Hydraulic Jack Question: Calculating Piston Movement for 2000 kg Car

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To calculate the movement of the small piston when a 2000 kg car is raised 1.5 meters by a hydraulic jack, the principle of work must be applied. The work done on the large piston equals the work done on the small piston, which can also be understood through the relationship of pressure and volume. Given the diameters of the pistons, the small piston will move a distance of 0.25 meters when the large piston moves 1.5 meters. This calculation is derived from the pressure relationship and the volume displacement between the two pistons. Understanding these hydraulic principles is essential for accurate calculations in hydraulic systems.
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A 2000 kg car is raised 1.5 meters high on a hydraulic jack, which has a large piston with a diameter of 30cm. The small piston has a diameter of 5cm (btw the reservoir is filled with oil)

d) When the car is raised by 1.5 m, what distance will the small piston move?

Is it simply that you have to take off 1.5 m on the side of the small piston in order for the big piston to raise by 1.5 m? Or is there something more complicated about this?
 
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You need to apply the idea of work. The work on one side will be the same as the other piston. Calculate the work for the large side and then use that amount for what the small piston must do.
 
Alternatively, you can remove Volume fom one side,
to put it on the other side.
(you get the same answer as using Work ...
W = F dh = P A dh = P dV)
 
is d = 0.25 m ?

I got this from the assumption 1/6th the work done at the small piston (from the pressure relationship).
 
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