A child stands on a weighing machine inside a | vedclass

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Question

(A) Let $m$ be the mass of the child and $N$ be the normal force (reading of the weighing machine).When the lift moves downwards with acceleration $a

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Let $m$ be the mass of the child and $N$ be the normal force (reading of the weighing machine).When the lift moves downwards with acceleration $a = \frac{g}{3}$,the equation of motion is $mg - N_1 = ma$.Given $N_1 = 20 \ N$,so $mg - 20 = m(\frac{g}{3})$.$mg - \frac{mg}{3} = 20 \implies \frac{2mg}{3} = 20 \implies mg = 30 \ N$.When the lift moves upwards with acceleration $a = \frac{g}{3}$,the equation of motion is $N_2 - mg = ma$.$N_2 = mg + ma = mg + m(\frac{g}{3}) = \frac{4mg}{3}$.Substituting $mg = 30 \ N$,we get $N_2 = \frac{4 	imes 30}{3} = 40 \ N$.

$A$ child stands on a weighing machine inside a lift. When the lift is going down with acceleration $\frac{g}{3}$,the machine shows a reading $20 \ N$. When the lift goes upwards with acceleration $\frac{g}{3}$,the reading would be ($g=$ gravitational acceleration). (in $N$)

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