Question

Two blocks are connected over a massless pulley as shown in fig. The mass of block $A$ is $10\, kg$ and the 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. 3.5
• B. 3.3
• C. 3
• D. 2.5

Answer 1

Answer: (B)

3.3

Answer 2

Considering the equilibrium of $A$, we get $10 a =10 g \sin 30^{\circ}-T$ - $mN$ where $N=10\, g \cos 30^{\circ}$ $\therefore 10 a=\frac{10}{2} g-T-\mu \times 10\, g \cos 30^{\circ}$ but $a=0, T =m_{B} g$ $0=5\, g-m_{B} g-\frac{0.2 \sqrt{3}}{2} \times 10\, g$ $\Rightarrow m_{B}=3.268 \approx 3.3\, kg$