91影视

Mass of block is A, ${\mathrm{m}}_{\mathrm{A}}=10\mathrm{kg}$

Coefficient of kinetic friction between the incline and Body A,${\mathrm{渭}}_{\mathrm{k}}=0.20$

Angle of incline, $\mathrm{胃}=30掳$

The problem is based on Newton鈥檚 second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. To find the acceleration of the body A, use Newton鈥檚 2nd law of motion.

Formula:

${\mathrm{F}}_{\mathrm{net}}=\underset{}{鈭�}\mathrm{ma}$

where, F is the net force, m is mass and a is an acceleration.

Free Body Diagram of the body A and body B:

Block A slides down the incline at constant speed.Thus, the acceleration of block A is 0.By using Newton鈥檚 2nd law of motion along the vertical direction (along y) to the system,(positive x axis along the incline and the positive y along the vertical direction),

$$\begin{gathered} \mathrm{N}={\mathrm{m}}_{\mathrm{A}}\mathrm{gcos}\left(30\right)=0 \\ \mathrm{N}={\mathrm{m}}_{\mathrm{A}}\mathrm{gcos}\left(30\right) \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{k}}={\mathrm{渭}}_{\mathrm{k}}\mathrm{N} \\ ={\mathrm{渭}}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{gcos}\left(30\right) \end{gathered}$$

Now, applying Newton鈥檚 2nd along horizontal direction to the Body A having massm,

$$\begin{gathered} \mathrm{T}+{\mathrm{f}}_{\mathrm{k}}鈥�{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)=0 \\ \mathrm{T}+{\mathrm{渭}}_{\mathrm{k}}鈥�{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)鈥�{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)=0 \end{gathered}$$ (1)Similarly, apply to the Body B having mass m鈥�,

${\mathrm{m}}_{\mathrm{B}}\mathrm{g}-\mathrm{T}=0$ (ii)

Adding equation (i) and (ii),

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}\mathrm{g}+{\mathrm{渭}}_{\mathrm{k}}{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)鈥�{\mathrm{m}}_{\mathrm{A}}\mathrm{g}\sin \left(30\right)=0 \\ {\mathrm{m}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{A}}\left(\sin \left(30\right)鈥�{\mathrm{渭}}_{\mathrm{k}}\cos \left(30\right)\right) \\ =3.3\mathrm{kg} \end{gathered}$$

Therefore, the mass of block B is 3.3 kg.