Question

A particle is released from rest at origin. It moves under the influence of a potential field, $U=x^2-3x$. Find the Kinetic energy of the particle at $x=2$.

• A. $2\text{J}$
• B. $1\text{J}$
• C. $1.5\text{J}$
• D. $0\text{J}$

Answer 1

The correct option is A $2\text{J}$

Given,
Potential energy function of the particle
$U=x^2-3x......\left(1\right)$
Since, $U$ is function of position only, we call the force associated with this potential as conservative force.
General relation between a conservative force and potential energy is given by
$\overrightarrow{F}=-(\frac{\partial U}{\partial x}\hat{i}+\frac{\partial U}{\partial y}\hat{j}+\frac{\partial U}{\partial z}\hat{k})$
From $\left(1\right)$ we can deduce that, $U$ is a function of $x$ only.
$\therefore$ The above formula can be written as,
$F=-\frac{dU}{dx}$
$\Rightarrow F=-2x+3.......\left(2\right)$
Since, the force is a variable force, From Newton's second law we can write that,
$F=ma=m\frac{dv}{dt}=m\frac{dv}{dx}.\frac{dx}{dt}$
$\Rightarrow F=mv\frac{dv}{dx}.........\left(3\right)$

From $\left(2\right)$ and $\left(3\right)$
$mv\frac{dv}{dx}=-2x+3$
Integrating,
$m{\int }_0^{v}vdv={\int }_0^x(-2x+3)dx$
$\Rightarrow \frac{m{v}^2}{2}=K.E=-x^2+3x$
$\therefore (K.E{)}_{x=2}=-2^2+6=2\text{J}$
Thus, option (a) is the correct answer.