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Work and force

  1. Nov 17, 2007 #1
    Hi,

    Could someone please help me with this problem?

    1. The problem statement, all variables and given/known data
    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?


    2. Relevant equations

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


    3. 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
     
  2. jcsd
  3. Nov 18, 2007 #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.
     
  4. Nov 18, 2007 #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
     
  5. Nov 19, 2007 #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
     
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