Help Me Spring Constant & Potential Energy

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Homework Help Overview

The discussion revolves around a problem involving a spring with a known spring constant and the potential energy stored in it. The original poster seeks assistance in determining how far the spring must be compressed to store a specific amount of potential energy.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the equation for potential energy in a spring and the values provided in the problem. There are questions about how to set up the equation and solve for the compression distance.

Discussion Status

Some participants have offered guidance on the setup of the equation and the relationship between the units involved. There is an ongoing exploration of the calculations, with some expressing confusion about the steps and seeking further clarification.

Contextual Notes

There is a noted uncertainty regarding the interpretation of units, specifically the distinction between spring constant and energy units. Participants are also navigating the expectations of the homework assignment and expressing a desire for more detailed guidance.

davidliebehar
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Here is a very basic question that I need help with.

Homework Statement


A spring has a spring constant (k) of 82 N/m. How far must the spring be compressed to store 35J of potential energy?


Homework Equations


I think it is: E = 1/2 kx^2

The Attempt at a Solution


I don't understand how to setup this problem. Please help me.
 
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davidliebehar said:
Here is a very basic question that I need help with.

Homework Statement


A spring has a spring constant (k) of 82 N/m. How far must the spring be compressed to store 35J of potential energy?


Homework Equations


I think it is: E = 1/2 kx^2

The Attempt at a Solution


I don't understand how to setup this problem. Please help me.

Welcome to PF.

Let's see. You have the equation. They tell you what E is. They tell you what k is ... so you're having trouble solving for x?
 
yes.
should the equation look like this?
35J = 1/2 (82 N/m) x^2
 
davidliebehar said:
yes.
should the equation look like this?
35J = 1/2 (82 N/m) x^2

Just solve for X.
 
I don't know how to.
is 82 N/m the same as 82 J?would I get

35J divided by 41 N/m = x^2? then x = square root of 35J divided by 41 N/m?
 
davidliebehar said:
I don't know how to.
is 82 N/m the same as 82 J?

would I get

35J divided by 41 N/m = x^2? then x = square root of 35J divided by 41 N/m?

No N/m is a spring constant. It moves 1 meter for every 82 Newtons of force.

Joules are units of what? N-m? and you are dividing by N/m? that yields m2 and x2 is m2 ... so everything is right.

As to your calculation ... that's right.
 
i don't know whAt to do
plaese show me exactly how to solve this problem and all the steps i am so lost
 
davidliebehar said:
35J divided by 41 N/m = x^2? then x = square root of 35J divided by 41 N/m?

Look. You've already solved it ... that is if you meant to divide 35 by 41 before you took the square root.
 
i just want an answer in meters.
 
  • #10
davidliebehar said:
i just want an answer in meters.

That gives you an answer in meters.
 
  • #11
(sqaure root of 35j divided by 41 N/m) meters?
 
  • #12
davidliebehar said:
(sqaure root of 35j divided by 41 N/m) meters?

sqaure root of (35j divided by 41 N/m) meters?

Yes. I already showed you that you get m2 before you take the square root.
 
  • #13
thanks for your help, but i still can't understand this, i'll have to ask my instructor.
 

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