Question

A Hooke's law bowstring is stretched x meters until a force of f newtons is applied, and then held. By what factor will the elastic potential energy of the bowstring be affected if the final force at which the bowstring is held is tripled?

Answer 1

The elastic potential energy increases by a factor of 9

Explanation:

The elastic potential energy of a bowstring is given by

$E=\frac{1}{2}kx^2$ (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:

$F=kx$

where

F is the force applied

x is the elongation

We can rewrite it as

$x=\frac{F}{k}$

And substituting into (1),

$E=\frac{1}{2}k(\frac{F}{k})^2=\frac{F^2}{2k}$

In  this problem, the force applied to the bowstring is tripled,

F' = 3F

So the final elastic potential energy is:

$E'=\frac{(3F)^2}{2k}=9(\frac{F^2}{2k})=9E$

So, the elastic potential energy increases by a factor of 9.

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