Question

In a hydraulic lift, the surface area of the input piston is 6 cm² and that of the output piston is 1500 cm². If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is kJ.

Answer 1

Correct Answer: 5

To calculate the work performed, Pascal's principle is utilized. This principle dictates that pressure applied to a contained fluid is uniformly distributed throughout the fluid.

Provided data:

• Input piston surface area, A₁ = 6 cm²
• Output piston surface area, A₂ = 1500 cm²
• Force on input piston, F₁ = 100 N
• Vertical displacement of output piston, h = 20 cm = 0.2 m

The pressure at the input piston is computed as:

P = F₁ / A₁ = 100 N / 6 cm²

Conversion factor: 1 cm² = 0.0001 m².

A₁ = 6 × 0.0001 = 0.0006 m²

P = 100 N / 0.0006 m² = 166666.67 N/m²

This pressure is transmitted to the output piston, resulting in the following force:

F₂ = P × A₂ = 166666.67 N/m² × (1500 cm² × 0.0001 m²/cm²) = 25000 N

Work done, W, is determined by:

W = F₂ × h = 25000 N × 0.2 m = 5000 J

Converting the work done from joules to kilojoules:

W = 5 kJ

The total work performed is 5 kJ.