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.

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{F}_1}{\mathrm{~A}_1}=\frac{\mathrm{F}_2}{\mathrm{~A}_2}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} \times 1500 \\ & \mathrm{~F}=50 \times 500=25 \times 10^3 \mathrm{~N} \\ & \omega=\overrightarrow{\mathrm{F}}. \overrightarrow{\mathrm{S}}=25 \times 10^3 \times \frac{20}{100} \\ & =5 \times 10^3=5 \mathrm{~kJ}\end{aligned}$

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

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