Question

If the coefficient of friction of a plane inclined at $45^{\circ }$ is $\mathrm{0.5.}$ Then acceleration of a body sliding freely on it will be:

• A. $\frac{9.8}{2\sqrt{2}}m/s^2$
• B. $\frac{9.8}{\sqrt{2}}m/s^2$
• C. $9.8m/s^2$
• D. $4.8m/s^2$

Answer 1

Answer: (A)

Here: Angle of plane is inclined $\theta =45^{\circ }$
Coefficient of friction $\mu =0.5$
The acceleration of the body sliding on the inclined plane is given by
$=g(\sin \theta -\mu \cos \theta )$
$=9.8\left(\sin 45^{\circ }-0.5\cos 45^{\circ }\right)$
$=9.8\left(\frac{1}{\sqrt{2}}-0.5\times \frac{1}{\sqrt{2}}\right)$
$=\frac{9.8}{2\sqrt{2}}m/s^2$