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(A) When the lift is stationary,the reading of the spring balance is equal to the weight of the bag,so $T = mg = 49 \, N$. Taking $g = 9.8 \, m/s^2$,t

Vedclass · Accepted Answer

$24.5$

Vedclass · Answer

(A) When the lift is stationary,the reading of the spring balance is equal to the weight of the bag,so $T = mg = 49 \, N$. 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 equation of motion for the bag is $mg - T' = ma$,where $T'$ is the new reading of the spring balance.Therefore,$T' = m(g - a)$.Substituting the values,$T' = 5 	imes (9.8 - 5) = 5 	imes 4.8 = 24 \, N$.

$A$ spring balance is attached to the ceiling of a lift. $A$ man hangs his bag on the spring and the spring 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 .......... $N$.

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