What is the stretch of the spring?

In summary: 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?
  • #1
KAC
16
0

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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  • #2
Where did you get that U = F from?

Chet
 
  • #3
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).
 
  • #4
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?
 
  • #5
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?
 
  • #6
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?
 
  • #7
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?
 
  • #8
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
 

1. What is the definition of "stretch" in relation to a spring?

The stretch of a spring refers to the change in length of the spring when a force is applied to it. This change in length is typically measured in units of distance, such as meters or centimeters.

2. How is the stretch of a spring calculated?

The stretch of a spring can be calculated using Hooke's Law, which states that the force applied to a spring is directly proportional to the amount of stretch it undergoes. The formula is F=kx, where F is the force applied, k is the spring constant, and x is the amount of stretch.

3. What factors affect the stretch of a spring?

The stretch of a spring is affected by several factors, including the material and thickness of the spring, the amount of force applied, and the length of the spring. The spring constant, which is unique to each spring, also plays a role in determining the amount of stretch.

4. How does the stretch of a spring change with different forces?

The stretch of a spring is directly proportional to the force applied to it. This means that as the force increases, the amount of stretch also increases. However, once the elastic limit of the spring is reached, the stretch will no longer be directly proportional to the force and the spring may permanently deform.

5. Can the stretch of a spring be reversed?

Yes, the stretch of a spring can be reversed by removing the force that was causing it to stretch. This is because springs exhibit elastic behavior, meaning they return to their original shape and length once the force is removed. However, if the spring is stretched beyond its elastic limit, it may not return to its original shape and length.

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