Question

A block of mass $5kg$ is kept against an accelerating wedge with a wedge angle of $45^{\circ }$ to the horizontal. The coefficient of friction between the block and the wedge is $\mu =0.4$. What is the minimum absolute value of the acceleration of the wedge to keep the block steady? [Assume, $g=10m/s^2]$
,

• A. $\frac{60}{7}m/s^2$
• B. $\frac{30}{7}m/s^2$
• C. $\frac{30}{\sqrt{7}}m/s^2$
• D. $\frac{60}{\sqrt{7}}m/s^2$

Answer 1

Answer: (B)

Let $a=$ acceleration of wedge, then force on block due to acceleration of wedge $=ma$.

Then, from FBD we have, For equilibrium of block,

$mg\sin \alpha =ma\cos \alpha +\mu (ma\cos \alpha +ma\sin \alpha )$
$\Rightarrow g=a+\mu (g+a)\Rightarrow \frac{0.6}{1.4}g=9$
$\Rightarrow a=\frac{30}{7}m{s}^{-2}$