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(C) When the lift is stationary,the reading of the spring balance is equal to the weight of the bag,$W = mg = 49 ~N$.Taking $g = 9.8 ~m/s^2$,the mass

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) When the lift is stationary,the reading of the spring balance is equal to the weight of the bag,$W = mg = 49 ~N$.Taking $g = 9.8 ~m/s^2$,the mass of the bag is $m = \frac{49}{9.8} = 5 ~kg$.When the lift moves downward with an acceleration $a = 5 ~m/s^2$,the apparent weight $W'$ is given by the formula $W' = m(g - a)$.Substituting the values,we get $W' = 5 	imes (9.8 - 5) = 5 	imes 4.8 = 24 ~N$.Therefore,the reading of the spring balance will be $24 ~N$.

$A$ spring balance is attached to the ceiling of a lift. $A$ man hangs his bag on the spring and the spring balance reads $49 ~N$,when the lift is stationary. If the lift moves downward with an acceleration of $5 ~m/s^2$,the reading of the spring balance will be (in $N$)

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