A raindrop of mass $1.00\, g$ falling from a height of $1\,km$ hits the ground with a speed of $50\,m s^{-1}$. Calculate
$(a)$ the loss of $PE$ of the drop
$(b)$ the gain in $KE$ of the drop
$(c)$ Is the gain in $KE$ equal to loss of $PE$ ? If not why ?
Take, $g = 10\, m s^{-2}$.
A raindrop of mass $1.00\, g$ falling from a height of $1\,km$ hits the ground with a speed of $50\,m s^{-1}$. Calculate
$(a)$ the loss of $PE$ of the drop
$(b)$ the gain in $KE$ of the drop
$(c)$ Is the gain in $KE$ equal to loss of $PE$ ? If not why ?
Take, $g = 10\, m s^{-2}$.
$m =1.00 \mathrm{~g}$
$=1 \times 10^{-3} \mathrm{~kg}$$h =1 \mathrm{~km}=10^{3} \mathrm{~m}g $
$=10 \mathrm{~m} / \mathrm{s}^{2}$
$v =50 \mathrm{~m} / \mathrm{s}$
$\text { (a) Loss in PE } =m g h$
$=1 \times 10^{-3} \times 10 \times 10^{3}$
$=10 \mathrm{~J}$
$(b)$ Gain in $\mathrm{KE}=\frac{1}{2} m v^{2}$
$=\frac{1}{2} \times 1 \times 10^{-3} \times(50)^{2}$
$=\frac{1}{2} \times 10^{-3} \times 2500$
$=1.250 \mathrm{~J}$
$(c)$ No, gain in $KE$ is not equal to the loss in its $PE$, because a part of $PE$ is used in doing work against the viscous drag of air.
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