91Ó°ÊÓ

Mass of the block hanging over a pulley, $m=2kg$

The angle of the plane,$\mathrm{ϕ}=20°$

Coefficient of static friction, ${\mathrm{μ}}_s=0.80$

Coefficient of kinematic friction,${\mathrm{μ}}_k=0.50$

Consider the equilibrium of the block$1$

localid="1649648629462" $$\begin{gathered} \left(2kg\right)\left(9.81m/s^2\right)-T=0 \\ T=19.62N \end{gathered}$$

Consider the equilibrium of the inclined box

localid="1649648590263" $$\begin{gathered} mg\sin 20+{\mathrm{μ}}_{s}mg\cos 20-T=0 \\ m×\left(9.81m/s^2\right)\sin \left(20°\right)+\left(0.8\right)m\left(9.81m/s^2\right)\cos \left(20°\right)=19.62N \\ m=1.8kg \end{gathered}$$

Thus, the mass of the block required to stick the system in the equilibrium is $1.8kg$.

Mass of the block hanging over a pulley, $m=2kg$

The angle of the plane,$\mathrm{ϕ}=20°$

Coefficient of static friction, ${\mathrm{μ}}_s=0.80$

Coefficient of kinematic friction, ${\mathrm{μ}}_k=0.50$

Consider the equilibrium of the block $2$

localid="1649648995886" $\left(2kg\right)\left(9.81m/s^2\right)-T=2a$

Consider the equilibrium of the block $1$

localid="1649648969079" role="math" $$\begin{gathered} T-mg\sin 20-{\mathrm{μ}}_{k}mg\cos 20=ma \\ T-\left[\left(1.8kg\right)×\left(9.81m/s^2\right)\sin \left(20°\right)\right]-\left[0.5\left(1.8kg\right)\left(9.81m/s^2\right)\cos \left(20°\right)\right]=\left(1.8kg\right)×a \\ \left(2kg+1.8kg\right)a=\left[\left(2kg\right)×\left(9.81m/s^2\right)-\left(1.8kg\right)×\left(9.81m/s^2\right)\left(\sin \left(20°\right)\right)\right]-\left[\left(0.5\right)×\left(1.8kg\right)×\left(9.81m/s^2\right)\cos (20°)\right] \\ a=1.39m/s^2 \end{gathered}$$

Acceleration of the block is$1.39m/s^2$.