The drawing shows a hydraulic chamber with a spring (spring constant: 1600 N/m) attached to the input piston and a rock of mass 40.0 kg resting on the output plunger.The piston and plunger are nearly at the same height, and each has a negligible mass. By how much is the spring compressed (in cm) from its unstrained position?

Let F1 and F2 are the magnitudes of the force the spring exerts on the piston and the rock exerts on the plunger respectively. Initially the piston and the plunger are at the same height, so the fluid pressure at the piston is equal to the fluid pressure at the plunger F1A1=F2A2 where A1 is the area of the piston and  A2 is the area of the plunger. Therefore, the magnitude of the force F2 that the rock exerts on the plunger is given by F2=F1A2A1Here F1= kx where k is the spring constant of the spring, and x is the amount by which the spring is compressed from its unstrained position. and the magnitude F2 of the force the rock exerts on the plunger is equal to the magnitude W = mg where m is the rock's mass and g is the magnitude of the acceleration due to gravity. Solving F2=F1A2A1 for F1 and subśtituting F2=mg yields F1=F2A1A2=mgA1A2......(1) Solving for x, we obtain, x=F1k.....(2)Substituting Eq. (1) into Eq. (2), we find thatx=F1k=mgkA1A2=(40kg)10m/s21600N/m15cm275cm2=5.0×10−2m