Find the distance travelled by a body before coming rest, if it is moving with a speed of `10ms^(-1)` and the coefficient of friction between the ground and the body is `0.4`.

Answer

To find the distance traveled by a body before coming to rest, we can follow these steps:

### Step 1: Identify the given values
- Initial speed of the body, \( u = 10 \, \text{m/s} \)
- Final speed of the body, \( v = 0 \, \text{m/s} \) (since it comes to rest)
- Coefficient of friction, \( \mu = 0.4 \)

### Step 2: Calculate the acceleration due to friction
The force of kinetic friction \( F_k \) can be expressed as:
\[ F_k = \mu \cdot N \]
where \( N \) is the normal force. For a body resting on a horizontal surface, \( N = mg \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

Thus, we have:
\[ F_k = \mu \cdot mg = 0.4 \cdot mg \]

The acceleration (or retardation, since it is acting opposite to the direction of motion) \( a \) can be calculated using Newton's second law:
\[ F = ma \]
So,
\[ a = \frac{F_k}{m} = \frac{0.4 \cdot mg}{m} = 0.4g \]

Since this is a retardation, we take it as negative:
\[ a = -0.4g \]

### Step 3: Substitute the value of \( g \)
Using \( g \approx 10 \, \text{m/s}^2 \):
\[ a = -0.4 \cdot 10 = -4 \, \text{m/s}^2 \]

### Step 4: Use the equation of motion to find the distance
We can use the equation of motion:
\[ v^2 = u^2 + 2aS \]
where \( S \) is the distance traveled.

Substituting the known values:
\[ 0 = (10)^2 + 2(-4)S \]
\[ 0 = 100 - 8S \]

### Step 5: Solve for \( S \)
Rearranging the equation gives:
\[ 8S = 100 \]
\[ S = \frac{100}{8} = 12.5 \, \text{m} \]

### Final Answer
The distance traveled by the body before coming to rest is:
\[ S = 12.5 \, \text{m} \]

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