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A person of mass $50 \mathrm{~kg}$ stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of $9 \mathrm{~ms}^{-2}$, what would be the reading of the weighing scale? $\left(g=10 \mathrm{~ms}^{-2}\right)$
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Verified Answer
When lift is descending with a acceleration, $a$ the apparent weight of a body of mass $m$ decreases on weighting scale is
$$
W=R=(m g-m a)=m(g-a)
$$
As given that,
Mass of the person, $m=50 \mathrm{~kg}$
Descending acceleration, $a=9 \mathrm{~m} / \mathrm{s}^2$
Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$
Apparent weight of the person due to reaction force by lift on weighing scale.
$W^{\prime}=m(g-a)=50(10-9)=50 \mathrm{~N}$
$\therefore$ Reading of the weighing scale
$$
=\frac{R}{g}=\frac{50}{10}=5 \mathrm{Kg} \text {. }
$$
$$
W=R=(m g-m a)=m(g-a)
$$
As given that,
Mass of the person, $m=50 \mathrm{~kg}$
Descending acceleration, $a=9 \mathrm{~m} / \mathrm{s}^2$
Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$
Apparent weight of the person due to reaction force by lift on weighing scale.
$W^{\prime}=m(g-a)=50(10-9)=50 \mathrm{~N}$
$\therefore$ Reading of the weighing scale
$$
=\frac{R}{g}=\frac{50}{10}=5 \mathrm{Kg} \text {. }
$$
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