Question

The upper half of an inclined plane of inclination θ is perfectly smooth while lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by

• A. $μ= \frac{1}{tan θ}$
• B. $μ= \frac2{tan θ}$
• C. μ = 2tan θ
• D. μ = tan θ

Answer 1

The Correct Option is C

To determine the correct expression for the coefficient of friction that will bring the block to rest at the bottom of the inclined plane, we need to analyze the problem using the principles of energy conservation and friction.

Let's break down the problem:

1. Consider the inclined plane divided into two equal halves - the upper half is smooth (frictionless), and the lower half is rough with a coefficient of friction μ.
2. The block starts from rest at the top of the inclined plane with height h and inclination angle θ.
3. As the block slides down the smooth surface, it converts potential energy to kinetic energy.
4. At the midpoint (beginning of the rough section), the potential energy is zero and all initial potential energy \((mgh)\) has been converted to kinetic energy \((\frac{1}{2}mv^2)\).
5. As the block enters the rough section, frictional force begins to act against the motion of the block. The work done by friction over the distance \(d\) (which is half of the inclined plane) can be expressed as: W_{\text{friction}} = μmg\cosθ \cdot d.

To ensure that the block comes to rest at the bottom, all the kinetic energy at the midpoint must be dissipated by the work done by friction:

\frac{1}{2}mv^2 = μmg \cosθ \cdot d

Since the plane is inclined, using \sinθ = \frac{h}{L} and \cosθ = \frac{d}{L} with \(d = \frac{L}{2}\), the potential energy at the top is mgh = mgL \sinθ.

At the midpoint (of length \frac{L}{2}), insert the kinetic energy equation:

mgL \sinθ = μmg \cosθ \cdot \frac{L}{2}

Cancel mg and L on both sides:

\sinθ = \frac{μ \cosθ}{2}

Rearranging gives:

μ = 2 \tanθ

Thus, the correct answer is μ = 2\tanθ, which justifies option 3.