Problem Involving Work Required for Stretching Spring

  • Thread starter chocolatelover
  • Start date
  • Tags
    Spring Work
In summary, the conversation is about a physics problem involving a spring with a natural length of 20 cm. The problem asks how much work is required to stretch the spring from 20 cm to 25 cm. The solution involves using the equation W=\frac{1}{2}kx^2, where k is the spring constant, and finding k by plugging in values for force and distance. The final answer is 1.56J.
  • #1
chocolatelover
239
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
  • #2
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.
 
  • #3
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
 
  • #4
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
 

1. What is the definition of "work required for stretching spring?"

The work required for stretching a spring is the amount of energy needed to extend the spring from its original length to a new length. This work is typically measured in units of joules (J).

2. How is the work required for stretching a spring calculated?

The work required for stretching a spring can be calculated using the formula W = 1/2 * k * (x^2 - x0^2), where W is the work, k is the spring constant, x is the final length of the spring, and x0 is the original length of the spring.

3. What factors affect the amount of work required for stretching a spring?

The amount of work required for stretching a spring is affected by the spring constant, the displacement of the spring, and the initial length of the spring. A higher spring constant or larger displacement will result in more work being required.

4. How does the work required for stretching a spring vary with different materials?

The work required for stretching a spring can vary depending on the material of the spring. Different materials have different spring constants, which will affect the amount of work required. For example, a steel spring will require more work to stretch than a rubber spring due to its higher spring constant.

5. Can the work required for stretching a spring ever be negative?

No, the work required for stretching a spring can never be negative. This is because work is defined as the transfer of energy, and stretching a spring always requires an input of energy. Therefore, the work required for stretching a spring will always be a positive value.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
944
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
449
  • Introductory Physics Homework Help
Replies
3
Views
715
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
3K
Back
Top