A body of mass m slides down a plane inclined at an angle `alpha`. The coefficient of friction is `mu`. Find the rate at which kinetic plus gravitational potential is dissipated at any time t.

Explanation

To solve the problem of finding the rate at which kinetic plus gravitational potential energy is dissipated when a body of mass \( m \) slides down an inclined plane at an angle \( \alpha \) with a coefficient of friction \( \mu \), we can follow these steps:

### Step 1: Identify the Forces Acting on the Body
The forces acting on the body include:
- The gravitational force \( mg \) acting vertically downward.
- The normal force \( N \) acting perpendicular to the inclined plane.
- The frictional force \( F_f \) acting opposite to the direction of motion along the plane.

### Step 2: Resolve the Gravitational Force
The gravitational force can be resolved into two components:
- Parallel to the incline: \( F_{\parallel} = mg \sin \alpha \)
- Perpendicular to the incline: \( F_{\perpendicular} = mg \cos \alpha \)

### Step 3: Calculate the Normal Force
The normal force \( N \) is equal to the perpendicular component of the gravitational force:
\[ N = mg \cos \alpha \]

### Step 4: Calculate the Frictional Force
The frictional force \( F_f \) can be expressed as:
\[ F_f = \mu N = \mu (mg \cos \alpha) = \mu mg \cos \alpha \]

### Step 5: Write the Net Force Acting on the Body
The net force \( F_{\text{net}} \) acting on the body along the incline is given by:
\[ F_{\text{net}} = F_{\parallel} - F_f = mg \sin \alpha - \mu mg \cos \alpha \]

### Step 6: Calculate the Acceleration of the Body
Using Newton's second law, we can find the acceleration \( a \):
\[ ma = mg \sin \alpha - \mu mg \cos \alpha \]
\[ a = g \sin \alpha - \mu g \cos \alpha \]

### Step 7: Determine the Rate of Work Done by Friction
The power \( P \) (rate of work done) due to the frictional force is given by:
\[ P = F_f \cdot v \]
where \( v \) is the velocity of the body down the incline.

### Step 8: Relate Velocity to Acceleration
Using the kinematic equation, we can express the velocity \( v \) in terms of acceleration and time:
\[ v = a \cdot t = (g \sin \alpha - \mu g \cos \alpha) t \]

### Step 9: Substitute into the Power Equation
Now substituting \( v \) into the power equation:
\[ P = \mu mg \cos \alpha \cdot v \]
\[ P = \mu mg \cos \alpha \cdot (g \sin \alpha - \mu g \cos \alpha) t \]

### Step 10: Final Expression for Power
Thus, the rate at which kinetic plus gravitational potential energy is dissipated is:
\[ P = \mu mg \cos \alpha (g \sin \alpha - \mu g \cos \alpha) t \]

### Summary
The rate at which kinetic plus gravitational potential energy is dissipated due to friction while sliding down the incline is given by the expression derived above.

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