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:
Show Hint
For inclined plane problems:
• Use force components along and perpendicular to the plane.
• Solve for friction using acceleration and the force equation.
- \(2\sqrt{3} - \frac{1}{2}\)
- \(\frac{1}{2\sqrt{3}}\)
- \(2\sqrt{3}+\frac{1}{2}\)
- \(\frac{\sqrt{3}}{2}\)
The Correct Option is B
Solution and Explanation
1. Force Equation:
\[mg \sin 30^{\circ} - \mu mg \cos 30^{\circ} = ma.\]
2. Substitute Values: - \(a = \frac{g}{4}\), \(\sin 30^{\circ} = \frac{1}{2}\), \(\cos 30^{\circ} = \frac{\sqrt{3}}{2}\).
\[mg \frac{1}{2} - \mu mg \frac{\sqrt{3}}{2} = m \frac{g}{4}.\]
3. Solve for \(\mu\):
\[\frac{\sqrt{3}}{2}\mu=\frac{1}{4}\]
\[\mu=\frac{1}{2\sqrt{3}}\]
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