Spring Constant (this is really easy, I'm just dumb haha)

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SUMMARY

The discussion centers on calculating the spring constant using Hooke's Law, represented by the formula F = kx. A spring measuring 150 cm extends to 155 cm under a load of 55 N, leading to the calculation of the spring constant. The extension, x, is determined as 0.05 m (5 cm), resulting in the equation 55 = 0.05k. The correct spring constant, k, is derived as 1100 N/m.

PREREQUISITES
  • Understanding of Hooke's Law and its formula F = kx
  • Basic knowledge of units of measurement (meters, newtons)
  • Ability to perform algebraic manipulations
  • Familiarity with the concept of spring extension
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  • Study the implications of different spring constants in mechanical systems
  • Explore the relationship between force, mass, and acceleration in spring dynamics
  • Learn about energy stored in springs and the potential energy formula
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pippintook
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A spring 150-cm long extends to a length of 155 cm when it supports a load of 55 N. Determine the spring constant.

F = kx

x1= 1.5 m
x2= 1.55 m
F = 55 N
k = ?

Now, "x" in F = kx is x2-x1, right? So the formula would be 55 = .05k?

Sorry, I know this is simple, I just am having a momentary lapse.
 
Last edited:
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pippintook said:
Now, "x" in F = kx is x2-x1, right? So the formula would be 55 = .05k?
Perfectly correct. :smile:
 
Awesome. That's what I thought, but when I was experimenting with springs the constant was always a lot smaller so I got confused :P
 

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