A 20 kg block is initially at rest on a rough horizontal surface. A horizontal force of 75 N is required to set the block in motion. After it is in motion, a horizontal force of 60 N is required to keep the block moving with constant speed. The coefficient of static friction is

Options

A. 0.38
B. 0.44
C. 0.52
D. 0.6

Answer

To find the coefficient of static friction (μs) for the given block, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the block (m) = 20 kg
- Force required to set the block in motion (F_s) = 75 N
- Force required to keep the block moving with constant speed (F_k) = 60 N

2. **Calculate the Weight of the Block:**
- The weight (W) of the block can be calculated using the formula:
\[ W = m \cdot g \]
- Here, \( g \) (acceleration due to gravity) is approximately \( 9.81 \, \text{m/s}^2 \).
- Therefore:
\[ W = 20 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 196.2 \, \text{N} \]

3. **Determine the Normal Force (N):**
- On a horizontal surface, the normal force (N) is equal to the weight of the block:
\[ N = W = 196.2 \, \text{N} \]

4. **Use the Static Friction Formula:**
- The force required to overcome static friction is equal to the maximum static friction force:
\[ F_s = \mu_s \cdot N \]
- Rearranging this formula to find the coefficient of static friction (μs):
\[ \mu_s = \frac{F_s}{N} \]

5. **Substitute the Values:**
- Now substituting the values we have:
\[ \mu_s = \frac{75 \, \text{N}}{196.2 \, \text{N}} \approx 0.382 \]

6. **Final Answer:**
- The coefficient of static friction (μs) is approximately:
\[ \mu_s \approx 0.38 \]