Find the distance that a force stretches a spring.

In summary, the conversation discusses a problem involving a mass hanging from a spring with a given force constant and unstretched length. The goal is to find the distance that the spring stretches when the mass is lowered until the system is in equilibrium. The conversation includes attempts at using different equations, but ultimately concludes that using Newton's 2nd law is the most accurate method. One person suggests using energy conservation, but another explains that this method only gives the lowest point of motion when the mass is released from rest. The correct equation to use is x = mg/k, where x is the displacement and k is the spring constant. One person makes a calculation error and realizes their mistake, leading to a discussion about units.
  • #1
student34
639
21

Homework Statement



A mass of 6.40 kg is hung from a spring with aforce constant of 7.80*10^3 N/m. The spring unstretched is 0.120m long. Find the distance that the spring stretches.

Homework Equations



F = k*x is the way that gives the answer in my textbook.

I tried W = 0.5*k*x^2.

The Attempt at a Solution



What is wrong with the following? I used W = 0.5*k*x^2 = 0.5*7800N/m*x^2 . Then made W = m*g*x. So, m*g*x = 0.5*7800N/m*x^2. Then I canceled out the x from both sides, and isolate the remaining x, so x = (m*g)/(0.5*78000N/m) = 0.136m. My answer is close to the correct answer of x = 0.128m, but I would really like to know what I don't understand which led me to the wrong answer.
 
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  • #2
Use Newton's 2nd law. Draw your FBD and correctly set up your terms for ƩF, knowing that the system is at static equilibirum. You'll be able to find the change in x of the spring from its unstretched length this way.

The x in your 'mgx' term (which in this case is also the gravitational potential energy) isn't the same as the x in your '.5mgx^2' term. Energy methods aren't always so convenient in these types of problems.
 
  • #3
hi student34! :smile:
student34 said:
A mass of 6.40 kg is hung from a spring with aforce constant of 7.80*10^3 N/m. The spring unstretched is 0.120m long. Find the distance that the spring stretches.

you're using conservation of energy, which (correctly) gives you the lowest point of the motion if the mass is released from rest in the unstretched position, and allowed to oscillate up and down

but the question is about carefully lowering the mass until the system is in equilibrium :wink:
 
  • #4
Thanks for both of these answers!
 
  • #5
If I know the spring constant, I can calculate the displacement given the mass,
mg=kx
x=mg/k

6400*.00980665=62.76256N
62.76256N/7800N=.008046482m.
This must be wrong if the textbook wants exactly .008m as displacement.
So I fail. A for effort.
 
  • #6
...and the units in the last expression all must be labeled N/m.
I am a disgrace to the profession.
 

What is Hooke's Law?

Hooke's Law is a principle in physics that states the force needed to stretch or compress a spring is proportional to the distance the spring is stretched or compressed. This relationship is expressed as F = -kx, where F is the force, k is the spring constant, and x is the distance the spring is stretched or compressed.

How do you calculate the distance a force stretches a spring?

The distance a force stretches a spring can be calculated using Hooke's Law and the equation x = F/k, where x is the distance, F is the force, and k is the spring constant. This equation assumes that the force is applied in a linear fashion and that the spring has not reached its elastic limit.

What is the spring constant?

The spring constant, denoted by k, is a measure of the stiffness of a spring. It is a constant value that relates the amount of force needed to stretch or compress a spring to the distance the spring is stretched or compressed. A higher spring constant indicates a stiffer spring, while a lower spring constant indicates a more flexible spring.

How does the distance a force stretches a spring relate to the potential energy stored in the spring?

The potential energy stored in a spring is directly related to the distance the spring is stretched or compressed. This relationship is expressed as U = 1/2kx^2, where U is the potential energy, k is the spring constant, and x is the distance the spring is stretched or compressed. As the distance increases, so does the potential energy stored in the spring.

What factors can affect the distance a force stretches a spring?

The distance a force stretches a spring can be affected by factors such as the magnitude of the force, the spring constant, the properties of the material the spring is made of, and the initial length of the spring. Additionally, external factors such as temperature and humidity can also affect the behavior of a spring.

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