In a hydraulic lift, the surface area of the input piston is $$6 cm^{2}$$ and that of the output piston is $$1500 cm^{2}$$. If 100 N force is applied to the input piston to raise the output piston by 20 cm, then the work done is _______ kJ.

Given a hydraulic lift with input piston area $$A_{1}=6\;\text{cm}^{2}$$ and output piston area $$A_{2}=1500\;\text{cm}^{2}$$. A force $$F_{1}=100\;\text{N}$$ is applied on the input piston to raise the output piston by $$h_{2}=20\;\text{cm}$$. We need to find the work done by the input force in kJ.

According to Pascal’s law, the pressure transmitted is the same throughout the fluid: $$\frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\quad-(1)$$ From $$(1)$$, the output force is $$F_{2}=\frac{A_{2}}{A_{1}}\;F_{1}=\frac{1500}{6}\times100=25000\;\text{N}.$$

By conservation of fluid volume, the volume displaced by the input piston equals the volume risen by the output piston: $$A_{1}\,h_{1}=A_{2}\,h_{2}\quad-(2)$$ Substituting values in $$(2)$$: $$6\;\text{cm}^{2}\times h_{1}=1500\;\text{cm}^{2}\times20\;\text{cm}$$ $$h_{1}=\frac{1500\times20}{6}=5000\;\text{cm}=50\;\text{m}.$$

The work done by the input force is $$W=F_{1}\,h_{1}=100\;\text{N}\times50\;\text{m}=5000\;\text{J}=5\;\text{kJ}.$$

Thus, the required work done is $$5\;\text{kJ}$$.