(adsbygoogle = window.adsbygoogle || []).push({}); what is wrong with this problem?

http://www.cramster.com//answers-ma...pring-spring-constant-360x104-mis_811494.aspx

In Fig. 14-36, a spring of spring constant 3.60x104 N/mis between a rigid beam and the output piston of a hydraulic lever.An empty container with negligible mass sits on the input piston.The input piston has area Ai, and the outputpiston has area 15.5Ai. Initially the spring isat its rest length. How many kilograms of sand must be (slowly)poured into the container to compress the spring by 5.80 cm?

See above link if you want to see image.

Basically I solved this same as was solved in the above link, but I do not understand why Hooke's law is used instead of the kx^2/2. What I did was:

work to compress spring=kx^2/2

work applied to piston 1=mgy_{1}, where m is mass of sand

the two works are equal by Pascal's Law, so kx^2/2=mgy_{1}

the volume of water displaced same at both pistons:

V=A_{i}y_{i}

V=18A_{i}x

Solving for y_{i}and x and plugging into my work equations gives:

(k/2)((V/18A_{i})^2)=mg(V/A_{i})

So the V does not cancel out. But I cannot see how Hooke's Law can be used. Yes, the spring will feel force of kx once it is compressed 5cm, but the work to acheive this compression is a variable force, so we must integrate or deal with the work...?

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# Work done by spring vs hookes law?

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