Question

In the figure block $A$ of mass $m$ is placed on block $B$ of mass $2m$. $B$ rests on floor. The coefficient of friction between $A$ and $B$ is $\mu$ and that between $B$ and floor $\frac{\mu }{2}$. Both blocks are given same initial velocity to the right. what is the acceleration of $A$ with respect to $B$.

• A. $\frac{\mu g}{4}$ towords right
• B. $\mu g$ towords right
• C. $\frac{\mu g}{2}$ towords right
• D. $\frac{3\mu g}{4}$ towords right

Answer 1

The correct option is D $\frac{3\mu g}{4}$ towords right

Let $N_1$ the normal reaction between blocks, $N_2$ be normal reaction between lower block and ground, $F_1$ be the friction between blocks and $F_2$ be the friction between lower block and ground
Since both the blocks are thrown with same initial speed, therefore lower block start decelerating due to friction provided by ground and in turn upper block also starts decelerating due to friction between blocks

Find relative acceleration of blocks.
For upper block
Since there is relative slipping between blocks hence kinetic frictiion will act
$F_1=\mu N_1=m{a}_1\therefore a_1=-\mu g$
Negative indicates declartion

For lower block
$F_1-F_2=2m{a}_{2}m\left(\mu g\right)-\frac{\mu }{2}\left(3mg\right)=2m{a}_2{a}_2=-\left(\frac{\mu g}{4}\right)$
Negative sign indicates declaeration.
Acceleration of $A$ with respect to block $B$
$\therefore a_{rel}=-\mu g-(-\frac{\mu g}{a})=-\frac{3\mu g}{4}$
Here, negative indicates towords left direction.