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, we need to find the coefficient of dynamic friction (μ) for a block that comes to rest due to friction over a distance of 40 m. The initial velocity of the block is given as 20 m/s, and we can use the equations of motion to find the required coefficient.

### Step-by-Step Solution:

1. **Identify the Known Values:**
- Initial velocity (u) = 20 m/s
- Final velocity (v) = 0 m/s (since the block comes to rest)
- Distance (s) = 40 m
- Acceleration due to gravity (g) = 10 m/s²

2. **Use the Third Equation of Motion:**
The third equation of motion relates initial velocity, final velocity, acceleration, and distance:
\[ v^2 = u^2 + 2as \]
Here, we can rearrange it to find acceleration (a):
\[ 0 = (20)^2 + 2a(40) \]
\[ 0 = 400 + 80a \]
\[ 80a = -400 \]
\[ a = -\frac{400}{80} = -5 \, \text{m/s}^2 \]

3. **Relate Acceleration to Friction:**
The acceleration due to friction can also be expressed as:
\[ a = -\mu g \]
Substituting the value of g:
\[ -5 = -\mu (10) \]
\[ \mu = \frac{5}{10} = 0.5 \]

4. **Conclusion:**
The coefficient of dynamic friction (μ) is 0.5.

### Final Answer:
The coefficient of dynamic friction is **0.5**.

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