Question

A unidirectional force $\vec{F}=\left(2 \alpha x-3 x^{2}\right) \hat{i}$ is acting on a block which is initially at rest on a smooth surface at position $x=0$. The minimum kinetic energy is found to be $4$ units. Find the positive numerical value of constant $\alpha$.

Answer 1

From work-energy theorem, $W=\Delta K E$
Kinetic energy,
$K E=\int\limits_{0}^{x} F_{x} d x=\int\limits_{0}^{x}\left(2 \alpha x-3 x^{2}\right) d x=\alpha x^{2}-x^{3}$
Kinetic energy will be minimum if potential energy is maximum.
At maximum potential energy condition, $F=0$
$\Rightarrow x=\frac{2 \alpha}{3}$
$K E_{\min }=\alpha\left(\frac{2 \alpha}{3}\right)^{2}-\left(\frac{2 \alpha}{3}\right)^{3}=4$
$\Rightarrow \frac{4}{9} \alpha^{3}-\frac{8}{27} \alpha^{3}=4$
$\Rightarrow \frac{4}{27} \alpha^{3}=4 $
$\Rightarrow \alpha^{3}=27 $
$\Rightarrow \alpha=3$