Question IV

IV) Two blocks are connected over a pulley as shown in the Figure below. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.20. The angle $\theta$ is 30⁰. Block A slides down the incline at constant speed. What is the mass of block B?

Solution:
We know that if we sum the forces on block A in the plane of the incline, we get
T - F_f - W
Where the force of friction is given by
$F_{f} = \mu_{k} N ~=~ \mu_{k} (m_{A}g \cos \theta)$
and the component of the weight in the direction of the incline is given by
$W = m_{A}g \sin \theta$
Solving for the tension, T, we find
$T = m_{A}g (\sin 30{}^{\circ}- \mu_{k} \cos 30{}^{\circ})$
and we know that
T - m_Bg = 0
so we find that
$m_{B} = m_{A} (\sin 30{}^{\circ}- \mu_{k}\cos 30{}^{\circ})$
or
m_B  =  3.3 kg