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A particle of mass $\mathrm{m}$ is at rest at the origin at time $\mathrm{t}=0$. It is subjected to a force $\mathrm{F}(\mathrm{t})=\mathrm{F}_0 \mathrm{e}^{-\mathrm{bt}}$ in the $x$ direction. Its speed $v(t)$ is depicted by which of the following curves?
Options:
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
$\mathrm{F}=\mathrm{F}_0 \mathrm{e}^{-\mathrm{bt}}$
$\Rightarrow \quad \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}}$
$\Rightarrow \quad \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}}$
$\quad\int \mathrm{dv}=\int_0^{\mathrm{t}} \frac{\mathrm{F}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$
$\Rightarrow v=\frac{F}{m}\left[\frac{-1}{b}\right]\left[e^{-b t}\right]_0^t$
$\Rightarrow \quad v=\frac{F}{m b}\left[e^{-b t}\right]$
$v=0 \text { at } t=0$
and $\quad v \rightarrow \frac{F}{m b}$ as $t \rightarrow \infty$
So, velocity increases continuously and attains a maximum value of $v=\frac{F}{m b}$ as $t \rightarrow \infty$.
$\Rightarrow \quad \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}}$
$\Rightarrow \quad \frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{F}_0}{\mathrm{~m}} \mathrm{e}^{-\mathrm{bt}}$
$\quad\int \mathrm{dv}=\int_0^{\mathrm{t}} \frac{\mathrm{F}}{\mathrm{m}} \mathrm{e}^{-\mathrm{bt}} \mathrm{dt}$
$\Rightarrow v=\frac{F}{m}\left[\frac{-1}{b}\right]\left[e^{-b t}\right]_0^t$
$\Rightarrow \quad v=\frac{F}{m b}\left[e^{-b t}\right]$
$v=0 \text { at } t=0$
and $\quad v \rightarrow \frac{F}{m b}$ as $t \rightarrow \infty$
So, velocity increases continuously and attains a maximum value of $v=\frac{F}{m b}$ as $t \rightarrow \infty$.
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