The upper half of an inclined plane with inclination $$\phi$$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Let the total length of the inclined plane be $$L$$. Hence, the upper half (smooth) has length $$\frac{L}{2}$$ and the lower half (rough) also has length $$\frac{L}{2}$$. The angle of inclination with the horizontal is $$\phi$$.

Case 1: Motion on the smooth upper half

• The only component of force along the plane is the gravitational component $$mg \sin \phi$$ (no friction).
• Acceleration along the plane: $$a_1 = g \sin \phi$$.
• Initial speed at the top is zero. Using the kinematic relation $$v^2 = u^2 + 2as$$ for the distance $$s = \frac{L}{2}$$, we get
$$v_1^2 = 0 + 2\bigl(g \sin \phi\bigr)\frac{L}{2} = gL \sin \phi$$.
Thus, the speed at the midpoint (start of the rough half) is $$v_1 = \sqrt{gL \sin \phi}$$.

Case 2: Motion on the rough lower half

• Forces along the plane:
  - Down the plane: $$mg \sin \phi$$.
  - Up the plane (friction): $$\mu mg \cos \phi$$, where $$\mu$$ is the coefficient of kinetic friction.
• Net force down the plane: $$mg(\sin \phi - \mu \cos \phi)$$.
• Hence, net acceleration along the plane:
$$a_2 = g(\sin \phi - \mu \cos \phi)$$.

The block starts this section with speed $$v_1$$ and travels the distance $$\frac{L}{2}$$. To come to rest exactly at the bottom, the final speed must be zero. Again using $$v^2 = u^2 + 2as$$:

$$0 = v_1^2 + 2a_2\left(\frac{L}{2}\right)$$ $$\Longrightarrow 0 = gL \sin \phi + g(\sin \phi - \mu \cos \phi)L$$ (dividing by $$gL$$ gives)

$$0 = 2\sin \phi - \mu \cos \phi$$ $$\Longrightarrow \mu = 2 \tan \phi$$.

Therefore, the coefficient of friction required on the lower half is $$\mu = 2 \tan \phi$$.

Option C which is: $$2 \tan \phi$$