If a block moving up at `theta = 30^(@)` with a velocity 5m/s, stops after 0.5 sec, then what is `mu`

Options

A. 0.5
B. 1.25
C. 0.6
D. None of the above

Explanation

To solve the problem, we need to find the coefficient of friction (μ) for a block moving up an incline at an angle of θ = 30° with an initial velocity of 5 m/s, which stops after 0.5 seconds.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Block:**
- The block experiences gravitational force acting downwards, which can be resolved into two components:
- **Parallel to the incline:** \( F_{\text{gravity, parallel}} = mg \sin \theta \)
- **Perpendicular to the incline:** \( F_{\text{gravity, perpendicular}} = mg \cos \theta \)

2. **Determine the Normal Force (N):**
- The normal force acting on the block is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \theta \]

3. **Frictional Force (F_f):**
- The frictional force acting on the block is given by:
\[ F_f = \mu N = \mu (mg \cos \theta) \]

4. **Apply Newton's Second Law:**
- According to Newton's second law, the net force acting on the block is equal to the mass times its acceleration (a):
\[ F_f + mg \sin \theta = ma \]
- Substituting the expressions for frictional force and gravitational force:
\[ \mu (mg \cos \theta) + mg \sin \theta = ma \]

5. **Cancel the Mass (m):**
- Since mass (m) appears in every term, we can cancel it out:
\[ \mu g \cos \theta + g \sin \theta = a \]

6. **Calculate the Acceleration (a):**
- The acceleration can be calculated using the formula:
\[ a = \frac{v_f - v_i}{t} \]
- Here, \( v_f = 0 \) m/s (final velocity), \( v_i = 5 \) m/s (initial velocity), and \( t = 0.5 \) s:
\[ a = \frac{0 - 5}{0.5} = -10 \, \text{m/s}^2 \]

7. **Substitute the Acceleration into the Equation:**
- Now, substituting \( a = -10 \) m/s² into the equation:
\[ -10 = \mu g \cos \theta + g \sin \theta \]

8. **Use Values for g, sin(30°), and cos(30°):**
- We know \( g \approx 10 \, \text{m/s}^2 \), \( \sin(30°) = \frac{1}{2} \), and \( \cos(30°) = \frac{\sqrt{3}}{2} \):
\[ -10 = \mu (10) \left(\frac{\sqrt{3}}{2}\right) + (10) \left(\frac{1}{2}\right) \]
- Simplifying:
\[ -10 = 5\sqrt{3} \mu + 5 \]

9. **Rearranging the Equation:**
- Move the constant term to the other side:
\[ -10 - 5 = 5\sqrt{3} \mu \]
\[ -15 = 5\sqrt{3} \mu \]

10. **Solve for μ:**
- Divide both sides by \( 5\sqrt{3} \):
\[ \mu = \frac{-15}{5\sqrt{3}} = \frac{-3}{\sqrt{3}} = -\sqrt{3} \]
- Since we are looking for the magnitude of the coefficient of friction, we take the positive value:
\[ \mu = \frac{1}{\sqrt{3}} \approx 0.577 \approx 0.6 \]

### Conclusion:
The coefficient of friction \( \mu \) is approximately 0.6. Thus, the correct answer is option C.