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/sec2 ...(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