An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of $0.25{ms}^{- 2}$ is $(g = 10{ms}^{- 2})$

Question

Dr. Abubacker · Accepted Answer

(a) On moving upward,
$weight,\ \ w = m(g + a) = 5(10 + 0.25)$
$= 5 	imes 10.25 = 51.25\ N$

Q. An object of mass 5 kg is attached to the hook of a spring balance and the balance is suspended vertically from the roof of a lift. The reading on the spring balance when the lift is going up with an acceleration of $0.25{ms}^{- 2}$ is $(g = 10{ms}^{- 2})$

Explanation

Reviewed by Dr. Abubacker

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