Calculating work done on nonlinear spring

Click For Summary

Homework Help Overview

The discussion revolves around calculating the work done on a nonlinear spring described by the force law F(x) = -10x + 3x^2. The original poster seeks assistance in determining the work done when stretching the spring from its equilibrium position at x = 0.0 m to x = +2.0 m, noting a discrepancy with the expected answer of -12 Nm.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the work formula W = F*d and the necessity of using calculus to accurately calculate work done on a nonlinear spring. There are differing opinions on whether calculus is required for this problem, with some asserting that it is essential due to the nature of the force being variable.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing guidance on the need for calculus to find the correct work done. Multiple interpretations of the problem's requirements are being discussed, particularly regarding the use of constant versus variable force in the calculations.

Contextual Notes

Some participants express confusion over the expected answer and the methods used to arrive at it, indicating a potential misunderstanding of the force's behavior throughout the spring's extension. The discussion highlights the importance of accurately representing the force as a function of position.

kivarocket
Messages
5
Reaction score
0
Can anyone help me solve this problem? It seemed straightforward at first, but I am not getting the correct answer of -12 Nm. Thank you!

A nonlinear spring is modeled by a force law given by F(x) = -10x + 3x^2, where F is measured in Newtons and x in meters. How much work is done stretching the spring to x = +2.0 m from its equilibrium position at x = 0.0 m?


2. Attempt: W = F*d; F = -8N, so -8N * 2m = -16J
 
Physics news on Phys.org
If a force 'F' moves its point of application a distance 'dx', then the incremental work done is dW=F dx, so over an interval x, the entire work done is

∫ dW = ∫ F dx or W = ∫F dx.
 
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.
 
kivarocket said:
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.
Contrary to what you said, this problem does require calculus. If you follow rock.freak667's advice, you will obtain -12 Nm using calculus.

Chet
 
kivarocket said:
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.

If you draw the graph of F vs x, you will get a parabolic shaped curve. The work done will be the area under that curve from x=0 to x=2 and in order to get the exact area, integral calculus is required otherwise, anything you do will be an approximation and may be inaccurate.
 
kivarocket said:
This problem doesn't require calculus, so I don't need to take the integral. However, if I use the equation provided to solve for F, I get -8 N. Since W = F*d (where d = distance traveled), then -8N * 2m = -16 Nm. The correct answer is supposedly -12 Nm, so I'm not sure what I could be missing in this scenario.

You're using the formula for the work of a constant force. The force produced by that spring is not constant. You cannot simply use the force obtained at the very end after stretching the spring and use it as if that had been the force produced by the spring all along throughout the whole expansion process. This problem indeed does require calculus.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
29
Views
3K