Question

A unidirectional force $\vec{F}=\left(2\alpha x-3x^2\right)\hat{i}$ 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

Answer: 3

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