**Calculate force transmitted by the man on the floor of lift:**

**A vertical lift having total mass 500Kg acquires an upward velocity of 2m/sec over a distance of** **3m of motion with the constant acceleration, starting from rest. Determine the tension in the cable supporting lift. If lift while stopping moves with the constant deceleration and comes to rest in 2sec, calculate force transmitted by the man having mass 75kg on the floor of lift during the interval.**

**Sol: **Given,

Mass of lift M_{L } = 500Kg

Final Velocity v = 2m/sec

Distance covered s = 3m

Initial velocity u = 0

Cable tension T = ?

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

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

As lift moves up, T > ML X g Net accelerating force = T - M_{L}g, and is equal to,T - M_{L}g = maT- 500 X 9.81 = 500 X 2/3

**T = 5238.5N .......ANS**

Let force transmitted by the man having mass of 75Kg, is FF - mg = ma

For finding acceleration, by sing relation v = u + at0 = 2 + a X 2

a = -1 m/sec^{2 } ..(*i**i*)

By putting value in equation, F - mg = ma

F - 75 X 9.81 = 75(-1)

**F = 660.75N .......ANS**