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 coeffecient of friction between the ground and the ring is large enough that rolling always occur and the coefficient of friction between the stick and the ring

Options

A. 0.4
B. 0.8
C. 0.2
D. 0.5

Explanation

To solve the problem, we will analyze the forces acting on the ring and apply Newton's laws for both translational and rotational motion.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the ring, \( m = 2 \, \text{kg} \)
- Radius of the ring, \( r = 0.5 \, \text{m} \)
- Applied force by the stick, \( F = 2 \, \text{N} \)
- Acceleration of the ring, \( a = 0.3 \, \text{m/s}^2 \)

2. **Draw the Free Body Diagram:**
- The forces acting on the ring include:
- Weight \( W = mg \) acting downwards.
- Normal force \( N \) acting upwards.
- Applied force \( F \) acting horizontally (to the left).
- Friction force \( F_G \) acting opposite to the direction of motion (to the right).
- Friction force \( F_S \) acting at the contact point between the stick and the ring (to the left).

3. **Apply Newton's Second Law for Translational Motion:**
- The net force in the horizontal direction is given by:
\[ F - F_G - F_S = ma \]
- Rearranging gives:
\[ 2 - F_G - F_S = 2 \times 0.3 \]
- This simplifies to:
\[ 2 - F_G - F_S = 0.6 \]
- Thus:
\[ F_G + F_S = 1.4 \quad (1) \]

4. **Apply Newton's Second Law for Rotational Motion:**
- The torque about the center of the ring due to friction forces is:
\[ \tau = F_G \cdot r - F_S \cdot r = I \alpha \]
- The moment of inertia \( I \) of a ring is given by \( I = mr^2 = 2 \times (0.5)^2 = 0.5 \, \text{kg m}^2 \).
- The angular acceleration \( \alpha \) can be related to linear acceleration \( a \) by \( \alpha = \frac{a}{r} = \frac{0.3}{0.5} = 0.6 \, \text{rad/s}^2 \).
- Substituting the values gives:
\[ F_G \cdot 0.5 - F_S \cdot 0.5 = 0.5 \cdot 0.6 \]
- This simplifies to:
\[ F_G - F_S = 0.6 \quad (2) \]

5. **Solve the System of Equations:**
- We have two equations:
1. \( F_G + F_S = 1.4 \)
2. \( F_G - F_S = 0.6 \)

- Adding these two equations:
\[ 2F_G = 2 \implies F_G = 1 \, \text{N} \]
- Substituting \( F_G \) back into equation (1):
\[ 1 + F_S = 1.4 \implies F_S = 0.4 \, \text{N} \]

6. **Calculate the Coefficient of Friction:**
- The friction force \( F_S \) is related to the normal force \( N \) (which is equal to the applied force since there is no vertical acceleration):
\[ F_S = \mu_s N \]
- Here, \( N = 2 \, \text{N} \) (the applied force).
- Thus:
\[ 0.4 = \mu_s \cdot 2 \]
- Solving for \( \mu_s \) gives:
\[ \mu_s = \frac{0.4}{2} = 0.2 \]

### Final Answer:
The coefficient of friction between the stick and the ring is \( \mu_s = 0.2 \).