Calculating Spring Constant: Why 1.5 cm Isn't the Correct Measurement

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Homework Help Overview

The discussion revolves around calculating the spring constant of a spring scale based on the force applied and the resulting stretch. Participants are examining the implications of their calculations and questioning the accuracy of a specific measurement related to the scale's markings.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between force, spring constant, and displacement, questioning the correctness of the calculated spring constant and the spacing of scale marks. There is a discussion on unit consistency and the implications of using different values for the spring constant.

Discussion Status

Some participants have offered insights into the calculations and unit considerations, while others have acknowledged misunderstandings in their initial reasoning. Multiple interpretations of the problem are being explored, particularly regarding the correct distance between scale marks.

Contextual Notes

There is an emphasis on ensuring unit consistency in calculations, and participants are reflecting on their assumptions regarding the spring constant and its application to the problem at hand.

RedBurns
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A spring scale is stretched 10 cm when a force of 15 N is applied to it. How far apart should adjacent 1.0 N marks be on the scale?

Why would 1.5 cm not be correct?

Force= Spring Constant* Change in spring from rest
15n=k*10 cm
k=1.5
 
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Wouldn't it be 0.66 cm? If there are 15 marks within 10 centimeters, there should be a mark every 2/3 of a centimeter. I may be reading this question wrong, though.
 
RedBurns said:
A spring scale is stretched 10 cm when a force of 15 N is applied to it. How far apart should adjacent 1.0 N marks be on the scale?

Why would 1.5 cm not be correct?

Force= Spring Constant* Change in spring from rest
15n=k*10 cm
k=1.5

For starters, watch the units! You are dividing N by cm! NOT a good habit to get into. The good news is, since you are using this k to calculate a distance in cm you can get away with it.

Otherwise:
F = kx
15 N = k(10 cm)
k = 15 N/10 cm = 1.5 N/cm.

Now, set F = 1 N. (Since F is linear in x, this will be the separation between 1 N marks.)

F = kx
1 N = (15 N/cm)x
x =1 N/(15 N/cm) = 0.66667 cm.

My tip to all my students, anyways, is to ALWAYS write out your work, even if it seems easy enough to do in your head. (Perhaps especially when!)

-Dan
 
your right. I had everything flipped in my head
 

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