Finding a spring constant

  • #1
Finding the spring constant of a spring...

Homework Statement


A mass is hung from a vertical spring. The mass is .200kg, and the spring stretches .086m. Find the spring constant.


Homework Equations


ΔE=ΔKE+ΔUg+ΔUs



The Attempt at a Solution


I first broke up the equation, to see which values would go to zero. I know ΔKE drops out, because the initial and final velocities would be zero. Eventually, I was left with this...
mgΔy=(1/2)K(x^2)

Then, I solved for K and got...
K=(2mgΔy)/(x^2)

Putting the numbers in, I get...
K=(2*.200kg*9.80m/s^2*.086m)/(.086m^2)

The final answer I got was approximately 45.6 N/M. I just want to make sure this is correct, along with one more quick thing. The next part of the problem asks, "If the mass is now lifted .086m back to the spring's unstretched position and released from rest, how far will it drop before recoiling back?"

I'm assuming that if this is a closed system, wouldn't the mass just drop down the same distance over and over again? (oscillate the same?) Or do I need to calculate the new distance?
 
Last edited:

Answers and Replies

  • #3
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
Please do not bumb.

Did you consider using Hook's law, F=-kx ?
 
  • #4
Please do not bumb.

Did you consider using Hook's law, F=-kx ?
No, didn't think of it at the time. Could I find the force with my given information, then use that to find K? Does this mean my answer is incorrect?
 
  • #5
1,039
2
When the mass is at equilibrium you know that one force is mg right, then apply newton's third law et al. That might help you find the force you need for the equation.
 
  • #6
Integral
Staff Emeritus
Science Advisor
Gold Member
7,201
56
You are given the mass, so you can compute the gravitational force.

Your result seems to big, done carefully your energy approach should work. I do not understand where your equations come from but you have lost something along the way. It would help if you gave more information. We need a definition of your variables... what is Δy?

Also a bit more history on how you arrived at your starting expression would be nice.
 
  • #7
You are given the mass, so you can compute the gravitational force.

Your result seems to big, done carefully your energy approach should work. I do not understand where your equations come from but you have lost something along the way. It would help if you gave more information. We need a definition of your variables... what is Δy?

Also a bit more history on how you arrived at your starting expression would be nice.
You have a point. I could have explained better. Okay, ΔY is change in height, which in this case, I said was the .086m the spring stretches. The x is the spring compression/stretch.

As for the equations. ΔUg=mgh(final) - mgh(initial). The mgh(final), I said mgh(final) cancels because it's final height is zero. ΔUs= (1/2)m*x^2 (Final) - (1/2)m*x^2 (initial). I said (1/2)m*x^2 (Initial) cancels, because initially, the spring is relaxed.

But okay, I think I see what I could of done. I should of used Hooke's Law. But I'm not exactly sure how to find the F, so I can solve for the K.
 
  • #8
1,039
2
But okay, I think I see what I could of done. I should of used Hooke's Law. But I'm not exactly sure how to find the F, so I can solve for the K.
Do a free body diagram of the mass on the end of the spring. It has a force mg pulling it down, what is stopping it from falling to the floor?
 
  • #9
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,165
507
Do not confuse the two parts of the problem.

In the first part, the mass is just hanging at rest in its equilibrium position. Identify the force the spring exerts on the mass, using Newton's First Law, and calculate the spring constant from Hooke's law.

In the 2nd part, when the mass is released from rest from the spring's unstretched position, use your energy method to calculate the max stretch of the spring.
 

Related Threads on Finding a spring constant

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
4
Views
5K
Replies
2
Views
3K
  • Last Post
Replies
22
Views
3K
  • Last Post
Replies
11
Views
10K
Replies
1
Views
468
  • Last Post
Replies
3
Views
2K
Replies
3
Views
3K
  • Last Post
Replies
3
Views
7K
Replies
2
Views
12K
Top