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Force acting on $a$ body of mass $1 kg$ is related to its position $x$ as $F = x^3 - 3x\, N$. It is at rest at $x = 1$. Its velocity at $x = 3$ can be ......... $m/s$
$4$
$3$
$2$
$5$
Solution
$F=\left(x^{3}-3 x\right) N$
from Newton's 2nd law, $F=$ ma
$\Rightarrow m a=\left(x^{3}-3 x\right)$
$\Rightarrow 1 kg \times a =\left( x ^{3}-3 x \right)[ mass$ of body $, m =1 kg ]$
$\Rightarrow a=\left(x^{3}-3 x\right)$
acceleration is the rate of change of velocity with respect to time.
i.e., $a=d v / d t=v d v / d x$
$\Rightarrow vdv / dx =\left( x ^{3}-3 x \right)$
$\underset{\rightarrow}{\int}_{u}^{v} v d v=\int_{1}^{3} x^{3}-3 x d x$
$\Rightarrow\left(v^{2}-u^{2}\right) / 2=\left[\frac{x^{4}}{4}-\frac{3}{2} x^{2}\right]_{1}^{3}$
as body is at rest at $x=1,$ so $u=0$
$\Rightarrow v ^{2} / 2-0=\left[3^{4} / 4-3 / 2(3)^{2}\right]-\left[1^{4} / 4-3 / 2(1)^{2}\right]$
$\Rightarrow v ^{2} / 2=[20.25-13.5]-[0.25-1.5]$
$\Rightarrow v ^{2} / 2=6.75+1.25=8$
$\Rightarrow v ^{2}=16$
$\Rightarrow v=4$
Therefore the velocity of the body at $x=3$ can be $4 m / s .$
