Spring Constants and compression

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Homework Help Overview

The problem involves a mass hanging from a spring scale that is lowered onto a vertical spring. It explores the relationship between the forces acting on the mass and the spring constant of the vertical spring.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the readings on the spring scale and how they relate to the forces acting on the mass and the vertical spring. There is uncertainty about how to incorporate the weight of the mass into the calculations for the spring constant and the conditions for the scale to read zero.

Discussion Status

Some participants have offered insights into the need to consider the weight of the mass when calculating the forces involved. There is ongoing exploration of how to correctly apply the equations of motion and the forces acting on the system.

Contextual Notes

Participants are grappling with the implications of the spring scale reading and the forces at play, including the weight of the mass and the forces exerted by the springs. There is a hint that multiple forces need to be accounted for in the calculations.

anotherperson
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Homework Statement


A 5.0kg mass is hanging from a spring scale and is slowly lowered onto a vertical spring. The scale reads in Newtons.

a) what does the scale read when the mass is not in contact with the spring?
b)the scale reads 20N when the lower spring is compressed by 2.0cm. show that the value of the spring constant of the lower spring is 1450 N/m
c) by how much must the lower spring be compressed for the scaled to read 0.0N


Homework Equations


Fnet=ma
Fsp=-kdeltax


The Attempt at a Solution



a)F=5X9.8
=49N

b) 20=-kX2.0

but i get the wrong answer??

c) is i do
0=1450x
then x=0
?
 
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Your answer to a) looks good.

Your answer to be would be correct if the spring were weightless, but you have calculated that it pulls down with a force of 49 N. How can you take that into account in b)?

For c), you must again allow for the 49 N that the scale reads when no weight is on it. I think you see you could make the scale read less if you push up of the spring. How hard would you ahve to push for it to read 0 Newtons?
 
for part b, I am still unsure how you factor in the 49N, i have a feeling you would add it but into what rule would you use these numbers, the rule i have used above?

so for part c you do 49N divided by the spring constant of 1450 and you get an answer of 3.38 cm which matches my answers, thanks!
 
anotherperson said:
b) 20=-kX2.0

but i get the wrong answer??
20 N is the reading on the spring scale, not the force compressing the vertical spring. If the spring scale reads 20 N, how much force must the vertical spring exert on the mass in order to support its weight?

(Hint: There are three forces acting on the mass. What are they? What must the net force be?)
 

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