Question

A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49N, when the lift is stationary. If the lift moves downward with an acceleration of 5m/s² , the reading of the spring balance will be

(a) 24N 

(b) 74N 

(c) 15N 

(d) 49N.

Answer 1

(a) : When lift is standing, W₁ = mg
When the lift descends with acceleration a, 

W₂ = m (g – a)

Therefore, w₂/w₁ = m(g - a)/mg = (9.8 - 5)/9.8 = 4.8/9.8

or w₂ = w₁ x 4.8/9.8 = (49 x 4.8)/9.8 = 24N.