Question

A block of mass $5 \,kg$ 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, $\left.g=10\,m / s ^{2}\right]$
,

• 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 $=m a$.

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

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