Question

A body of mass 15 kg is hung by a spring balance in a lift what would be the reading of the balance when 1) the lift is ascending with an acceleration of 2m/s squared 2)descending with the same acceleration 3) descending with a constant velocity of 2m/s in sl arora

Answer 1

1)

m = mass of the body hanged from spring balance = 15 kg

W = actual weight of the body hanged in downward direction = mg = 15 x 9.8 = 147 N

a = acceleration of the lift in upward direction = 2 m/s²

F = reading of the spring balance

force equation for the motion of the body hanged in the lift is given as

F - W = ma

F - 147 = 15 (2)

F = 177 N

2)

a = acceleration of the lift in downward direction = 2 m/s²

force equation for the motion of the body hanged in the lift is given as

W - F = ma

147 - F = 15 x 2

F = 117 N

3)

Since the lift is going at constant velocity

a = acceleration of the lift = 0 m/s²

W - F = 0

F = W

F = 147 N

Answer 2

it will help u guys...Live live laugh