Problem Involving Work Required for Stretching Spring

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A spring with a natural length of 20 cm requires a 25 N force to stretch to 30 cm, prompting a calculation of work needed to stretch it from 20 cm to 25 cm. The user initially attempts to find the spring constant (k) using the force and displacement but receives feedback to ensure all units are in SI. The correct approach involves integrating the force function, leading to the formula for work done as W = (1/2)kx^2. After recalculating with the correct limits and values, the user arrives at a work value of approximately 1.56 J. The discussion emphasizes the importance of unit consistency and proper application of work formulas in non-constant force scenarios.
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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
 
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No, its not right. You've almost hit it. You know F=kx 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 W=\int F(x)dx.
This gives you W=\frac{1}{2}kx^2 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
 
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