A block moving on a horizontal surface with velocity `20 ms^(-1)` comes to rest because of surface friction over a distance of 40 m. Taking `g = 10 ms^(-2)` , the coefficient of dynamic friction is :

Options

A. 0.5
B. 0.3
C. 0.2
D. 0.1

Explanation

To solve the problem of finding the coefficient of dynamic friction for a block that comes to rest due to friction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Initial velocity, \( u = 20 \, \text{m/s} \)
- Final velocity, \( v = 0 \, \text{m/s} \) (since the block comes to rest)
- Distance, \( s = 40 \, \text{m} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Use the Third Equation of Motion**:
The third equation of motion relates the initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Here, \( a \) is the retardation (negative acceleration) due to friction. Since the block comes to rest, we can rewrite the equation as:
\[ 0 = (20)^2 + 2(-a)(40) \]
Simplifying this gives:
\[ 0 = 400 - 80a \]
Rearranging gives:
\[ 80a = 400 \implies a = \frac{400}{80} = 5 \, \text{m/s}^2 \]

3. **Relate Retardation to Friction**:
The retardation \( a \) due to friction can be expressed in terms of the coefficient of dynamic friction \( \mu_k \):
\[ a = \mu_k g \]
Substituting the value of \( g \):
\[ 5 = \mu_k \times 10 \]

4. **Solve for the Coefficient of Dynamic Friction**:
Rearranging the equation gives:
\[ \mu_k = \frac{5}{10} = 0.5 \]

5. **Conclusion**:
The coefficient of dynamic friction \( \mu_k \) is \( 0.5 \).

### Final Answer:
The coefficient of dynamic friction is \( \mu_k = 0.5 \).

---