Question

Two blocks are connected over a massless pulley as shown in the figure. The mass of block $A$ is $10\text{kg}$ and coefficient of kinetic friction is $0.2$. Block $A$ slides down the incline at constant speed. The mass of the block $B$ (in $\text{kg}$) is

• A. $5.4$
• B. $3.3$
• C. $4.2$
• D. $6.8$

Answer 1

The correct option is B $3.3$

As it is given in the problem the the mass $A$ slides down with the constant velocity, hence net force on the system will be zero.

By the FBD of both the blocks, ,

Balancing the force on block $B$ we get $T=m_{B}g...\left(i\right)$,
Net force on block $A$,
$m_{A}g\sin {30}^{\text{o}}-T-{\mu }_{k}N=0...\left(ii\right)$, putting the value of $T$ in equation $\left(i\right)$ we get,
$m_{B}g=m_{A}g\sin {30}^{\text{o}}-{\mu }_k\times m_{A}g\cos {30}^{\text{o}}$, (As $N=m_{A}g\cos {30}^{\text{o}}$)

Solving for the above equation we get,

$m_B=\frac{m_A}{2}-\frac{{\mu }_k{m}_A\sqrt{3}}{2}$
$\Rightarrow m_B=10(\frac{1}{2}-0.2\times \frac{\sqrt{3}}{2})=5-\sqrt{3}$
$m_B=3.3\text{kg}$, as $\sqrt{3}\approx 1.7$

Hence option B is the correct answer