A marble block of mass 2 kg lying on ice when given a velocity of `6m//s` is stopped by friction in 10s. Then the coefficient of friction is

Options

A. 0.01
B. 0.02
C. 0.03
D. 0.06

Explanation

To solve the problem step by step, we will use the concepts of kinematics and friction.

### Step 1: Identify the given values
- Mass of the marble block, \( m = 2 \, \text{kg} \)
- Initial velocity, \( u = 6 \, \text{m/s} \)
- Final velocity, \( v = 0 \, \text{m/s} \) (since the block stops)
- Time taken to stop, \( t = 10 \, \text{s} \)

### Step 2: Use the kinematic equation
We can use the kinematic equation:
\[ v = u + at \]
where \( a \) is the acceleration. Rearranging this gives:
\[ a = \frac{v - u}{t} \]

### Step 3: Substitute the known values into the equation
Substituting the values we have:
\[ a = \frac{0 - 6}{10} = \frac{-6}{10} = -0.6 \, \text{m/s}^2 \]
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (deceleration).

### Step 4: Relate acceleration to friction
The acceleration due to friction can also be expressed as:
\[ a = -\mu g \]
where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 5: Set the two expressions for acceleration equal
From the previous steps, we have:
\[ -0.6 = -\mu g \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ -0.6 = -\mu \cdot 10 \]

### Step 6: Solve for the coefficient of friction \( \mu \)
Rearranging gives:
\[ \mu = \frac{0.6}{10} = 0.06 \]

### Conclusion
The coefficient of friction \( \mu \) is \( 0.06 \).

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