A body is moving along a rough horizontal surface with an initial velocity 6m/s m If the body comes to rest after travelling 9 m , then the coefficient of sliding friction will be

Options

A. 0.5
B. 0.4
C. 0.6
D. 0.2

Explanation

To find the coefficient of sliding friction for a body moving along a rough horizontal surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial velocity (u) = 6 m/s
- Final velocity (v) = 0 m/s (the body comes to rest)
- Distance traveled (s) = 9 m

2. **Use the Kinematic Equation:**
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Rearranging gives:
\[ a = \frac{v^2 - u^2}{2s} \]

3. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ a = \frac{0^2 - (6)^2}{2 \times 9} \]
\[ a = \frac{0 - 36}{18} \]
\[ a = \frac{-36}{18} = -2 \, \text{m/s}^2 \]
The negative sign indicates that the acceleration is in the opposite direction of the motion (retardation).

4. **Relate Acceleration to Friction:**
The acceleration (a) is caused by the frictional force acting against the motion of the body. According to Newton's second law:
\[ F = ma \]
The frictional force (F_friction) can also be expressed as:
\[ F_friction = \mu N \]
where \( N \) is the normal force. For a horizontal surface, \( N = mg \).

5. **Set Up the Equation:**
Equating the two expressions for force gives:
\[ \mu mg = ma \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \mu g = a \]

6. **Solve for the Coefficient of Friction:**
Rearranging gives:
\[ \mu = \frac{a}{g} \]
Substituting \( a = -2 \, \text{m/s}^2 \) and \( g = 10 \, \text{m/s}^2 \):
\[ \mu = \frac{-2}{10} = -0.2 \]
Since we are interested in the magnitude of the coefficient of friction:
\[ \mu = 0.2 \]

### Final Answer:
The coefficient of sliding friction is \( \mu = 0.2 \).

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