm >T
Net accelerating force = (WL + Wm)- T
Using the relation F = ma, we get (WL + Wm)- T = ma = [(WL + Wm)/g]a(8000 + 600) – T = [(8000 + 600)/9.81] X 1.67
T = 7135.98N
Calculation for force exerted by the man Consider only the weight of the man,
Fm – Wm = maFm – 600 = (600/9.81) X 1.67Fm = 714.37N
Since Newton’s third law i.e The force of action and reaction between interacting bodies are equal in magnitude, opposite in direction and have the same line of action.
i.e., Force exerted by the man = F = 714.37N