Question

The graph below shows the force required to stretch a spring various distances. Review the graph below.

A graph of Force versus Elongation. Elongation (m) is on the x axis and Force (N) is on the y axis. A straight line passes through the points 0 comma 0 and 10 comma 15.

What is the best approximate value for the elastic potential energy (EPE) of the spring elongated by 7.0 meters?

1.5 J
11 J
37 J
74 J

Answer 1

Answer:

$EPE=37\ J$

Explanation:

Elastic Potential Energy

It's the term that expresses the amount of work an object can do due to its deformation. It's commonly applied to springs where the compression distance or elongation (x) is related to the Force (F) needed to stretch it. It can be found as

$\displaystyle EPE=\frac{kx^2}{2}$

Where k is the spring constant.

In this problem, we don't know the value of k, but it can be determined with the data provided. Though there is not a graph as suggested in the question, we have two points in the form (x, F) to find k. The points are (0,0) (10 m,15 N). Applying the Hooke's law

$F=k.x$

we can find the value of k as the slope of the line:

$\displaystyle k=\frac{F_2-F_1}{x_2-x_1}=\frac{15-0}{10-0}=1.5$

$k=1.5\ N/m$

Now we compute the EPE when x=7 m

$\displaystyle EPE=\frac{(1.5)7^2}{2}=36.75\ J$

The best approximate value is the third option:

$\boxed{EPE=37\ J}$