# A vertical lift of total mass 500Kg acquires an upward velocity of 2m/sec over a distance of 3m of motion with

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

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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 v2 = u2 + 2as

22 = 0 + 2a X 3a = 2/3 m/sec...(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/sec2

Putting the value in equation, F – mg = ma

F – 75 X 9.81 = 75(–1) ...(ii)

F = 660.75N