`phi` is the angle of the incline when a block of mass m just starts slipping down. The distance covered by the block if thrown up the incline with an initial speed `v_(0)` is `:`

Options

A. `(v_(0)^(2))/( 4g sin phi)`
B. `(4v_(0)^(2))/( g sin phi)`
C. `(v_(0)^(2)sin phi)/( 4 g ) `
D. `(4v_(0)^(2)sin phi )/( g)`

Answer

To solve the problem of finding the distance covered by a block of mass \( m \) thrown up an incline with an initial speed \( v_0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Forces Acting on the Block:**
- The block is on an incline at an angle \( \phi \).
- The gravitational force acting on the block can be resolved into two components:
- Perpendicular to the incline: \( mg \cos \phi \)
- Parallel to the incline (downward): \( mg \sin \phi \)
- The frictional force \( f \) acts up the incline when the block is thrown upwards.

2. **Determine the Condition for Motion:**
- When the block just starts slipping down the incline, the frictional force \( f \) is equal to the component of gravitational force acting down the incline:
\[ f = \mu mg \cos \phi = mg \sin \phi \]
- From this, we can derive the coefficient of friction:
\[ \mu = \tan \phi \]

3. **Calculate the Net Force When Thrown Up:**
- When the block is thrown up with an initial speed \( v_0 \), the net force acting on it is:
\[ F_{\text{net}} = -mg \sin \phi - f \]
- Since the friction acts in the opposite direction to the motion, we can express the frictional force as:
\[ f = \mu mg \cos \phi = mg \sin \phi \]
- Thus, the net force becomes:
\[ F_{\text{net}} = -mg \sin \phi - mg \sin \phi = -2mg \sin \phi \]

4. **Determine the Acceleration:**
- Using Newton's second law, the acceleration \( a \) of the block is:
\[ a = \frac{F_{\text{net}}}{m} = -2g \sin \phi \]

5. **Apply Kinematic Equation:**
- We use the kinematic equation to find the distance \( s \) covered by the block until it comes to rest:
\[ v^2 = u^2 + 2as \]
- Here, \( v = 0 \) (final velocity), \( u = v_0 \) (initial velocity), and \( a = -2g \sin \phi \):
\[ 0 = v_0^2 + 2(-2g \sin \phi)s \]
- Rearranging gives:
\[ v_0^2 = 4g \sin \phi \cdot s \]
- Solving for \( s \):
\[ s = \frac{v_0^2}{4g \sin \phi} \]

### Final Answer:
The distance covered by the block when thrown up the incline with an initial speed \( v_0 \) is:
\[ s = \frac{v_0^2}{4g \sin \phi} \]