Question

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

• A. 5.4
• B. 3.3
• C. 4.2
• D. 6.8

Answer 1

The correct option is B 3.3

Given that,

Mass of block $A=10$ kg

Kinetic friction $\mu =0.2$

Now, according to diagram

Now, the component of the weight in the direction of the incline

$W=m_{A}g\sin \theta$

$T=F_f-W....\left(I\right)$

When the force of friction

$F_f={\mu }_{k}N$

$F_f={\mu }_k\left(m_{A}g\cos \theta \right)$

Now, put the value in equation (I)

The tension is

$T=m_{A}g\sin {30}^0-{\mu }_k{m}_{A}g\cos {30}^0$

We know that,

$T-m_{B}g=0$

$T=m_{B}g$

So,

$m_{B}g=m_{A}g(\sin {30}^0-0.2\times \cos {30}^0)$

$m_B=10\times (\frac{1}{2}-0.2\times \frac{\sqrt{3}}{2})$

$m_B=10\times (0.5-0.1732)$

$m_B=10\times 0.3268$

$m_B=3.268$

$m_B=3.3kg$

Hence, the mass of block B is $3.3$ kg