How Is Spring Stiffness Calculated in Physics Problems?

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    Spring Stiffness
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Spring stiffness, or the spring constant (K), is calculated using Hooke's law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position. In the given problem, a 4 kg rock is attached to a spring and moves in a circular path with a constant force of 720 N. The speed of the rock was calculated to be 32.8 m/s using the formula F=mv^2/r. To find the stiffness of the spring, the change in length (x) must be determined, which is the difference between the relaxed length of the spring (5.7 m) and the stretched length (6 m). The discussion highlights the need for clarity on how to apply these concepts to solve for spring stiffness effectively.
ohheytai
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what is the spring stiffness?!??!

In outer space a rock of mass 4 kg is attached to a long spring and swung at constant speed in a circle of radius 6 m. The spring exerts a force of constant magnitude 720 N. What is the speed of the rock?

i found the speed it is 32.8 m/s

part 2:
The relaxed length of the spring is 5.7 m. What is the stiffness of this spring?

thats what i have no idea about could someone help lead me in the direction of solving it or show me how? please and thanks!



Homework Equations


i used F=mv^2/r to get speed
 
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Hooke's law:
F=-Kx

x is the change in length of the spring due to a force F.
K is the the spring constant or "stiffness"
 


thanks!
 


what are you talking about it changes from 6m to 5.7m, there is no material in the world with no elasticity.
 
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