The elastic potential energy increases by a factor of 9
Explanation:
The elastic potential energy of a bowstring is given by
(1)
where
k is the spring constant
x is the elongation of the bowstring
Hooke's law states the relationship between the force applied and the elongation of an elastic object:
![F=kx](https://tex.z-dn.net/?f=F%3Dkx)
where
F is the force applied
x is the elongation
We can rewrite it as
![x=\frac{F}{k}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7BF%7D%7Bk%7D)
And substituting into (1),
![E=\frac{1}{2}k(\frac{F}{k})^2=\frac{F^2}{2k}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1%7D%7B2%7Dk%28%5Cfrac%7BF%7D%7Bk%7D%29%5E2%3D%5Cfrac%7BF%5E2%7D%7B2k%7D)
In this problem, the force applied to the bowstring is tripled,
F' = 3F
So the final elastic potential energy is:
![E'=\frac{(3F)^2}{2k}=9(\frac{F^2}{2k})=9E](https://tex.z-dn.net/?f=E%27%3D%5Cfrac%7B%283F%29%5E2%7D%7B2k%7D%3D9%28%5Cfrac%7BF%5E2%7D%7B2k%7D%29%3D9E)
So, the elastic potential energy increases by a factor of 9.
Learn more about potential energy:
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