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  4. If the coefficient of friction of a plane inclined at 45?, is 0.5, then acceleration of a body sliding freely on it is:

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Q. If the coefficient of friction of a plane inclined at 45?, is 0.5, then acceleration of a body sliding freely on it is:

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Solution:

Acceleration of the body on inclined plane. $ a=g(\sin \theta -\mu \cos \theta ) $ $ =g(\sin {{45}^{o}}-\mu \cos {{45}^{o}}) $ $ =g\left( \frac{1}{\sqrt{2}}-0.5\frac{1}{\sqrt{2}} \right) $ $ =\frac{g}{\sqrt{2}}(1-0.5)=\frac{g}{2\sqrt{2}} $ $ =\frac{9.8}{2\sqrt{2}}m/{{s}^{2}} $