Question

The potential energy of a particle of mass $0.1$ kg, moving along the x-axis, is given by $U=5x(x-4)$J, where $x$ is in meters. It can be concluded that

• A. The particle is acted upon by a variable force.
• B. The minimum potential energy during motion is $-20$ J.
• C. The speed of the particle is maximum at $x=2$ m.
• D. The period of oscillation of the particle is ($\pi$$/5$) sec.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The particle is acted upon by a variable force.
B The minimum potential energy during motion is $-20$ J.
C The period of oscillation of the particle is ($\pi$$/5$) sec.
D The speed of the particle is maximum at $x=2$ m.

Potential energy, $U=5x(x-4)=5x^2-20x$
Force, $F=-\frac{dU}{dx}=-(10x-20)=-10x+20$
This force is dependent on $x$, hence particle is acted upon by a variable force. (Option A)
Now, at equilibrium position potential energy is minimum and kinetic energy is maximum. To determine equilibrium position:
$F=0-10x+20=0x=2$
So, minimum potential energy will be $U_{min}=5\left(2\right)(2-4)=-20J$ (Option B)
Since kinetic energy is maximum at equilibrium position $x=2$, the speed of the particle will also be maximum. (Option C)
$F=-10x+20=-10(x+2)$
Define a new variable y such that $y=x+2$. Now,
$F=-10y$
Comparing with $F=-kx$, $k=10N/m$
Now, time period of oscillation, $T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.1}{10}}=\frac{\pi }{5}sec.$ (Option D)