Question

Two bodies are connected over a massless pulley.the mass of block A is 10kg 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

Answer 1

Explanation:

We know that if we sum the forces on block A in the plane of the incline, we get

T - Ff - 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 - mBg = 0

so we find that

$m_{B} = m_{A} (\sin 30{}^{\circ}- \mu_{k}\cos 30{}^{\circ})$

or

mB = 3.3 kg