A body is moving up an inclined plane of angle `theta` with an anitial kinetic energy E. The coefficient of friction between the plane and body is `mu`. The work done against friction before the body comes to rest is

Options

A. `(muE costheta)/(costheta+sintheta)`
B. E
C. `(muE costheta)/(mucostheta-sintheta)`
D. `(muE costheta)/(mucostheta+sintheta)`

Answer

To solve the problem step by step, we will use the work-energy theorem and the forces acting on the body moving up the inclined plane.

### Step 1: Understand the Problem
We have a body moving up an inclined plane at an angle \( \theta \) with an initial kinetic energy \( E \). The coefficient of friction between the body and the plane is \( \mu \). We need to find the work done against friction before the body comes to rest.

### Step 2: Apply the Work-Energy Theorem
According to the work-energy theorem, the work done on an object is equal to the change in its kinetic energy. The equation can be expressed as:
\[ W_{\text{friction}} + W_{\text{gravity}} = K_f - K_i \]
where:
- \( W_{\text{friction}} \) is the work done against friction,
- \( W_{\text{gravity}} \) is the work done by gravity,
- \( K_f \) is the final kinetic energy (which is 0 since the body comes to rest),
- \( K_i \) is the initial kinetic energy \( E \).

Thus, we have:
\[ W_{\text{friction}} + W_{\text{gravity}} = 0 - E \]
or
\[ W_{\text{friction}} + W_{\text{gravity}} = -E \]

### Step 3: Calculate Work Done by Gravity
The work done by gravity can be calculated as:
\[ W_{\text{gravity}} = -Mg \cdot s \sin \theta \]
where \( M \) is the mass of the body, \( g \) is the acceleration due to gravity, and \( s \) is the distance moved along the incline.

### Step 4: Substitute Work Done by Gravity
Substituting the expression for \( W_{\text{gravity}} \) into the work-energy equation gives:
\[ W_{\text{friction}} - Mg s \sin \theta = -E \]
Rearranging this, we find:
\[ W_{\text{friction}} = -E + Mg s \sin \theta \]

### Step 5: Determine the Distance \( s \)
To find the distance \( s \), we need to consider the forces acting on the body. The net force acting on the body while it is moving up the incline includes the gravitational force and the frictional force. The net deceleration \( a \) can be expressed as:
\[ a = g \sin \theta + \mu g \cos \theta \]

Using the kinematic equation \( v^2 = u^2 + 2as \) where \( v = 0 \) (final velocity), \( u^2 = \frac{2E}{M} \) (initial velocity), we have:
\[ 0 = \frac{2E}{M} - 2(g \sin \theta + \mu g \cos \theta)s \]
Solving for \( s \):
\[ s = \frac{E}{M(g \sin \theta + \mu g \cos \theta)} \]

### Step 6: Substitute \( s \) Back into Work Done by Friction
Now substituting \( s \) back into the expression for \( W_{\text{friction}} \):
\[ W_{\text{friction}} = -E + Mg \left(\frac{E}{M(g \sin \theta + \mu g \cos \theta)}\right) \sin \theta \]
This simplifies to:
\[ W_{\text{friction}} = -E + \frac{E g \sin \theta}{g \sin \theta + \mu g \cos \theta} \]
Factoring out \( E \):
\[ W_{\text{friction}} = E \left(-1 + \frac{g \sin \theta}{g \sin \theta + \mu g \cos \theta}\right) \]
This can be simplified further to:
\[ W_{\text{friction}} = -\frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta} \]

### Final Answer
Thus, the work done against friction before the body comes to rest is:
\[ W_{\text{friction}} = -\frac{\mu E \cos \theta}{\sin \theta + \mu \cos \theta} \]