A solid spherical ball is released from rest on an incline of inclination angle `theta` (which can be varied ) but through a fixed vertical height `h`. The coefficient of static and kinetic friction are both equal to `mu`. If `E` represents the total kinetic energy of the ball at the bottom of the incline as a function of the angle of inclination `theta`. `W` represents the work done by friction for the whole time of motion as a function of the angle of inclination `theta`. Choose of correct graph (s)

Options

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Explanation

To solve the problem, we need to analyze the motion of a solid spherical ball released from rest on an incline. We will derive expressions for the total kinetic energy \( E \) at the bottom of the incline and the work done by friction \( W \) as functions of the angle of inclination \( \theta \).

### Step 1: Analyze Forces Acting on the Ball
When the ball is on the incline, the forces acting on it are:
- The weight of the ball \( mg \), which can be resolved into two components:
- Parallel to the incline: \( mg \sin \theta \)
- Perpendicular to the incline: \( mg \cos \theta \)
- The normal force \( N \) acting perpendicular to the incline.
- The frictional force \( f \), which opposes the motion and is given by \( f = \mu N \).

### Step 2: Condition for Motion
The ball will start moving when the component of gravitational force down the incline exceeds the frictional force:
\[ mg \sin \theta > \mu mg \cos \theta \]
This simplifies to:
\[ \tan \theta > \mu \]
Thus, the ball will not move for \( \theta < \tan^{-1}(\mu) \).

### Step 3: Net Force and Acceleration
Once the ball starts moving, the net force acting on the ball along the incline can be expressed as:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
Using Newton's second law, we have:
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
Thus, the acceleration \( a \) of the ball is:
\[ a = g \sin \theta - \mu g \cos \theta \]

### Step 4: Work Done by Gravity and Friction
The work done by gravity when the ball moves a distance \( s \) down the incline is:
\[ W_{\text{gravity}} = mg h \]
Where \( h \) is the fixed vertical height.

The work done by friction is:
\[ W_{\text{friction}} = -f \cdot s = -\mu mg \cos \theta \cdot s \]

### Step 5: Kinetic Energy at the Bottom of the Incline
Using the work-energy theorem:
\[ W_{\text{gravity}} + W_{\text{friction}} = \Delta KE \]
Since the initial kinetic energy is zero (the ball starts from rest), we have:
\[ mg h - \mu mg \cos \theta \cdot s = \frac{1}{2} mv^2 \]
Where \( v \) is the final velocity of the ball at the bottom of the incline.

### Step 6: Relating Distance \( s \) to Height \( h \)
From geometry, we have:
\[ h = s \sin \theta \]
Thus, we can express \( s \) as:
\[ s = \frac{h}{\sin \theta} \]

### Step 7: Total Kinetic Energy \( E \)
Substituting \( s \) into the kinetic energy equation:
\[ E = mg h - \mu mg \cos \theta \cdot \frac{h}{\sin \theta} \]
This simplifies to:
\[ E = mg h \left(1 - \frac{\mu \cos \theta}{\sin \theta}\right) \]
\[ E = mg h \left(1 - \mu \cot \theta\right) \]

### Step 8: Work Done by Friction \( W \)
The work done by friction can also be expressed as:
\[ W = -\mu mg \cos \theta \cdot \frac{h}{\sin \theta} \]
This simplifies to:
\[ W = -\mu mg h \cot \theta \]

### Conclusion
- The total kinetic energy \( E \) as a function of \( \theta \) is given by:
\[ E = mg h \left(1 - \mu \cot \theta\right) \]
- The work done by friction \( W \) as a function of \( \theta \) is given by:
\[ W = -\mu mg h \cot \theta \]