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Question

(A) The acceleration $a$ of a body sliding down an inclined plane with friction is given by the formula: $a = g(\sin 	heta - \mu \cos 	heta)$.Given:

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$\frac{4.9}{\sqrt{2}}$

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(A) The acceleration $a$ of a body sliding down an inclined plane with friction is given by the formula: $a = g(\sin 	heta - \mu \cos 	heta)$.Given: $g = 9.8\,m/s^2$,$	heta = 45^o$,and $\mu = 0.5$.Substituting the values:$a = 9.8(\sin 45^o - 0.5 \cos 45^o)$Since $\sin 45^o = \cos 45^o = \frac{1}{\sqrt{2}}$:$a = 9.8 \left( \frac{1}{\sqrt{2}} - 0.5 	imes \frac{1}{\sqrt{2}} ight)$$a = 9.8 \left( \frac{1 - 0.5}{\sqrt{2}} ight)$$a = 9.8 \left( \frac{0.5}{\sqrt{2}} ight)$$a = \frac{4.9}{\sqrt{2}}\,m/s^2$.

The coefficient of friction between a body and the surface of an inclined plane at $45^o$ is $0.5$. If $g = 9.8\,m/s^2$,the acceleration of the body downwards in $m/s^2$ is

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