Leg Presses Work Energy Theorem Confusion

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SUMMARY

The discussion centers on the application of the Work Energy Theorem in a scenario involving two parallel springs during a leg press exercise. The user performs 80.0 J of work to compress the springs by 0.200 m. The confusion arises from the calculation of force and work done, where the user initially assumes the total work is distributed between the two springs, while the correct approach uses the effective spring constant for both springs combined. Both methods yield the same results for the required force and additional work needed.

PREREQUISITES
  • Understanding of the Work Energy Theorem
  • Knowledge of spring mechanics, specifically Hooke's Law
  • Familiarity with calculating work done in physics
  • Basic algebra for solving equations
NEXT STEPS
  • Review the principles of Hooke's Law and effective spring constants
  • Study the Work Energy Theorem in various mechanical systems
  • Practice problems involving multiple springs in parallel
  • Explore advanced applications of energy conservation in mechanical systems
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators seeking to clarify concepts related to work and energy in spring systems.

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Homework Statement



As part of your daily workout, you lie on
your back and push with your feet against a platform attached to
two stiff springs arranged side by side so that they are parallel to
each other. When you push the platform, you compress the springs.
You do 80.0 J of work when you compress the springs 0.200 m
from their uncompressed length. (a) What magnitude of force must
you apply to hold the platform in this position? (b) How much
additional work must you do to move the platform 0.200 m
farther, and what maximum force must you apply?

Homework Equations



Work Energy theorem ONLY!

The Attempt at a Solution


I assumed that the 80 J would be equal to the sum of work done to compress each spring... So 80 J = 2* (.5 K x^2) ... However, the book had an attempt directly the 80 J = .5 Kx^2... Kindly Clarify why is it so if the force is applied on two springs? THANKS A LOT!
 
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Perhaps they are letting K represent the effective spring constant of both springs together. Your approach is also fine. Both methods should give the same answers for (a) and (b).
 

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