Given data;

Weight of empty lift W_{L} = 8KN = 8000N

Weight of man W_{m} = 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 F_{m} =?

Apply the relation v^{2} = u^{2} + 2as, for finding acceleration.

(25)^{2} = 0 + 2a(187.5)a = 1.67m/sec^{2}

Cage moves down only when W_{L} + W_{m} >T

Net accelerating force = (W_{L} + W_{m})- T

Using the relation F = ma, we get (W_{L} + W_{m})- T = ma = [(W_{L} + W_{m})/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,

F_{m} – W_{m} = maF_{m} – 600 = (600/9.81) X 1.67F_{m} **= 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**