What is the stretch of the spring?

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SUMMARY

The discussion revolves around calculating the stretch of a spring when different masses are applied. Initially, a 1.2 kg mass causes a stretch of 6.3 cm, leading to the determination of the spring constant (k) as 5925.9 N/m. When a 1.4 kg mass is used, the calculated stretch is 6.8 cm, which is identified as incorrect. The participants clarify the relationship between spring force and potential energy, emphasizing that the spring force varies linearly with displacement while potential energy varies with the square of displacement.

PREREQUISITES
  • Understanding of Hooke's Law and spring constants
  • Knowledge of potential energy equations for springs
  • Familiarity with unit conversions (cm to m)
  • Basic physics concepts of force and mass (F=mg)
NEXT STEPS
  • Study the derivation and application of Hooke's Law in various contexts
  • Learn about energy conservation in spring systems
  • Explore the implications of oscillation on spring force calculations
  • Investigate the units and dimensions of physical quantities in mechanics
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Students studying physics, particularly those focusing on mechanics and spring dynamics, as well as educators looking for problem-solving strategies in spring-related scenarios.

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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