Question

The potential energy of a particle that is free to move along $x-\text{axis}$ is given by $U=(x^3-\frac{x^2}{2})\text{J}$, where $x$ is in $\text{meter}$. The particle is initially at $x=0$. Find the magnitude of force at $x=2\text{m}$.

• A. $2\text{N}$
• B. $6\text{N}$
• C. $8\text{N}$
• D. $10\text{N}$

Answer 1

The correct option is D $10\text{N}$

Given: $U=x^3-\frac{x^2}{2}$

Relation between conservative force $\left(F\right)$ and potential energy $\left(U\right)$, corresponding to a conservative potential field is given by

$F=-\frac{dU}{dx}$

$F=-\frac{dU}{dx}=-(3x^2-x)$

$\therefore F=x-3x^2$

Putting $x=2$, we get

$F=2-12=-10\text{N}$

The magnitude of force at $x=2\text{m}$ is $|F|=10\text{N}$.

Hence, option $\left(d\right)$ is the correct answer.