Work done by spring vs hookes law?

AI Thread Summary
The discussion revolves around a physics problem involving a spring and a hydraulic lever, questioning the application of Hooke's Law versus the work-energy principle. The user expresses confusion over why Hooke's Law (F = kx) is used instead of the work formula (W = kx^2/2) for calculating the work done to compress the spring. They argue that the force exerted by the spring changes as it compresses, suggesting the need for integration to find the total work done. The user also highlights the importance of considering the movement of the beam and the relationship between the distances moved by the spring and beam. Ultimately, the conversation emphasizes the complexities of applying these principles in fluid mechanics and spring dynamics.
kahwawashay1
Messages
95
Reaction score
0
what is wrong with this problem?

http://www.cramster.com//answers-mar-10/physics/buoyancy-force-fig-14-36-spring-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=mgy1, where m is mass of sand
the two works are equal by Pascal's Law, so kx^2/2=mgy1

the volume of water displaced same at both pistons:
V=Aiyi
V=18Aix
Solving for yi and x and plugging into my work equations gives:

(k/2)((V/18Ai)^2)=mg(V/Ai)

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 achieve this compression is a variable force, so we must integrate or deal with the work...?
 
Last edited by a moderator:
Physics news on Phys.org
From what I can gather, if you use conservation of energy, then you'd have to take into account the fact that the beam might move up some distance y2 while the spring will compress a distance x.

Whereas if you were using Hooke's law, the pressure transmitted throughout the fluid is constant. So the force on the beam is the force that the spring is seeing.
 
rock.freak667 said:
From what I can gather, if you use conservation of energy, then you'd have to take into account the fact that the beam might move up some distance y2 while the spring will compress a distance x.

y2=x

The distance the spring compresses is obviously equal to the distance the beam moves up...so that doesn't solve it..
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top