A person of mass 50 , kg stands on a weighing | vedclass

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(A) Given:Mass of the person,$m = 50 \, kg$Acceleration of the lift,$a = 9 \, m/s^2$ (downwards)Acceleration due to gravity,$g = 10 \, m/s^2$When a li

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$5$

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(A) Given:Mass of the person,$m = 50 \, kg$Acceleration of the lift,$a = 9 \, m/s^2$ (downwards)Acceleration due to gravity,$g = 10 \, m/s^2$When a lift descends with an acceleration $a$,the apparent weight $R$ of a person of mass $m$ is given by:$R = m(g - a)$Substituting the given values:$R = 50 	imes (10 - 9)$$R = 50 	imes 1 = 50 \, N$The reading of the weighing scale is usually calibrated in kilograms,which represents the apparent mass $m_{app} = R/g$.$m_{app} = \frac{50 \, N}{10 \, m/s^2} = 5 \, kg$Therefore,the reading of the weighing scale is $5 \, kg$.

$A$ person of mass $50 \, kg$ stands on a weighing scale in a lift. If the lift is descending with a downward acceleration of $9 \, m/s^2$,what would be the reading of the weighing scale (in $, kg$)? (Take $g = 10 \, m/s^2$)

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