Work Energy And Power Ques 1
1. Consider a drop of rain water having mass $1$ $ \mathrm{g}$ falling from a height of $1$ $ \mathrm{km}$. It hits the ground with a speed of $50$ $ \mathrm{m} / \mathrm{s}$. Take ‘g’ constant with a value $10 $ $\mathrm{m} / \mathrm{s}^2$. The work done by the (i) gravitational force and the (ii) resistive force of air is
[2017]
(a) $\quad $ (i) $1.25 $ $\mathrm{J} \quad $ (ii) $-8.25 $ $\mathrm{J}$
(b) $\quad $ (i) $100 $ $\mathrm{J} \quad $ (ii) $8.75 $ $\mathrm{J}$
(c) $\quad $ (i) $10 $ $\mathrm{J} \quad\quad $ (ii) $-8.75 $ $\mathrm{J}$
(d) $\quad $ (i) $-10 $ $\mathrm{J} \quad $ (ii) $-8.25 $ $\mathrm{J}$
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Answer:
Correct Answer: 1.(c)
Solution: (c) From work-energy theorem,
$ \mathrm{W}_g+\mathrm{W}_a=\Delta \mathrm{K} . \mathrm{E} \text {. } $
or, $m g h+\mathrm{W}_a=\frac{1}{2} m v^2-0$
$10^{-3} \times 10 \times 10^3+\mathrm{W}_a=\frac{1}{2} \times 10^{-3} \times(50)^2$
$ \Rightarrow \mathrm{W}_a=-8.75 $ $\mathrm{J} $
which is the work done due to air resistance
Work done due to gravity $=\mathrm{mgh}$
$ =10^{-3} \times 10 \times 10^3=10 $ $\mathrm{J}$
BETA
