A block of mass 2kg lying on ice when given a velocity of `6ms^(-1)` is stopped by friction in 5s. The coefficient of friction between the block and ice is (g=10 `ms^(-2)`)

Options

A. 0.5
B. 0.2
C. 0.12
D. 0.4

Answer

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

### Step 1: Identify the given data
- Mass of the 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, \( t = 5 \, \text{s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the acceleration
Using the equation of motion:
\[ v = u + at \]
Substituting the known values:
\[ 0 = 6 + a \cdot 5 \]
Rearranging gives:
\[ a \cdot 5 = -6 \implies a = -\frac{6}{5} = -1.2 \, \text{m/s}^2 \]

### Step 3: Relate acceleration to friction
The acceleration due to friction can be expressed as:
\[ a = -\mu g \]
where \( \mu \) is the coefficient of friction. Substituting \( g = 10 \, \text{m/s}^2 \):
\[ -1.2 = -\mu \cdot 10 \]
Removing the negative signs gives:
\[ 1.2 = \mu \cdot 10 \]

### Step 4: Solve for the coefficient of friction
Now, we can solve for \( \mu \):
\[ \mu = \frac{1.2}{10} = 0.12 \]

### Conclusion
The coefficient of friction between the block and the ice is:
\[ \mu = 0.12 \]

---