Explain the operation of a vapor absorption refrigeration system and list two methods of starting a synchronous motor.

Step 1: Resolve the pulling force $F$ into components.
$F_x = F \cos \theta = 20 \cos 30^\circ \, \mathrm{N}$
$\cos 30^\circ = \frac{\sqrt{3}}{2}$
$F_x = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \mathrm{N}$

$F_y = F \sin \theta = 20 \sin 30^\circ \, \mathrm{N}$
$\sin 30^\circ = \frac{1}{2}$
$F_y = 20 \times \frac{1}{2} = 10 \, \mathrm{N}$ (upward)

Step 2: Calculate the normal force $N$.
Vertical equilibrium: $N + F_y = mg$
$N = mg - F_y$
$mg = 4 \times 10 = 40 \, \mathrm{N}$
$N = 40 - 10 = 30 \, \mathrm{N}$

Step 3: Calculate the frictional force $f_k$.
$f_k = \mu N = 0.2 \times 30 = 6 \, \mathrm{N}$

Step 4: Net horizontal force $F_\mathrm{net}$.
$F_net = F_x - f_k = 10\sqrt{3} - 6 \, \mathrm{N}$
$\sqrt{3} \approx 1.732$
$10\sqrt{3} \approx 17.32 \, \mathrm{N}$
$F_net \approx 17.32 - 6 = 11.32 \, \mathrm{N}$

Step 5: Acceleration $a$.
$a = \frac{F_net}{m} = \frac{11.32}{4} = 2.83 \, \mathrm{m/s^2}$

Or exactly:
$a = \frac{10\sqrt{3} - 6}{4} \, \mathrm{m/s^2}$

Final answer: $a \approx 2.83 \, \mathrm{m/s^2}$
2.83 m/s^{2}