A block of mass m is released on the top of a smooth inclined plane of length x and inclination `theta` as shown in figure. Horizontal surface is rough. If block comes to rest after moving a distabce d on the horizontal surface, then coefficient of friction between block and surface is :-

Options

A. `(xsintheta)/(2d)`
B. `(xcostheta)/(2d)`
C. `(xsintheta)/(d)`
D. `(xcostheta)/(d)`

Answer

To solve the problem step by step, we will analyze the motion of the block on the inclined plane and the horizontal surface.

### Step 1: Understand the Energy Conservation
When the block is released from the top of the inclined plane, it will convert its potential energy into kinetic energy as it moves down the incline. The potential energy at the top is given by:
\[ PE = mgh \]
where \( h \) is the height of the incline.

### Step 2: Determine the Height
The height \( h \) can be expressed in terms of the length of the incline \( x \) and the angle \( \theta \):
\[ h = x \sin \theta \]

### Step 3: Write the Kinetic Energy Equation
As the block reaches the bottom of the incline, all the potential energy will convert into kinetic energy:
\[ KE = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy gives:
\[ mgh = \frac{1}{2} mv^2 \]

### Step 4: Substitute for Height
Substituting \( h \) into the equation:
\[ mg(x \sin \theta) = \frac{1}{2} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ g(x \sin \theta) = \frac{1}{2} v^2 \]

### Step 5: Solve for Velocity
Rearranging gives us the velocity \( v \) at the bottom of the incline:
\[ v^2 = 2g(x \sin \theta) \]
\[ v = \sqrt{2g(x \sin \theta)} \]

### Step 6: Analyze Motion on the Horizontal Surface
When the block reaches the horizontal surface, it will experience friction, which will decelerate it until it comes to rest after traveling a distance \( d \). The frictional force \( f \) acting on the block is given by:
\[ f = \mu mg \]
where \( \mu \) is the coefficient of friction.

### Step 7: Apply Newton's Second Law
Using Newton's second law, the deceleration \( a \) due to friction can be expressed as:
\[ ma = \mu mg \]
Thus, the deceleration \( a \) is:
\[ a = \mu g \]

### Step 8: Use Kinematic Equation
Using the kinematic equation for motion:
\[ v^2 = u^2 + 2ad \]
where \( u \) is the initial velocity (which we found earlier), \( v = 0 \) (final velocity), and \( d \) is the distance traveled on the horizontal surface:
\[ 0 = v^2 - 2(\mu g)d \]
Substituting for \( v^2 \):
\[ 0 = 2g(x \sin \theta) - 2(\mu g)d \]

### Step 9: Rearrange to Find Coefficient of Friction
Rearranging gives:
\[ 2(\mu g)d = 2g(x \sin \theta) \]
Dividing both sides by \( 2g \):
\[ \mu d = x \sin \theta \]
Thus, the coefficient of friction \( \mu \) is:
\[ \mu = \frac{x \sin \theta}{d} \]

### Final Answer
The coefficient of friction between the block and the surface is:
\[ \mu = \frac{x \sin \theta}{d} \]