Question

If the coefficient of friction of a plane inclined at $ 45^{\circ}$ is $0.5.$ 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. $ 9.8\,m/s^{2}$
• D. $ 4.8\,m/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
$=g(\sin \theta-\mu \cos \theta)$
$=9.8\left(\sin 45^{\circ}-0.5 \cos 45^{\circ}\right)$
$=9.8\left(\frac{1}{\sqrt{2}}-0.5 \times \frac{1}{\sqrt{2}}\right)$
$=\frac{9.8}{2 \sqrt{2}} m / s ^{2}$