Question

A block of mass m slides down the plane inclined at angle \(30\degree\) with an acceleration \(\frac{g}{4}\). The value of the coefficient of kinetic friction will be:

• A. \(2\sqrt{3} - \frac{1}{2}\)
• B. \(\frac{1}{2\sqrt{3}}\)
• C. \(2\sqrt{3}+\frac{1}{2}\)
• D. \(\frac{\sqrt{3}}{2}\)

Hint:

For inclined plane problems:
• Use force components along and perpendicular to the plane.
• Solve for friction using acceleration and the force equation.

Answer 1

The Correct Option is B

To solve this problem, we need to find the coefficient of kinetic friction (\(\mu_k\)) for a block sliding down an inclined plane.

Let's denote the given values:

• Mass of the block, \(m\)
• Angle of inclination, \(\theta = 30\degree\)
• Acceleration of the block, \(a = \frac{g}{4}\)
• Gravitational acceleration, \(g\)

The forces acting on the block are:

1. Component of gravitational force parallel to the incline: \(mg \sin \theta\)
2. Normal force (\(N\)), perpendicular to the incline: \(mg \cos \theta\)
3. Kinetic friction force (\(f_k\)): \(\mu_k N = \mu_k mg \cos \theta\)

According to Newton's second law along the incline:

mg \sin \theta - \mu_k mg \cos \theta = ma

Substituting the given acceleration and simplifying for \(\mu_k\):

mg \sin 30\degree - \mu_k mg \cos 30\degree = m\frac{g}{4}

Cancel out \(m\) and \(g\) from all terms:

\sin 30\degree - \mu_k \cos 30\degree = \frac{1}{4}

Substituting the trigonometric values, \(\sin 30\degree = \frac{1}{2}\) and \(\cos 30\degree = \frac{\sqrt{3}}{2}\):

\frac{1}{2} - \mu_k \frac{\sqrt{3}}{2} = \frac{1}{4}

Rearrange to solve for \(\mu_k\):

\frac{1}{2} - \frac{1}{4} = \mu_k \frac{\sqrt{3}}{2} \frac{1}{4} = \mu_k \frac{\sqrt{3}}{2}

Multiplying both sides by \(\frac{2}{\sqrt{3}}\) to isolate \(\mu_k\):

\mu_k = \frac{1}{4} \times \frac{2}{\sqrt{3}} = \frac{1}{2\sqrt{3}}

Thus, the coefficient of kinetic friction is:

\(\frac{1}{2\sqrt{3}}\)