The Change in Elastic Potential Energy with Different Spring Stretches

Click For Summary

Homework Help Overview

The discussion revolves around the change in elastic potential energy of a spring with a given spring constant when subjected to different stretches and compressions. The problem involves calculating the change in energy based on initial and final states of the spring's stretch.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to calculate the change in elastic potential energy using the formula provided, questioning the correctness of their results for different scenarios of stretching and compressing the spring.

Discussion Status

Some participants express uncertainty regarding the calculations, particularly for the compression scenarios, and question whether the potential energy should be considered negative when the spring is compressed. There is an ongoing exploration of the mathematical implications of squaring negative values.

Contextual Notes

Participants are working under the assumption that the initial potential energy is defined as zero for no stretch, and they are trying to reconcile their results with the expected outcomes based on the equations provided.

lion_
Messages
18
Reaction score
0

Homework Statement


A spring constant of ##3200 N/m## is initially streched until the elastic potential energy is ##1.44## J (##U=0## for no stretch). What is the change in elastic potential energy if the initial stretch is changed to (a) a stretch of ##2.0## cm, (b) a compression of ##2.0## cm, (c) a compression of ##4.0## cm.

Homework Equations


##\Delta U=U_f-U_i##
##U=\frac{1}{2}kx^2##

The Attempt at a Solution



(a) ##\Delta U=U_f-U_i=\frac{1}{2}(3200)(0.02)^2-1.44=-0.8 J##

(b) ##\Delta U=U_f-U_i=-\frac{1}{2}(3200)(0.02)^2-1.44=-2.08 J##

(c) ##\Delta U=U_f-U_i=-\frac{1}{2}(3200)(0.04)^2-1.44=-4J##

The first answer (a) is correct but the last 2 they get (b) ##-0.8J## and (c) ##1.1J## respectively. How?
 
Physics news on Phys.org
lion_ said:

Homework Statement


A spring constant of ##3200 N/m## is initially streched until the elastic potential energy is ##1.44## J (##U=0## for no stretch). What is the change in elastic potential energy if the initial stretch is changed to (a) a stretch of ##2.0## cm, (b) a compression of ##2.0## cm, (c) a compression of ##4.0## cm.


Homework Equations


##\Delta U=U_f-U_i##
##U=\frac{1}{2}kx^2##

The Attempt at a Solution



(a) ##\Delta U=U_f-U_i=\frac{1}{2}(3200)(0.02)^2-1.44=-0.8 J##

(b) ##\Delta U=U_f-U_i=-\frac{1}{2}(3200)(0.02)^2-1.44=-2.08 J##

(c) ##\Delta U=U_f-U_i=-\frac{1}{2}(3200)(0.04)^2-1.44=-4J##

The first answer (a) is correct but the last 2 they get (b) ##-0.8J## and (c) ##1.1J## respectively. How?
You should check your math...
 
PhanthomJay said:
You should check your math...
If the spring is compressed, isn't potential energy for the spring negative?
 
square a negative and you get a ?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
15
Views
2K
Replies
29
Views
3K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K