Question

A particle of mass $\frac{10}{7}\text{kg}$ is moving along the positive $x$ direction. Its initial position is $x=0$ and initial velocity is $1\text{m/s}$. The graph representing the power delivered the by force vs distance travelled for the straight line motion of the particle is given below.


The velocity of particle at $x=10\text{m}$ is:

• A. $\frac{100}{3}\text{m/s}$
• B. $2\text{m/s}$
• C. $4\text{m/s}$
• D. $3\sqrt{2}\text{m/s}$

Answer 1

The correct option is C $4\text{m/s}$

From the given graph, the slope of the straight line is:

$m=\frac{4-2}{10-0}=0.2$ ($\Rightarrow$ slope is $+$ve)

Intercept on Power $P$-axis is $+2$, thus the equation of the line showing $P$ vs $x$ is,

$P=mx+c$

$\Rightarrow P=0.2x+2.....\left(i\right)$

Here the particle started from rest and moves along the $+\text{ve}x$-axis.

$\Rightarrow P=\overrightarrow{F}.\overrightarrow{V}=FV\cos 0^{\circ }=FV$

or, $P=mav$

using relation $a=v\frac{dv}{dx}$

$\Rightarrow P=\left(mv\frac{dv}{dx}\right)v=m{v}^2\frac{dv}{dx}...\left(ii\right)$

From Equation $\left(i\right)$ & $\left(ii\right)$,

$m{v}^2\frac{dv}{dx}=0.2x+2$

or, $m{v}^{2}dv=(0.2x+2)dx$

Substituting limits for integration as $v=1$ at $x=0$ and $v=v$ at $x=10\text{m}$.

$\Rightarrow m{\int }_1^v{v}^{2}dv={\int }_0^{10}(0.2x+2)dx$

$\frac{10}{7}\left[\frac{v^3-1^3}{3}\right]=\frac{0.2}{2}(10{)}^2+2(10)-0$

or, $\frac{10}{21}(v^3-1)=30$

or, $v^3=63+1=64$

$\therefore v=(64{)}^{1/3}=4\text{m/s}$

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: The variation of power w.r.t 'x' from the graph}\\ \text{serves as a clue that force is variable.}\\ \text{In order to obtain velocity, power and displacement}\\ \text{should be linked through}P=FV=\left(mv\frac{dv}{dx}\right)v\end{array}$