A body of mass m is pushed with the initial velocity `v_(0)` up an inclined plane of angle of inclination `theta`. The co-efficient or friction between the body and the plane is `mu`. What is the net work done by fnction during the ascent of the body before it comes to stop ?

Answer

To find the net work done by friction during the ascent of a body of mass \( m \) pushed up an inclined plane with an initial velocity \( v_0 \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Body
The forces acting on the body as it moves up the incline include:
1. The gravitational force acting downwards, which can be resolved into two components:
- \( mg \sin \theta \) (parallel to the incline)
- \( mg \cos \theta \) (perpendicular to the incline)
2. The frictional force, which opposes the motion of the body. The frictional force \( f \) is given by:
\[ f = \mu N = \mu (mg \cos \theta) \]

### Step 2: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy:
\[ W_{\text{net}} = K_f - K_i \]
Where:
- \( K_f \) is the final kinetic energy (0, since the body comes to rest)
- \( K_i \) is the initial kinetic energy, given by:
\[ K_i = \frac{1}{2} m v_0^2 \]

Thus, the net work done is:
\[ W_{\text{net}} = 0 - \frac{1}{2} m v_0^2 = -\frac{1}{2} m v_0^2 \]

### Step 3: Calculate Work Done by Gravity
The work done by gravity as the body moves a distance \( L \) up the incline is:
\[ W_{\text{gravity}} = -mg \sin \theta \cdot L \]

### Step 4: Calculate Work Done by Friction
The work done by friction can be expressed as:
\[ W_{\text{friction}} = W_{\text{net}} - W_{\text{gravity}} \]
Substituting the expressions we have:
\[ W_{\text{friction}} = -\frac{1}{2} m v_0^2 - (-mg \sin \theta \cdot L) \]
This simplifies to:
\[ W_{\text{friction}} = -\frac{1}{2} m v_0^2 + mg \sin \theta \cdot L \]

### Step 5: Find the Distance \( L \)
Using the kinematic equation, we can relate the distance \( L \) to the initial velocity \( v_0 \) and the acceleration \( a \) (which is negative due to deceleration):
\[ v^2 = u^2 + 2aL \]
Setting \( v = 0 \) (final velocity), \( u = v_0 \), and \( a = -g \sin \theta - \mu g \cos \theta \):
\[ 0 = v_0^2 - 2(g \sin \theta + \mu g \cos \theta)L \]
From this, we can solve for \( L \):
\[ L = \frac{v_0^2}{2(g \sin \theta + \mu g \cos \theta)} \]

### Step 6: Substitute \( L \) Back into the Work Done by Friction
Now substituting \( L \) back into the work done by friction:
\[ W_{\text{friction}} = -\frac{1}{2} m v_0^2 + mg \sin \theta \cdot \frac{v_0^2}{2(g \sin \theta + \mu g \cos \theta)} \]

### Step 7: Simplify the Expression
Factoring out \( \frac{1}{2} m v_0^2 \):
\[ W_{\text{friction}} = -\frac{1}{2} m v_0^2 \left( 1 - \frac{g \sin \theta}{g \sin \theta + \mu g \cos \theta} \right) \]
This leads to:
\[ W_{\text{friction}} = -\frac{1}{2} m v_0^2 \left( \frac{\mu g \cos \theta}{g \sin \theta + \mu g \cos \theta} \right) \]

### Final Result
The net work done by friction during the ascent of the body before it comes to stop is:
\[ W_{\text{friction}} = -\frac{\mu m v_0^2}{2 \cos \theta ( \tan \theta + \mu )} \]