Terdapat satu dongkrak hidrolik memiliki dua piston dengan luas penampang masing-masing 10 cm² dan 200 cm². Jika gaya sebesar 50 N diberikan pada piston yang lebih kecil, berapakah gaya yang dihasilkan pada piston yang lebih besar?

Given:
Area of smaller piston ($A_1$) = $10 \text{ cm}^2$
Area of larger piston ($A_2$) = $200 \text{ cm}^2$
Force on smaller piston ($F_1$) = $50 \text{ N}$
Pascal's Principle states that pressure is transmitted equally throughout an enclosed fluid:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

$$\frac{50 \text{ N}}{10 \text{ cm}^2} = \frac{F_2}{200 \text{ cm}^2}$$ $$F_2 = \frac{50 \text{ N} \times 200 \text{ cm}^2}{10 \text{ cm}^2}$$ $$F_2 = 5 \text{ N} \times 200$$ $$F_2 = 1000 \text{ N}$$

The force produced on the larger piston is $\boxed{1000 \text{ N}}$.

2. Mengangkat Mobil dengan Hidrolik (Lifting a Car with Hydraulics)

Given:
Weight of the car ($F_2$) = $5000 \text{ N}$
Force on smaller piston ($F_1$) = $100 \text{ N}$
Area of smaller piston ($A_1$) = $0.5 \text{ m}^2$
Formula:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

$$\frac{100 \text{ N}}{0.5 \text{ m}^2} = \frac{5000 \text{ N}}{A_2}$$ $$A_2 = \frac{5000 \text{ N} \times 0.5 \text{ m}^2}{100 \text{ N}}$$ $$A_2 = 50 \times 0.5 \text{ m}^2$$ $$A_2 = 25 \text{ m}^2$$

The area of the larger piston is $\boxed{25 \text{ m}^2}$.

3. Mengangkat Mobil dengan Dongkrak Hidrolik (Lifting a Car with a Hydraulic Jack)

Given:
Weight of the car ($F_2$) = $12000 \text{ N}$
Area of smaller piston ($A_1$) = $25 \text{ cm}^2$
Area of larger piston ($A_2$) = $500 \text{ cm}^2$
Formula:

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$

$$\frac{F_1}{25 \text{ cm}^2} = \frac{12000 \text{ N}}{500 \text{ cm}^2}$$ $$F_1 = \frac{12000 \text{ N} \times 25 \text{ cm}^2}{500 \text{ cm}^2}$$ $$F_1 = \frac{12000 \text{ N}}{20}$$ $$F_1 = 600 \text{ N}$$

The force that must be applied to the smaller piston is $\boxed{600 \text{ N}}$.

4. Dongkrak Hidrolik dengan Jari-jari Piston (Hydraulic Jack with Piston Radii)

Given:
Radius of smaller piston ($r_1$) = $2 \text{ cm}$
Radius of larger piston ($r_2$) = $10 \text{ cm}$
Force on smaller piston ($F_1$) = $100 \text{ N}$
Area of a circle ($A$) = $\pi r^2$

$$A_1 = \pi (2 \text{ cm})^2 = 4\pi \text{ cm}^2$$ $$A_2 = \pi (10 \text{ cm})^2 = 100\pi \text{ cm}^2$$

$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$ $$\frac{100 \text{ N}}{4\pi \text{ cm}^2} = \frac{F_2}{100\pi \text{ cm}^2}$$ $$F_2 = \frac{100 \text{ N} \times 100\pi \text{ cm}^2}{4\pi \text{ cm}^2}$$ $$F_2 = \frac{100 \text{ N} \times 100}{4}$$ $$F_2 = 100 \text{ N} \times 25$$ $$F_2 = 2500 \text{ N}$$

The force produced on the larger piston is $\boxed{2500 \text{ N}}$.

5. Sistem Rem Mobil (Car Brake System)

Given:
Area of main piston ($A_{pedal}$) = $10 \text{ cm}^2$
Force on pedal ($F_{pedal}$) = $500 \text{ N}$
Pressure ($P$) = $\frac{F}{A}$

$$P = \frac{500 \text{ N}}{10 \text{ cm}^2}$$ $$P = 50 \text{ N/cm}^2$$

The pressure generated in the brake fluid is $\boxed{50 \text{ N/cm}^2}$.

Given:
Total area of wheel pistons ($A_{total\_wheel}$) = $200 \text{ cm}^2$
Number of wheels = 4
The pressure is transmitted equally, so the pressure on the wheel pistons is also $50 \text{ N/cm}^2$.
Total force on wheel pistons ($F_{total\_wheel}$) = $P \times A_{total\_wheel}$

$$F_{total\_wheel} = 50 \text{ N/cm}^2 \times 200 \text{ cm}^2$$ $$F_{total\_wheel} = 10000 \text{ N}$$

Force on each wheel ($F_{each\_wheel}$) = $\frac{F_{total\_wheel}}{\text{Number of wheels}}$

$$F_{each\_wheel} = \frac{10000 \text{ N}}{4}$$ $$F_{each\_wheel} = 2500 \text{ N}$$

The force generated on each wheel is $\boxed{2500 \text{ N}}$.