A block of mass m slides down a rough inclined plane of inclination `theta` with horizontal with zero initial velocity. The coefficient of friction between the block and the plane is `mu` with `theta gt tan^(-1)(mu)`. Rate of work done by the force of friction at time t is
A.`mumg^(2)tsintheta` B.`mg^(2)t(sintheta-mucostheta)` C.`mumg^(2)tcostheta(sintheta-mucostheta)` D.`mumg^(2)tcostheta`

Options

A. `mumg^(2)tsintheta`
B. `mg^(2)t(sintheta-mucostheta)`
C. `mumg^(2)tcostheta(sintheta-mucostheta)`
D. `mumg^(2)tcostheta`

Answer

To solve the problem step by step, we will analyze the forces acting on the block, find the acceleration, determine the velocity at time \( t \), and finally calculate the rate of work done by the force of friction.

### Step 1: Identify the Forces Acting on the Block
When the block is on the inclined plane, the forces acting on it are:
- The gravitational force acting downwards: \( mg \)
- The component of gravitational force acting down the incline: \( mg \sin \theta \)
- The normal force acting perpendicular to the incline: \( N = mg \cos \theta \)
- The frictional force acting up the incline: \( F_f = \mu N = \mu mg \cos \theta \)

### Step 2: Write the Equation of Motion
The net force acting on the block along the incline can be expressed as:
\[ F_{\text{net}} = mg \sin \theta - F_f \]
Substituting the frictional force:
\[ F_{\text{net}} = mg \sin \theta - \mu mg \cos \theta \]
Using Newton's second law \( F = ma \):
\[ ma = mg \sin \theta - \mu mg \cos \theta \]
Dividing through by \( m \):
\[ a = g \sin \theta - \mu g \cos \theta \]

### Step 3: Find the Velocity at Time \( t \)
Using the first equation of motion:
\[ v = u + at \]
Since the initial velocity \( u = 0 \):
\[ v = (g \sin \theta - \mu g \cos \theta) t \]

### Step 4: Calculate the Rate of Work Done by the Force of Friction
The power (rate of work done) due to the frictional force can be calculated using the formula:
\[ P = F_f \cdot v \]
Substituting for \( F_f \) and \( v \):
\[ P = \mu mg \cos \theta \cdot \left( (g \sin \theta - \mu g \cos \theta) t \right) \]
Expanding this:
\[ P = \mu mg \cos \theta \cdot (g \sin \theta - \mu g \cos \theta) t \]
\[ P = \mu mg \cos \theta \cdot g \sin \theta t - \mu^2 mg \cos^2 \theta \cdot g t \]
\[ P = \mu mg^2 t \cos \theta (\sin \theta - \mu \cos \theta) \]

### Final Answer
The rate of work done by the force of friction at time \( t \) is:
\[ P = \mu mg^2 t \cos \theta (\sin \theta - \mu \cos \theta) \]