A20 kg block is initally at rest. A 75 N force is required to set the block in motion. After the motion starts, a force of 60 N is required to keep the block moving with constant speed. The coefficient of static friction is

Options

A. ` 0.62`
B. `0.52`
C. `0.44`
D. `0.375`

Answer

To find the coefficient of static friction (\( \mu_s \)) for the given problem, we can follow these steps:

### Step 1: Identify the forces acting on the block
The block has a weight (\( W \)) acting downwards, which is given by the formula:
\[ W = mg \]
where \( m = 20 \, \text{kg} \) (mass of the block) and \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

### Step 2: Calculate the weight of the block
Substituting the values:
\[ W = 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 3: Determine the normal force
Since the block is resting on a horizontal surface and there are no vertical forces other than its weight and the normal force, the normal force (\( N \)) is equal to the weight of the block:
\[ N = W = 200 \, \text{N} \]

### Step 4: Relate the static friction force to the applied force
The maximum static friction force (\( f_s \)) that can act on the block before it starts moving is given as 75 N. This can be expressed using the coefficient of static friction:
\[ f_s = \mu_s \cdot N \]
Substituting the known values:
\[ 75 \, \text{N} = \mu_s \cdot 200 \, \text{N} \]

### Step 5: Solve for the coefficient of static friction
Rearranging the equation to find \( \mu_s \):
\[ \mu_s = \frac{75 \, \text{N}}{200 \, \text{N}} = 0.375 \]

### Conclusion
The coefficient of static friction (\( \mu_s \)) is:
\[ \mu_s = 0.375 \]