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

Options

A. 0.6
B. 0.52
C. 0.44
D. 0.35

Explanation

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

### Step 1: Understand the forces acting on the block
When the block is at rest, the force of static friction (f_s) opposes the applied force. The block will start moving when the applied force exceeds the maximum static friction force.

### Step 2: Write the equation for static friction
The maximum static friction force can be expressed as:
\[ f_s = \mu_s \cdot N \]
where:
- \( f_s \) is the force of static friction,
- \( \mu_s \) is the coefficient of static friction,
- \( N \) is the normal force.

### Step 3: Determine the normal force (N)
For a block resting on a horizontal surface, the normal force (N) is equal to the weight of the block:
\[ N = m \cdot g \]
where:
- \( m = 20 \, \text{kg} \) (mass of the block),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

Calculating the normal force:
\[ N = 20 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 200 \, \text{N} \]

### Step 4: Set up the equation using the given force
According to the problem, a force of 75 N is required to set the block in motion. This means:
\[ f_s = 75 \, \text{N} \]

### Step 5: Substitute into the static friction equation
Now we can substitute the values into the static friction equation:
\[ 75 \, \text{N} = \mu_s \cdot 200 \, \text{N} \]

### Step 6: Solve for the coefficient of static friction (μ_s)
Rearranging the equation to solve for μ_s gives:
\[ \mu_s = \frac{75 \, \text{N}}{200 \, \text{N}} \]
\[ \mu_s = 0.375 \]

### Final Answer
The coefficient of static friction \( \mu_s \) is **0.375**.

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