Answer:
37 J
Explanation:
Given:
A graph of force versus elongation in which a line passes through the points (0, 0) and (10, 15).
The x-axis represents elongation and y-axis represents force.
Elongation (x) = 7.0 m
Now, we know that the slope of a line of force versus elongation graph is the value of the spring constant.
Therefore, finding the slope of the line passing through (0, 0) and (10, 15), we get:
[tex]k=\frac{15-0}{10-0}=1.5\ N/m[/tex]
Therefore, the spring constant (k) = 1.5 N/m.
Now, the elastic potential energy (EPE) of the spring is given as:
[tex]EPE=\frac{1}{2}kx^2[/tex]
Plug in 1.5 for 'k' and 7.0 for 'x'. Solve for EPE.
[tex]EPE=\frac{1}{2}\times 1.5\times (7.0)^2\\\\EPE=\frac{1.5\times 49}{2}\\\\EPE=36.75\approx 37\ J[/tex]
Therefore, the best approximate value for the elastic potential energy (EPE) of the spring elongated by 7.0 meters is 37 J.
