Calculating work using hookes law

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SUMMARY

The discussion focuses on calculating the work done on a muscle modeled as a spring, using Hooke's Law. The muscle has a cross-sectional area of 1 cm² and an initial length of 10 cm, which is stretched to 11 cm by a hanging mass. The spring constant (k) is determined to be 1000 N/m, calculated from the force exerted (5.00 N) over the stretch (0.005 m). The integral setup for calculating work is correctly initiated, with the expression |work| = ∫ (from 0.10 m to 0.11 m) -1000 (x - x0) dx.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Basic calculus, specifically integration techniques
  • Knowledge of units in physics, particularly in the context of force and work
  • Familiarity with the concept of work done in mechanical systems
NEXT STEPS
  • Review the derivation of Hooke's Law and its applications in biomechanics
  • Practice integration techniques for calculating work done in variable force scenarios
  • Explore the relationship between force, displacement, and energy in elastic materials
  • Investigate the implications of muscle elasticity in physiological contexts
USEFUL FOR

This discussion is beneficial for physics students, biomechanics researchers, and anyone interested in the mechanical properties of biological tissues and their applications in health sciences.

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Homework Statement


Calculate the work done, in Joules for a system in which a muscle of 1cm^2 cross section and 10cm length is stretched to 11cm by hanging a mass on it. The muscle behaves like a spring. The spring constant for the muscle was determined by finding that the muscle exerts a force of 5.00N when it is stretched from 10.0cm to 10.5cm



Homework Equations


Is spring constant (-k) calculated simply by dividing .005m into 5.00N? (because F= -kx, and -k = F/x) = -1000.


The Attempt at a Solution



i don't think my answer will be correct because i have a feeling that I am calculating k incorrectly??
OK so assuming I am doing it right though...

i set up the integral using hookes law:

|work| = int (from .10m to .11m) -k (x - x0)dx

= int (from .10m to .11m) -1000 (x - x0)dx
then i took constants out of integral and integrated...

|work| = -1000/2 * |from .10m to .11m [ (xf-x0)^2 - (xi - x0)^2]

i don't want to proceed further unless i am on right track (which i doubt) thanks for any help!


 
Physics news on Phys.org
Right. k=1000*kg/s^2. Put units on the these numbers, ok? They aren't dimensionless. Now continue with the second part.
 

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