Asked by Kid
Why do I see some sources define the work on a spring as W = (1/2)(k)(xf^2 - xi^2) and some sources define it as W = (1/2)(k)(x^2) ?
For example (I'll test out both equations):
Let xi = 3, xf = 7, and k = 3
So by using W = (1/2)(k)(xf^2 - xi^2)
= (1/2)(3)(7^2 - 3^2)
= (1/2)(3)(49 - 9)
= (1/2)(3)(40)
= 60
By using W = (1/2)(k)(x^2)
= (1/2)(k)(xf - xi)^2
= (1/2)(3)(7-3)^2
= (1/2)(3)(4)^2
= (1/2)(3)(16)
= 24
I'm confused since I didn't get the same result for both equations, so they are not the same then? Am I miscalculating something?
For example (I'll test out both equations):
Let xi = 3, xf = 7, and k = 3
So by using W = (1/2)(k)(xf^2 - xi^2)
= (1/2)(3)(7^2 - 3^2)
= (1/2)(3)(49 - 9)
= (1/2)(3)(40)
= 60
By using W = (1/2)(k)(x^2)
= (1/2)(k)(xf - xi)^2
= (1/2)(3)(7-3)^2
= (1/2)(3)(4)^2
= (1/2)(3)(16)
= 24
I'm confused since I didn't get the same result for both equations, so they are not the same then? Am I miscalculating something?
Answers
Answered by
bobpursley
In the second equation, it assumes the initial x is zero.
Work= INtegral force(x)*dx= INT k*x*dx over limits or
work= 1/2 k (xf^2-xi^2)
Work= INtegral force(x)*dx= INT k*x*dx over limits or
work= 1/2 k (xf^2-xi^2)
Answered by
Kid
Oh I think I get it now. So if xi = 0, then both versions of the equation will give the same result. Otherwise, if the initial position is not zero, then W = (1/2)(k)(xf^2 - xi^2) MUST be used. Correct?
Answered by
bobpursley
correct. The second equation is for beginners (xi=0)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.