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Work and force 
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#1
Nov1707, 10:47 PM

P: 239

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
Nov1807, 12:00 AM

P: 1,017

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
Nov1807, 11:25 AM

P: 239

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
Nov1907, 09:59 PM

P: 239

Work and force
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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