Problem Involving Work Required for Stretching Spring

  • Thread starter Thread starter chocolatelover
  • Start date Start date
  • Tags Tags
    Spring Work
Click For Summary

Homework Help Overview

The problem involves calculating the work required to stretch a spring from its natural length to a specified length, focusing on the relationship between force, spring constant, and work done in the context of Hooke's Law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the spring constant using given forces and lengths, and question the limits for integration in the work calculation. There is also consideration of whether different formulas for work apply due to the non-constant nature of force.

Discussion Status

Some participants have provided guidance on finding the spring constant and applying the work formula, while others are exploring different interpretations of the limits for integration and the formulas to use. There is no explicit consensus on the correctness of the attempts presented.

Contextual Notes

Participants are reminded to ensure unit consistency and to convert measurements to SI units before calculations. The problem is framed within the constraints of a homework assignment, which may influence the approaches taken.

chocolatelover
Messages
238
Reaction score
0
Hi,

Could someone please help me with this problem?

Homework Statement


A spring has a natural length of 20 cm. If a 25 N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?


Homework Equations



w=lim a to b f(x)dx
work=(force)(distance)


The Attempt at a Solution



int. from 0 to 10 kxdx=25N
25=[1/2kx^2]0 to 10
25=1/2k(10)^2
25=50k
k=.5

int. from .05 to .10 (.5)xdx

Could someone please tell me if this is correct so far?

Thank you very much
 
Physics news on Phys.org
No, its not right. You've almost hit it. You know [tex]F=kx[/tex] Now, you know 25N force is required to hold it at 30 cm. Plug these values in and find k from here. Be careful about the unit and convert everything to SI before plugging the values in.

The second part asks you to find the work done, you know [tex]W=\int F(x)dx[/tex].
This gives you [tex]W=\frac{1}{2}kx^2[/tex] Apply the limits as you did, and use the value of k found above, and you have your answer.
 
Thank you very much

Would the limit be from 0 to 10?

Also, I was thinking that maybe I should have used the formulas f/d=k and w=1/2kd^2

Would that also work, since the force isn't constant?

Thank you very much
 
Does this look right?

.30-.20=.10m
f(.10)=25
k=250m

.25-.10=.15
.20-.10=.10

w=int. .10 to .15(250xdx)
=1.56J

Thank you very much
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
1
Views
3K
Replies
2
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
15K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K