What is the stretch of the spring?

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Homework Help Overview

The problem involves a spring scale and its behavior under different weights, specifically examining the stretch of the spring when different masses are applied. The subject area includes concepts from mechanics, particularly related to forces, spring constants, and energy stored in springs.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to determine the spring constant using the force exerted by a mass and its corresponding displacement. Some participants question the validity of equating potential energy and force, while others suggest reconsidering the equations used for energy and force in the context of spring mechanics.

Discussion Status

The discussion is ongoing, with participants exploring the relationships between force, displacement, and energy in springs. There is a focus on clarifying the correct equations to use and the implications of oscillation on the forces involved. Some guidance has been offered regarding the units of energy and force, but no consensus has been reached.

Contextual Notes

Participants are navigating potential misunderstandings regarding the application of energy equations in dynamic situations, as well as the implications of oscillation on force balance. There is also a discussion about the units of various quantities involved in the problem.

KAC
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Homework Statement


A spring scale hung from the ceiling stretches by 6.3 cm when a 1.2 kg mass is hung from it. The 1.2 kg mass is removed and replaced with a 1.4 kg mass. What is the stretch of the spring?

F=mg
U (spring force) = 1/2kx^2

Homework Equations


The units of cm need to be converted into m, so 6.3cm becomes 0.063m. Units of kg are kept the same.
To solve the problem we need to find k, the spring constant.

The Attempt at a Solution


First, I'm going to use the first situation with the displacement and mass given to determine the spring constant.

F = mg; F = (1.2kg)(9.8m/s^2); F = 11.76N
F = U;
U = 1/2kx^2
F = 1/2kx^2
k = 2F/x^2
k = 2(11.76N)/0.063m^2
k = 5925.9

From there I plugged the spring constant into the second situation.

F = mg; F = (1.4kg)(9.8m/s^2) = 13.72N
U = 1/2kx^2
x = sqrt of 2*F/k
x = sqrt 2(13.72N)/5925.9
x = 0.068 m or 6.8 cm.

This final answer for displacement is incorrect and I am unsure of my mistake. Did I make a calculation error or set the problem up incorrectly?

Thank you in advance for your help!
 
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Where did you get that U = F from?

Chet
 
Chestermiller said:
Where did you get that U = F from?

Chet

Chet,

I used U = F as the force; the equation for U that I used is the potential energy of the spring. Since it is being pulled down and there is no normal force in this case, only the weight force is acting on the object which is = mg. (mass*gravity).
 
KAC said:
Chet,

I used U = F as the force; the equation for U that I used is the potential energy of the spring. Since it is being pulled down and there is no normal force in this case, only the weight force is acting on the object which is = mg. (mass*gravity).
Spring force varies linearly with displacement. Spring energy varies as the square of the displacement. Did you choose the wrong equation?
 
jbriggs444 said:
Spring force varies linearly with displacement. Spring energy varies as the square of the displacement. Did you choose the wrong equation?

Two other formulas I may be able to use are Force of the spring = -kx (where x is displacement) and L(change in length) = mg/k. But both of these equations have two variables I do not know in them. When you say that spring energy varies as the square of the displacement, this leads me to believe that U = 1/2kx2 is the correct equation since it is varying as the square of whatever the displacement is. But since the only other force acting on the mass is it's weight force since it is oscillating vertically, I am not quite understanding why it is incorrect to use mg = 1/2kx2. Is it incorrect because the weight is moving constantly (oscillating) and isn't in a set position? So therefore the weight force does not always equal the spring force?
 
KAC said:
Two other formulas I may be able to use are Force of the spring = -kx (where x is displacement) and L(change in length) = mg/k. But both of these equations have two variables I do not know in them. When you say that spring energy varies as the square of the displacement, this leads me to believe that U = 1/2kx2 is the correct equation since it is varying as the square of whatever the displacement is. But since the only other force acting on the mass is it's weight force since it is oscillating vertically, I am not quite understanding why it is incorrect to use mg = 1/2kx2. Is it incorrect because the weight is moving constantly (oscillating) and isn't in a set position? So therefore the weight force does not always equal the spring force?
What are the units of the spring constant? What are the units of mg? What are the units of 1/2 kx2?
 
jbriggs444 said:
What are the units of the spring constant? What are the units of mg? What are the units of 1/2 kx2?
The spring constant, k, has units of N/m; mg = F which is N (N = kg*m/s2), and U = N/m * m = N as well since it is the specific force in this case. So setting F = U in this case should be acceptable since the units are the same and U is the specific force that is working on the object, yes?
 
What would you say if I told you that the units of U are Nm? All energy has units of Nm=Joules. You must have learned this previously, correct?

Chet
 

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