How Does Hooke's Law Determine Spring Constant and Extension?

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SUMMARY

This discussion focuses on the application of Hooke's Law to determine the spring constant and extension of a light spring. The spring, with an unstretched length of 33.0 cm, extends to 42.50 cm when a 5.80-kg mass is applied, resulting in a calculated spring constant of 598.3 N/m. In the second part, the incorrect assumption of summing two opposing forces of 170 N each leads to a miscalculation of the spring's extension, as the forces cancel each other out. The correct approach requires considering the weight of the object alone for determining the extension.

PREREQUISITES
  • Understanding of Hooke's Law and its formula, F = -kx
  • Basic knowledge of Newton's second law, F = ma
  • Familiarity with units of force (Newtons) and spring constant (N/m)
  • Concept of forces acting in opposite directions and their effects on equilibrium
NEXT STEPS
  • Explore the derivation and applications of Hooke's Law in different contexts
  • Learn about the relationship between mass, weight, and force in physics
  • Investigate the effects of multiple forces acting on a spring system
  • Study real-world applications of spring constants in engineering and design
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and elasticity, as well as educators looking to enhance their understanding of Hooke's Law and its practical implications.

mandy9008
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Homework Statement


Hooke's law describes a certain light spring of unstretched length 33.0 cm. When one end is attached to the top of a door frame, and a 5.80-kg object is hung from the other end, the length of the spring is 42.50 cm.
(a) Find its spring constant.
(b) The load and the spring are taken down. Two people pull in opposite directions on the ends of the spring, each with a force of 170 N. Find the length of the spring in this situation.


Homework Equations


F=ma
F=-kx


The Attempt at a Solution


a. F=ma
F=(5.8kg)(9.8m/s2)
F=56.84N

F=-kx
56.84N = -k (.425m - .330m)
k=598.3 N/m or 0.598 kN/m

b. F=-kx
170N = -598.3 N/m x
x= 284.3m
 
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I did the first part using work done by the mass and the spring and got a spring constant of 1196.63 N/m. For part b since each is pulling I believe you should sum both forces, good luck.
 
a. F=ma
F=(5.8kg)(9.8m/s2)
F=56.84N

F=-kx
56.84N = -k (.425m - .330m)
k=598.3 N/m or 0.598 kN/m

b. F=-kx
170N = -598.3 N/m x
x= 284.3m[/QUOTE]


I think part (a) is correct. Part (b) is in correct. Two applied forces are same magnitude and opposite directions. It will cancel each other. You need to consider weight of the object. So extension is previous extension.
 

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