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  • The time taken by an object to slide down rough inclined plane is n times as it takes to slide down

The time taken by an object to slide down 45^{\circ} rough inclined plane is n times as it takes to slide down a perfectly smooth 45^{\circ} incline plane. The coefficient of kinetic friction between the object and the incline plane is:

Option: 1

\sqrt{1-\frac{1}{n^2}}


Option: 2

1+\frac{1}{n^2}


Option: 3

1-\frac{1}{n^2}


Option: 4

\sqrt{\frac{1}{1-n^2}}


Answers (1)

best_answer

Acceleration on the smooth inclined plane

\mathrm{a}_1=\mathrm{g} \sin \theta=\frac{\mathrm{g}}{\sqrt{2}}

Acceleration on the rough inclined plane

\mathrm{a}_2=\mathrm{g} \sin \theta-\mu \mathrm{g} \cos \theta=\frac{\mathrm{g}}{\sqrt{2}}-\frac{\mathrm{Kg}}{\sqrt{2}}(\mathrm{~K}=\mu)

Given that:

\begin{aligned} & \mathrm{t}_2=\mathrm{nt}_1 \quad \text { and } \quad \frac{1}{2} \mathrm{a}_1 \mathrm{t}_1^2=\frac{1}{2} \mathrm{a}_2 \mathrm{t}_2^2 \\ & \mathrm{a}_1 \mathrm{t}_1^2=\mathrm{a}_2 \mathrm{t}_2^2 \\ & \frac{\mathrm{g}}{\sqrt{2}} \mathrm{t}_1^2=\left(\frac{\mathrm{g}}{\sqrt{2}}-\frac{\mathrm{Kg}}{\sqrt{2}}\right)\left(\mathrm{n}^2 \mathrm{t}_1^2\right) \\ & \frac{\mathrm{g}}{\sqrt{2}}=\mathrm{n}^2\left(\frac{\mathrm{g}}{\sqrt{2}}-\frac{\mathrm{Kg}}{\sqrt{2}}\right) \\ & \mathrm{K}=1-\frac{1}{\mathrm{n}^2} \end{aligned}

Posted by

Ritika Harsh

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