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. (a) 0.02
B. (b) 0.03
C. (c) 0.04
D. (d) 0.06

Explanation

To find the coefficient of friction between the marble block and the ice, we can follow these steps:

### 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: Calculate the acceleration
Using the equation of motion:
\[ v = u + at \]
We can rearrange this to find the acceleration \( a \):
\[ 0 = 6 + a(10) \]
\[ a = \frac{-6}{10} = -0.6 \, \text{m/s}^2 \]

### Step 3: Relate acceleration to friction
The frictional force acting on the block can be expressed as:
\[ F = \mu mg \]
where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

The net force acting on the block is also equal to \( ma \):
\[ ma = -\mu mg \]
Substituting \( a \):
\[ m(-0.6) = -\mu mg \]

### Step 4: Simplify the equation
Since the mass \( m \) appears on both sides, we can cancel it out (as long as \( m \neq 0 \)):
\[ -0.6 = -\mu g \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ -0.6 = -\mu(10) \]

### Step 5: Solve for \( \mu \)
\[ 0.6 = 10\mu \]
\[ \mu = \frac{0.6}{10} = 0.06 \]

### Conclusion
The coefficient of friction \( \mu \) between the marble block and the ice is \( 0.06 \).