Question

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 3.0 meters?

Answer 1

The elastic potential energy of the spring is 6.8 J

Explanation:

The elastic potential energy of a compressed/stretched spring is given by the equation:

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

where

k is the spring constant

x is the elongation of the spring

The spring constant of the spring in this problem can be found by keeping in mind the relationship between force (F) and elongation (x) (Hooke's law):

$F=kx$

By looking at the graph and comparing it with the formula, we realize that the slope of the force-elongation graph corresponds to the spring constant. Therefore in this case,

$k=\frac{15.0-0}{10.0-0}=1.5 N/m$

Therefore when the spring has a elongation of $x=3.0 m$, its potential energy is

$E=\frac{1}{2}(1.5)(3.0)^2=6.8 J$

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