A body is rolling on the ground with a velocity of `1ms^(-1)` . After travelling a distance of 5 m, it comes to rest. Find the coefficient of friction. Take `g=10ms^(-2)`.

Answer

To solve the problem step by step, we will follow the principles of motion and friction.

### Step 1: Identify the known values
- Initial velocity (u) = 1 m/s
- Final velocity (v) = 0 m/s (the body comes to rest)
- Distance traveled (s) = 5 m
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Use the equation of motion to find acceleration
We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
Here, \( a \) is the acceleration (which will be negative since it's retardation).

Substituting the known values:
\[ 0 = (1)^2 + 2a(5) \]
\[ 0 = 1 + 10a \]
Now, solving for \( a \):
\[ 10a = -1 \]
\[ a = -\frac{1}{10} \, \text{m/s}^2 \]

### Step 3: Relate friction to acceleration
The frictional force \( F_f \) acting on the body is responsible for this retardation. According to Newton's second law:
\[ F_f = m \cdot a \]
Where \( m \) is the mass of the body.

Substituting the value of \( a \):
\[ F_f = m \left(-\frac{1}{10}\right) \]
\[ F_f = -\frac{m}{10} \, \text{N} \]

### Step 4: Relate friction to the normal force
The frictional force can also be expressed in terms of the coefficient of friction \( \mu \):
\[ F_f = \mu \cdot N \]
Where \( N \) is the normal force. For a body resting on a horizontal surface, \( N = mg \):
\[ F_f = \mu \cdot mg \]

### Step 5: Set the equations for friction equal
Since both expressions represent the frictional force, we can set them equal to each other:
\[ -\frac{m}{10} = \mu \cdot mg \]

### Step 6: Solve for the coefficient of friction \( \mu \)
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ -\frac{1}{10} = \mu \cdot g \]
Now substituting \( g = 10 \, \text{m/s}^2 \):
\[ -\frac{1}{10} = \mu \cdot 10 \]
\[ \mu = -\frac{1}{10} \cdot \frac{1}{10} \]
\[ \mu = -\frac{1}{100} \]

Since we are interested in the magnitude of the coefficient of friction:
\[ \mu = 0.01 \]

### Conclusion
The coefficient of friction \( \mu \) is \( 0.01 \).

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