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.
Correct Answer: 5
Approach Solution - 1
To determine the work done, we start with Pascal's principle, which states that the pressure applied to a confined fluid is transmitted undiminished throughout the fluid.
Given:
- Surface area of input piston, A1 = 6 cm²
- Surface area of output piston, A2 = 1500 cm²
- Force on input piston, F1 = 100 N
- Output piston raised by h = 20 cm = 0.2 m
The pressure applied at the input piston is:
P = F1 / A1 = 100 N / 6 cm²
Since 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:
F2 = P × A2 = 166666.67 N/m² × (1500 cm² × 0.0001 m²/cm²) = 25000 N
The work done, W, is given by:
W = F2 × h = 25000 N × 0.2 m = 5000 J
Converting 5000 J to kJ:
W = 5 kJ
The work done is 5 kJ.
Approach Solution -2
Area of input piston, \( A_1 = 6 \, \text{cm}^2 \)
Area of output piston, \( A_2 = 1500 \, \text{cm}^2 \)
Force applied on input piston, \( F_1 = 100 \, \text{N} \)
Displacement of output piston, \( h_2 = 20 \, \text{cm} = 0.2 \, \text{m} \)
Step 2: Apply Pascal’s law.
According to Pascal’s law, the pressure is transmitted equally in all directions in an enclosed fluid. Therefore,
\[ \frac{F_1}{A_1} = \frac{F_2}{A_2} \] Hence, the force on the output piston is:
\[ F_2 = F_1 \times \frac{A_2}{A_1} \] Substitute the given values:
\[ F_2 = 100 \times \frac{1500}{6} = 100 \times 250 = 25000 \, \text{N} \]
Step 3: Relation between displacements.
Since the volume of the fluid displaced remains the same,
\[ A_1 h_1 = A_2 h_2 \] Therefore, \[ h_1 = \frac{A_2 h_2}{A_1} = \frac{1500 \times 0.2}{6} = 50 \, \text{m} \]
Step 4: Calculate the work done.
Work done on the input side = \( F_1 \times h_1 \)
\[ W = 100 \times 50 = 5000 \, \text{J} = 5 \, \text{kJ} \]
Final Answer:
\[ \boxed{5 \, \text{kJ}} \]
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