In a hydraulic lift, the surface area of the input piston is 6 cm^2 and that of the output piston is 1500 cm^2.

Question

In a hydraulic lift, the surface area of the input piston is \(6 \mathrm{~cm}^{2}\) and that of the output piston is \(1500 \mathrm{~cm}^{2}.\) 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

Answer is: 5

\(\frac{\mathrm{F}_{1}}{\mathrm{~A}_{1}}=\frac{\mathrm{F}_{2}}{\mathrm{~A}_{2}}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500\)

\(\mathrm{F}=50 \times 500=25 \times 10^{3} \mathrm{~N}\)

\(\omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{S}}=25 \times 10^{3} \times \frac{20}{100}\)

\(=5 \times 10^{3}=5 \mathrm{~kJ}\)