A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of $0.3 m/s^{2}.$ The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is $(P/10).$ The value of P is

Step 1: Identify the given parameters.

The ring has mass $m$, radius $R$, and rolls without slipping with linear acceleration $a$ under a horizontal force $F$ applied by the stick.

$$m = 2\text{ kg}$$ $$R = 0.5\text{ m}$$ $$F = 2\text{ N}$$ $$a = 0.3\text{ m/s}^2$$

Step 2: Relate the linear and angular motion.

For rolling without slipping, the angular acceleration $\alpha$ is related to the linear acceleration $a$.

$$a = \alpha R$$

The moment of inertia of the ring about its center of mass is:

$$I = mR^2$$

Step 3: Apply Newton's second law for horizontal motion.

The horizontal forces on the ring are the stick force $F$ acting to the left and the ground friction $f$ acting to the right.

$$F - f = ma$$

Solve for the ground friction $f$.

$$f = F - ma$$ $$f = 2 - 2(0.3) = 1.4\text{ N}$$

Step 4: Apply the torque equation about the center of mass.

The torque is produced by the ground friction $f$ and the stick friction $f_s$ acting downwards at the rightmost point.

$$fR - f_sR = I\alpha$$

Substitute $I = mR^2$ and $\alpha = \frac{a}{R}$ into the torque equation.

$$fR - f_sR = mR^2\left(\frac{a}{R}\right)$$ $$f - f_s = ma$$

Step 5: Calculate the friction force from the stick.

Substitute the expression for $f$ from Step 3.

$$f_s = F - 2ma$$ $$f_s = 2 - 2(2)(0.3) = 0.8\text{ N}$$

Step 6: Determine the coefficient of friction and the value of $P$.

The normal force exerted by the stick is $N_s = F = 2\text{ N}$.

$$\mu = \frac{f_s}{N_s}$$ $$\mu = \frac{0.8}{2} = 0.4$$

Equate this to the given expression for the coefficient of friction.

$$\frac{P}{10} = 0.4$$ $$P = 4$$

Answer: The value of $P$ is 4.