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:

(1) 24N 

(2) 74N 

(3) 15N 

(4) 49N

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1 Answers

(1) The reading of the spring balance is 24N.

Explanation:

When lift is stationary, 

W1 = mg

When the lift descends with acceleration a,

W2 = m(g – a)

W2/W1 = (g – a)/g

= (9.8 - 5)/9.8 = 4.8/9.8

W2 = W1 x (4.8/9.8)

= 49 x (4.8/9.8)

= 24 N

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