Question

A particle moving on x-axis has potential energy $U=2-20x+5{{x}^{2}}\text{ Joule}$ along x-axis. The particle is released at $x=-3$. The maximum value of $x$ will be ($x$ is in meter)
$\begin{align}
  & \text{A}\text{. }5m \\
 & \text{B}\text{. }3m \\
 & \text{C}\text{. }7m \\
 & \text{D}\text{. 8}m \\
\end{align}$

Answer 1

- Hint: Potential energy is the energy that is stored in a system of particles. The formula for potential energy depends upon the force acting on the two particles. The energy stored in a particle due to its position is known as its potential energy, while the energy that a moving particle has due to its motion is known as its kinetic energy. When force is applied on a particle, or say work is done on a particle, it is stored in the particle as its potential energy.

Formula used:
Relationship between force and potential energy,
$F=-\dfrac{dU}{dx}$

Complete step-by-step solution
Potential energy is described as the energy held by a particle because of its position relative to another particle. Potential energy arises in a system with particles that exert forces on each other of a magnitude dependent on the configuration, or relative position, of the particles. The potential energy of a system of particles depends only on their initial and final configurations and independent of the paths travelled by the particles.
If the potential energy function $U\left( x \right)$ of a particle is known, then the force on the particle at any position can be obtained by taking the derivative of the potential with a negative sign.
$F=-\dfrac{dU}{dx}$
Graphically, this implies that if we have a potential energy versus position graph of a particle, then the force on the particle is the negative of the slope of the function at some point.
The negative sign in the formula implies that the applied force on the particle will try to push the particle back to lower potential and in the process the potential energy of the particle rises.
Force on particle is given by,
$F=-\dfrac{dU}{dx}$
Given that,
$U=2-20x+5{{x}^{2}}$
Therefore,
$\begin{align}
  & F=-\dfrac{d\left( 2-20x+5{{x}^{2}} \right)}{dx} \\
 & F=20-10x \\
\end{align}$
Force on particle is zero at,
$\begin{align}
  & F=0 \\
 & 20-10x=0 \\
 & x=2 \\
\end{align}$
So, from $x=-3$ to $x=2$, the amplitude of the particle is $x=5$
Therefore,
The other extreme position of the particle is given as,
$x=2+5=7$
The maximum value of $x$ is $7m$
Hence, the correct option is C.

Note: Potential energy, also known as the stored energy, depends upon the relative position of various parts of a system. For example, a spring has more potential energy when it is compressed or stretched rather than when it is present in its mean length, or a steel ball has more potential energy raised above the ground than it has after falling to the surface of the earth. Potential energy is a property of a system of particles and not of an individual body or particle.