Question

A particle of mass $0.1\text{kg}$ is subjected to a force which varies with distance as shown in the figure. If it starts its journey from rest at $x=0$, then its velocity at $x=12\text{m}$ is:

• A. $0\text{m/s}$
• B. $20\sqrt{2}\text{m/s}$
• C. $20\sqrt{3}\text{m/s}$
• D. $40\text{m/s}$

Answer 1

The correct option is D $40\text{m/s}$

$m=0.1\text{kg}$
At $t=0$ particle starts from rest i.e $(v=0)$
At $t=t\text{s},v=v\text{m/s}$

Applying Newton's $2^{\text{nd}}$ law:
$F=ma$
$\Rightarrow F=mv\frac{dv}{dx}$
$\Rightarrow Fdx=mvdv...\left(i\right)$

From graph:


Area under the graph $=$ Area $\text{I}$ $+$ Area $\text{II}$ $+$ Area $\text{III}$
$=(\frac{1}{2}\times 4\times 10)+(4\times 10)+(\frac{1}{2}\times 4\times 10)$
$=80$

Integrating Eq. $\left(i\right)$ on both sides with limits $[x=0\text{to}x=12]$ and $[v=0$ to $v=v]$

${\int }_0^{12}Fdx=m{\int }_0^{v}vdv$
where ${\int }_0^{12}Fdx$ represents area under graph.
$\Rightarrow$ Area under graph $=\left(0.1\right)\frac{v^2}{2}$
$\Rightarrow 80=\left(0.1\right)\frac{v^2}{2}$

$\therefore v=\sqrt{1600}=40\text{m/s}$