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  4. 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:

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Q. 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:Physics Question Image

BITSATBITSAT 2010

Solution:

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 $a=0, T =m_{B} g$
$0=5\, g-m_{B} g-\frac{0.2 \sqrt{3}}{2} \times 10\, g$
$\Rightarrow m_{B}=3.268 \approx 3.3\, kg$

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