Question 216: Work, Energy and Power & Circular Motion

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.

Accepted Answer

$$\begin{aligned}
& \frac{\mathrm{F}_1}{\mathrm{~A}_1}=\frac{\mathrm{F}_2}{A_2}, \frac{100}{6}=\frac{\mathrm{F}}{1500}, \mathrm{~F}=\frac{50}{3} 	imes 1500 \
& \mathrm{~F}=50 	imes 500=25 	imes 10^3 \mathrm{~N} \
& \omega=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{~S}}=25 	imes 10^3 	imes \frac{20}{100} \
& =5 	imes 10^3=5 \mathrm{~kJ}
\end{aligned}$$

Work, Energy and Power & Circular Motion

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.

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