Calculating Spring Constant from 52N Force and 0.73m Stretch

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Homework Help Overview

The discussion revolves around calculating the spring constant using a force of 52N and a stretch of 0.73m. The context involves applying Hooke's law, which relates the force exerted by a spring to its displacement from equilibrium.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between force and displacement in Hooke's law, questioning the signs associated with these quantities. There is a focus on the directionality of forces and displacements, with attempts to clarify the correct interpretation of the spring constant.

Discussion Status

Participants are actively discussing the signs of the spring constant and the forces involved. Some have provided guidance on the correct interpretation of Hooke's law, emphasizing the need to consider the directions of forces and displacements. There is an acknowledgment of differing views on the sign of the spring constant.

Contextual Notes

There is an ongoing discussion about the definitions of force in the context of Hooke's law, particularly distinguishing between the external force applied to the spring and the force exerted by the spring itself. Participants are also addressing the implications of vector directions in their calculations.

StephenDoty
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A force of 52N stretches a spring 0.73m from equalibrium. What is the value of the spring constant?

F=-k * delta s
52N= -k * 0.73m
52N/0.73N = -k
k= -71.2 N/m or -71 N/m with sig figs.
The negative sign means that the force is opposite of the stretch.

Is this right?
 
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Of course it's right.
 
The spring constant k will be a positive number.

The force F in Hooke's law is the force of the spring, not the external force that pulls on the spring. Also, F and (delta s) are vectors so we need to keep track of their directions.

So, just to pick a specific direction, let the spring be pulled to the right, and also choose rightwards to be the positive direction. Then (delta s)=+0.73m. But to get the sign of F we need to find which direction the spring is pulling.
 
So:
both the force and the displacement are in the same direction since the sping is being stretched, but the k constant is pulling in the opposite direction so k = 71 N/m pulling the spring in the opposite direction of the stretch.
 
Last edited:
If you pull a spring to the right, the spring pulls back against you to the left. So if delta s is positive, then F is negative.
 
ok..

-52N=-k*0.73m
-52N/0.73m = -k
-71 N/m = -k
71 N/m = k

Which is what I got before.
 
If my posts made it seem like I was saying the magnitude of your answer was wrong, I definitely did not intend that. I was just pointing out that the sign of your original answer was wrong, and that it might be coming about because maybe you were thinking that F was the external, applied force when it is really the force of the spring.
 

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