Answer:
P.E = 4.398 Joules.
Explanation:
Given the following data;
Spring constant, k = 71.8N/m
Displacement, x = 0.35m
To find the elastic potential energy;
The elastic potential energy of an object is given by the formula;
[tex] P.E = \frac {1}{2}kx^{2}[/tex]
Substituting into the equation, we have;
[tex] P.E = \frac {1}{2}*71.8 *(0.35)^{2}[/tex]
[tex] P.E = 35.9 * 0.1225 [/tex]
Elastic potential energy = 4.398 Joules.
Therefore, the elastic potential energy of the spring is 4.398 Joules.
