The upper portion of an inclined plane of inclination `alpha` is smooth and the lower portion is rough. A particle slides down from rest from the top and just comesto rest at the foot.If the ratio ofthe smooth length to rough length is w : find the coefficient of friction.

Answer

To solve the problem step-by-step, we will analyze the forces acting on the particle as it slides down the inclined plane and apply the work-energy principle.

### Step 1: Identify the components of the inclined plane
- The inclined plane has two sections: a smooth upper portion of length \( L \) and a rough lower portion of length \( L' \).
- The angle of inclination is given as \( \alpha \).
- The ratio of the smooth length to the rough length is given as \( \frac{L}{L'} = w \).

### Step 2: Set up the work-energy principle
- The particle starts from rest at the top and comes to rest at the bottom. Therefore, the initial kinetic energy (KE_initial) and final kinetic energy (KE_final) are both zero.
- According to the work-energy principle, the total work done by all forces is equal to the change in kinetic energy:
\[ W_{\text{total}} = KE_{\text{final}} - KE_{\text{initial}} = 0 - 0 = 0 \]

### Step 3: Calculate the work done by gravity
- The work done by gravity as the particle moves down the incline can be calculated using the vertical height:
\[ W_{\text{gravity}} = mgh \]
- The vertical height \( h \) can be expressed in terms of the lengths of the inclined plane:
\[ h = (L + L') \sin \alpha \]
- Thus, the work done by gravity becomes:
\[ W_{\text{gravity}} = mg(L + L') \sin \alpha \]

### Step 4: Calculate the work done by friction
- The frictional force acting on the particle while it moves down the rough portion is given by:
\[ F_{\text{friction}} = \mu mg \cos \alpha \]
- The work done by friction as the particle moves down the rough portion (length \( L' \)) is:
\[ W_{\text{friction}} = -F_{\text{friction}} \cdot L' = -\mu mg \cos \alpha \cdot L' \]

### Step 5: Set up the equation for total work done
- According to the work-energy principle, the total work done is zero:
\[ W_{\text{gravity}} + W_{\text{friction}} = 0 \]
- Substituting the expressions for work done:
\[ mg(L + L') \sin \alpha - \mu mg \cos \alpha \cdot L' = 0 \]

### Step 6: Simplify the equation
- Cancel \( mg \) from both sides:
\[ (L + L') \sin \alpha = \mu \cos \alpha \cdot L' \]
- Rearranging gives:
\[ \mu = \frac{(L + L') \sin \alpha}{L' \cos \alpha} \]

### Step 7: Substitute the ratio \( \frac{L}{L'} = w \)
- From the ratio \( \frac{L}{L'} = w \), we can express \( L \) as \( L' \cdot w \):
\[ L = wL' \]
- Substitute \( L \) into the equation:
\[ \mu = \frac{(wL' + L') \sin \alpha}{L' \cos \alpha} = \frac{(w + 1)L' \sin \alpha}{L' \cos \alpha} \]
- Simplifying further:
\[ \mu = (w + 1) \frac{\sin \alpha}{\cos \alpha} = (w + 1) \tan \alpha \]

### Final Result
The coefficient of friction \( \mu \) is given by:
\[ \mu = (w + 1) \tan \alpha \]