work and force


by chocolatelover
Tags: force, work
chocolatelover
chocolatelover is offline
#1
Nov17-07, 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
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chaoseverlasting
chaoseverlasting is offline
#2
Nov18-07, 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.
chocolatelover
chocolatelover is offline
#3
Nov18-07, 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

chocolatelover
chocolatelover is offline
#4
Nov19-07, 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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