A block slides down a rough inclined plane of slope angle `theta` with a constnat velocity. It is then projected up the same plane with an intial velocity v the distance travelled by the block up the plane coming to rest is .

Options

A. `(v^(2))/(4sintheta)`
B. `(v^(2))/(2sintheta)`
C. `(v^(2))/(gsintheta)`
D. `(4gv^(2))/(sintheta)`

Explanation

To solve the problem step by step, we will analyze the forces acting on the block as it moves down and then up the inclined plane.

### Step 1: Analyze the block sliding down the inclined plane
When the block slides down the inclined plane with constant velocity, the net force acting on it is zero. The forces acting on the block are:
- The gravitational force component acting down the incline: \( F_{\text{gravity}} = mg \sin \theta \)
- The frictional force acting up the incline: \( F_{\text{friction}} = \mu mg \cos \theta \)

Since the block is moving with constant velocity, we can set up the equation:
\[ mg \sin \theta = \mu mg \cos \theta \]
From this, we can simplify to find the coefficient of friction:
\[ \mu = \tan \theta \]

### Step 2: Analyze the block projected up the inclined plane
When the block is projected up the incline with an initial velocity \( v \), the forces acting on it are:
- The gravitational force component acting down the incline: \( mg \sin \theta \)
- The frictional force acting down the incline: \( F_{\text{friction}} = \mu mg \cos \theta \)

The total force acting on the block while it moves up the incline is:
\[ F_{\text{net}} = -mg \sin \theta - \mu mg \cos \theta \]
This results in a net acceleration \( a \) acting down the incline:
\[ a = g \sin \theta + \mu g \cos \theta \]
Substituting \( \mu = \tan \theta \) into the equation gives:
\[ a = g \sin \theta + \tan \theta \cdot g \cos \theta = g \sin \theta + g \sin \theta = 2g \sin \theta \]

### Step 3: Use kinematic equations to find the distance traveled
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 when the block comes to rest)
- \( u \) is the initial velocity (given as \( v \))
- \( a \) is the acceleration (which we found to be \( -2g \sin \theta \))
- \( s \) is the distance traveled up the incline

Substituting the values into the equation:
\[ 0 = v^2 - 2(2g \sin \theta)s \]
Rearranging gives:
\[ v^2 = 4g \sin \theta s \]
Thus, solving for \( s \):
\[ s = \frac{v^2}{4g \sin \theta} \]

### Final Answer
The distance traveled by the block up the inclined plane before coming to rest is:
\[ s = \frac{v^2}{4g \sin \theta} \]