Work done by spring vs hookes law?

Click For Summary
SUMMARY

The discussion revolves around the application of Hooke's Law in a physics problem involving a spring with a spring constant of 3.60x104 N/m, a hydraulic lever, and the compression of the spring due to added mass. The user questions the use of Hooke's Law (F = kx) versus the work-energy principle (W = kx2/2) in calculating the work done to compress the spring by 5.80 cm. The user correctly identifies that the force exerted by the spring is variable during compression and emphasizes the need for integration to accurately compute the work done.

PREREQUISITES
  • Understanding of Hooke's Law and its application in spring mechanics
  • Familiarity with Pascal's Law and hydraulic systems
  • Knowledge of work-energy principles in physics
  • Basic calculus for integrating variable forces
NEXT STEPS
  • Study the derivation and application of Hooke's Law in various scenarios
  • Learn about the principles of hydraulic systems and Pascal's Law
  • Explore the work-energy theorem and its implications in mechanical systems
  • Practice integration techniques for variable forces in physics problems
USEFUL FOR

Students and professionals in physics, mechanical engineering, and anyone interested in understanding the principles of spring mechanics and hydraulic systems.

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..
 

Similar threads

Replies
3
Views
2K
Replies
17
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
20
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K