Given data,

Mass of lift ML = 500Kg

Final Velocity v = 2m/sec

Distance covered s = 3m

Initial velocity u = 0 Cable tension T = ?

Apply the relation v^{2} = u^{2} + 2as

2^{2} = 0 + 2a X 3a = 2/3 m/sec^{2 }...(i)

Since lift moves up, T > ML X gNet accelerating force = T – MLg, and it is equal to,T – MLg = maT – 500 X 9.81 = 500 X 2/3

**T = 5238.5N**

Let force transmitted by man of mass of 75Kg, is FF – mg = ma For finding the acceleration, using the relation v = u + at0 = 2 + a X 2

a = –1 m/sec^{2}

Putting the value in equation, F – mg = ma

F – 75 X 9.81 = 75(–1) ...(ii)

**F = 660.75N**