Question

The upper \( \left(\frac{1}{n} \right)^{\text{th}} \) part of an inclined plane is smooth, and the remaining lower part is rough with a coefficient of friction \( \mu_k \). If a body starting from rest at the top of the inclined plane will again come to rest at the bottom of the plane, then the angle of inclination of the inclined plane is:

• A. \( \sin^{-1} \left[ \left( \frac{n}{n-1} \right) \mu_k \right] \)
• B. \( \sin^{-1} \left[ \left( \frac{n-1}{n} \right) \mu_k \right] \)
• C. \( \tan^{-1} \left[ \left( \frac{n}{n-1} \right) \mu_k \right] \)
• D. \( \tan^{-1} \left[ \left( \frac{n-1}{n} \right) \mu_k \right] \)

Hint:

When dealing with friction on an inclined plane, use energy conservation principles to equate the work done by friction with the change in kinetic energy.

Answer 1

The Correct Option is D

Step 1: Understanding the given problem
- The upper \( \frac{1}{n} \)th part of the incline is smooth.
- The lower remaining \( \frac{n-1}{n} \)th part has friction with a coefficient \( \mu_k \).
- The object starts from rest and returns to rest at the bottom.
- This implies that the energy lost due to friction in the rough region exactly cancels out the kinetic energy gained in the smooth region.
Step 2: Using Energy Conservation
The potential energy at the top of the plane is: \[ PE = mg h. \] Since the upper part is smooth, all of this potential energy converts to kinetic energy at the boundary: \[ KE = mg h. \] As the body moves through the rough region, work done against friction is: \[ W = {Friction force} \times {Distance}. \] \[ = mg \cos \theta \cdot \mu_k \cdot \frac{(n-1) L}{n}. \] For the object to stop, energy balance gives: \[ mg h = mg \cos \theta \cdot \mu_k \cdot \frac{(n-1) L}{n}. \] Step 3: Expressing in terms of \( \theta \)
Since \( h = L \sin \theta \), we substitute: \[ mg L \sin \theta = mg \cos \theta \cdot \mu_k \cdot \frac{(n-1) L}{n}. \] Canceling \( mg L \): \[ \sin \theta = \frac{(n-1)}{n} \mu_k \cos \theta. \] Dividing both sides by \( \cos \theta \): \[ \tan \theta = \frac{(n-1)}{n} \mu_k. \] Step 4: Conclusion
\[ \theta = \tan^{-1} \left[ \left( \frac{n-1}{n} \right) \mu_k \right]. \] Thus, the correct answer is: \[ \tan^{-1} \left[ \left( \frac{n-1}{n} \right) \mu_k \right]. \]

Answer 2

The problem involves a body sliding down an inclined plane with its upper \( \left(\frac{1}{n}\right)^{\text{th}} \) part being smooth and the remaining part being rough with a coefficient of friction \( \mu_k \). The body comes to rest again at the bottom. The question is to find the angle of inclination \( \theta \) of the plane. Let's solve it step-by-step.

1. Initial Energy and Work Done: The body starts from rest, so its initial kinetic energy is zero. As it moves down the smooth part, it gains kinetic energy which is then dissipated by the work done against friction on the rough part.

2. Energy Gain on Smooth Part: Let the length of the smooth part be \( \frac{L}{n} \) and the rest be \( L - \frac{L}{n} \). The height of the plane is \( h = L \sin \theta \).

3. Kinetic Energy at End of Smooth Part: By energy conservation, the kinetic energy at the end of the smooth part is given by \( \text{Potential Energy} = mgh = mg\left(\frac{L}{n}\right) \sin \theta \).

4. Work Done on Rough Part: On the rough part, the work done against friction (which equals the gain in kinetic energy on the smooth part) is given by:

\( f \cdot (L - \frac{L}{n}) = \text{Kinetic Energy at the end of smooth part} \)

\( \Rightarrow \mu_k \cdot mg \cos \theta \cdot (L - \frac{L}{n}) = mg\left(\frac{L}{n}\right) \sin \theta \)

\( \Rightarrow \mu_k \cos \theta \cdot (L - \frac{L}{n}) = \left(\frac{L}{n}\right) \sin \theta \)

5. Simplify and Solve for \( \theta \): Divide throughout by \( L \) and rearrange the terms:

\( \mu_k \cos \theta (1 - \frac{1}{n}) = \frac{1}{n} \sin \theta \)

\( \Rightarrow (n-1) \mu_k \cos \theta = \sin \theta \)

\( \Rightarrow \frac{\sin \theta}{\cos \theta} = \frac{(n-1)}{n} \mu_k \)

\( \Rightarrow \tan \theta = \frac{(n-1)}{n} \mu_k \)

6. Solution: Thus, the angle of inclination is:

\( \theta = \tan^{-1} \left[ \left( \frac{n-1}{n} \right) \mu_k \right] \)

The correct option is:

\( \tan^{-1} \left[ \left( \frac{n-1}{n} \right) \mu_k \right] \)