What is the force of the spring constant?

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Homework Help Overview

The discussion revolves around the concept of spring constants and the application of Hooke's law in the context of a spring launching a block. The original poster presents a scenario involving a compressed spring and a block of known mass, seeking to understand the force associated with the spring constant and its implications in different setups.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the spring constant using Hooke's law and raises a follow-up question regarding the displacement of the spring when a different mass is suspended from it. Some participants question whether the calculations change with different mass values and explore the relationship between the velocity of the block and the spring's displacement.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem. There is a mix of attempts to clarify the application of Hooke's law and the conditions under which it is applied, particularly regarding vertical versus horizontal setups. Guidance has been offered to refine the problem description and clarify assumptions.

Contextual Notes

Participants are considering the implications of using Hooke's law in different contexts, such as vertical versus horizontal spring setups, and the assumptions related to equilibrium states. There is a noted need for more precise problem description to facilitate further discussion.

tgould43
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If a spring is originally compressed by 6cm, what is the force of the spring constant?
A block of inertia m=0.2kg is launched from a spring onto frictionless surface. The velocity is 2.5m/s upon leaving the spring.
I used Hooke's law and got
k=mg/x
0.2kg x 9.8m/s / 0.06m = 32.7 N/m
The next question is if this spring was used to suspend a 3kg block from ceiling what would the displacement of the spring be?
I need some help with the ratio of what is needed to solve this problem
 
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would the equation change by the new equation of
3.0kg x 9.8 ms / .06 = 490 N/m
 
My mind keeps thinking that there is some relationship between the displacement of the 2.5 m/s velocity of the first block after it left the spring. Is this something that is valid?
 
I moved your posts to a separate thread.
tgould43 said:
If a spring is originally compressed by 6cm, what is the force of the spring constant?
A block of inertia m=0.2kg is launched from a spring onto frictionless surface. The velocity is 2.5m/s upon leaving the spring.
You'll need to describe the problem more carefully. Is the mass launched horizontally?
I used Hooke's law and got
k=mg/x
0.2kg x 9.8m/s / 0.06m = 32.7 N/m
That would be true if the mass were placed on a vertical spring and allowed to come to equilibrium. Is that what's going on here?
 

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