Question:
On an incline plane of angle \(45^\circ\), time taken by an object is \(t\) if it is smooth and \(2t\) if it is rough. Find value of \(\alpha\) if friction coefficient is \(\frac{\alpha}{100}\).
On an incline plane of angle \(45^\circ\), time taken by an object is \(t\) if it is smooth and \(2t\) if it is rough. Find value of \(\alpha\) if friction coefficient is \(\frac{\alpha}{100}\).
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When an object slides down an inclined plane with friction, the time taken can be compared using the relationship between acceleration and frictional forces. Use the equation \(\frac{t_{\text{rough}}}{t_{\text{smooth}}}\) to relate the two times and solve for the friction coefficient.
Updated On: Apr 7, 2026
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Correct Answer: 75
Solution and Explanation
For an object sliding down an incline, the time taken to reach the bottom is given by the equation:
\[
t = \sqrt{\frac{2h}{g \sin 45^\circ}}
\]
For the rough surface, the acceleration is reduced due to friction, and the time taken is given by:
\[
2t = \sqrt{\frac{2h}{g \sin 45^\circ - \mu g \cos 45^\circ}}
\]
Now, dividing the two equations:
\[
\frac{2t}{t} = \frac{\sqrt{\frac{2h}{g \sin 45^\circ - \mu g \cos 45^\circ}}}{\sqrt{\frac{2h}{g \sin 45^\circ}}}
\]
Simplifying:
\[
2 = \frac{\sqrt{g \sin 45^\circ}}{\sqrt{g \sin 45^\circ - \mu g \cos 45^\circ}}
\]
\[
\frac{1}{2} = \frac{1}{1 - \mu}
\]
Solving for \(\mu\):
\[
\mu = 1 - \frac{1}{4} = \frac{3}{4}
\]
Thus, the friction coefficient \(\mu = \frac{\alpha}{100}\), so:
\[
\frac{\alpha}{100} = \frac{3}{4}
\]
\[
\alpha = 75
\]
Final Answer: 75.
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