A block is sliding on a rough horizontal surface. If the contact force on the block is `sqrt2` times the frictional force, the coefficient of friction is

Options

A. `0.25`
B. `1/sqrt2`
C. `sqrt2`
D. 1

Explanation

To solve the problem, we need to find the coefficient of friction (μ) given that the contact force (R) on the block is √2 times the frictional force (F). Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Block
The block is sliding on a rough horizontal surface, which means there are two main forces acting on it:
- The normal force (N) acting perpendicular to the surface.
- The frictional force (F) acting opposite to the direction of motion.

### Step 2: Relate the Contact Force and Frictional Force
According to the problem, the contact force (R) is given as:
\[ R = \sqrt{2} F \]
where F is the frictional force.

### Step 3: Express the Frictional Force in Terms of the Coefficient of Friction
The frictional force can be expressed as:
\[ F = \mu N \]
where μ is the coefficient of friction and N is the normal force.

### Step 4: Substitute the Frictional Force into the Contact Force Equation
Substituting \( F = \mu N \) into the equation for R gives:
\[ R = \sqrt{2} (\mu N) \]
Thus,
\[ R = \sqrt{2} \mu N \]

### Step 5: Use the Pythagorean Theorem for the Contact Force
The contact force R can also be expressed using the Pythagorean theorem since the frictional force and normal force are perpendicular:
\[ R = \sqrt{F^2 + N^2} \]
Substituting \( F = \mu N \) into this equation gives:
\[ R = \sqrt{(\mu N)^2 + N^2} \]
\[ R = \sqrt{\mu^2 N^2 + N^2} \]
\[ R = \sqrt{N^2 (\mu^2 + 1)} \]
\[ R = N \sqrt{\mu^2 + 1} \]

### Step 6: Set the Two Expressions for R Equal to Each Other
Now we have two expressions for R:
1. \( R = \sqrt{2} \mu N \)
2. \( R = N \sqrt{\mu^2 + 1} \)

Setting them equal gives:
\[ \sqrt{2} \mu N = N \sqrt{\mu^2 + 1} \]

### Step 7: Simplify the Equation
Assuming \( N \neq 0 \), we can divide both sides by N:
\[ \sqrt{2} \mu = \sqrt{\mu^2 + 1} \]

### Step 8: Square Both Sides
Squaring both sides results in:
\[ 2 \mu^2 = \mu^2 + 1 \]

### Step 9: Rearrange the Equation
Rearranging gives:
\[ 2 \mu^2 - \mu^2 = 1 \]
\[ \mu^2 = 1 \]

### Step 10: Solve for μ
Taking the square root of both sides gives:
\[ \mu = 1 \]

Thus, the coefficient of friction is:
\[ \mu = 1 \]

### Final Answer
The coefficient of friction (μ) is **1**.

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