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. 0.02
B. 0.03
C. 0.06
D. 0.01

Answer

To find the coefficient of friction for the marble block, we can follow these steps:

### Step 1: Identify the known 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 first equation of motion
We can use the first equation of motion to find the acceleration (or deceleration in this case):
\[ v = u + at \]
Substituting the known values:
\[ 0 = 6 + a \cdot 10 \]
Rearranging to solve for \( a \):
\[ a \cdot 10 = -6 \implies a = -\frac{6}{10} = -0.6 \, \text{m/s}^2 \]
The negative sign indicates that this is a deceleration (retardation).

### Step 3: Apply the force balance equation
The frictional force acting on the block can be expressed as:
\[ F_{\text{friction}} = \mu mg \]
where \( \mu \) is the coefficient of friction and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

According to Newton's second law, the net force acting on the block is also equal to:
\[ F_{\text{net}} = ma \]
Since the only horizontal force acting on the block is the frictional force, we can set these equal:
\[ \mu mg = ma \]

### Step 4: Solve for the coefficient of friction \( \mu \)
We can cancel \( m \) from both sides (as long as \( m \neq 0 \)):
\[ \mu g = a \]
Now, substituting the values we have:
\[ \mu \cdot 9.8 = -0.6 \]
Solving for \( \mu \):
\[ \mu = \frac{-0.6}{9.8} \approx -0.0612 \]
Since the coefficient of friction cannot be negative, we take the absolute value:
\[ \mu \approx 0.0612 \]

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