A spring of spring constant $k$ is placed horizontally on a rough horizontal surface. It is compressed against a block of mass $m$ which is placed on a rough surface, so as to store maximum energy in the spring. If the coefficient of friction between the block and the surface is $\mu$, the potential energy stored in the spring is: (block does not slide due to force of spring.)

Step 1: Given data

The spring constant of the spring is $k$.

The mass of the block is $m$.

The coefficient of friction between the block and the surface is $\mu$.

Step 2: Frictional force

1. The frictional force is the force that resists the motion of a body over a surface contact with the body.
2. The frictional force is defined by the form, $F_f=\mu N$, where, N is the normal force exerted on a body and $\mu$ is the coefficient of friction.

Step 3: Force on a spring and spring constant

1. The force exerted on a spring due to some extension is $F=Kx$, where, K is the spring constant and x is the amount of extension.
2. The potential energy stored in the spring when it is extended at a length x is defined by the form, $P=\frac{1}{2}K{x}^2$.

Step 4: Diagram

The diagrammatic representation of the given situation is:

Step 5: Finding the forces

In this question, the body of mass is placed at the plane. So, the downward gravitational force (weight) is equal to the upward normal force N.

So, normal force, $N=mg$

Now, frictional force,

$F_f=\mu mg..................\left(1\right)$

The restoring force on the body of mass m is

$F=Kx..................\left(2\right)$

According to the question, the block does not slide due to the force of the spring, it means the net force on the mass is zero. Equating equations 1 and 2 we get,

$$\begin{gathered} \mu mg=Kx \\ orx=\frac{\mu mg}{K}...................\left(3\right) \end{gathered}$$

Step6: Finding the potential energy

The potential energy stored in the spring when it is extended at a length x is defined by the form, $P=\frac{1}{2}K{x}^2$.

So,

$$\begin{gathered} P=\frac{1}{2}K{x}^2=\frac{1}{2}K.{\left(\frac{\mu mg}{K}\right)}^2=\frac{1}{2}\frac{{\mu }^2{m}^2{g}^2}{K} \\ orP=\frac{1}{2}\frac{{\mu }^2{m}^2{g}^2}{K} \end{gathered}$$

Therefore, the potential energy stored in the spring is $P=\frac{1}{2}\frac{{\mu }^2{m}^2{g}^2}{K}$.