Answer:
d. 20J
Explanation:
An ideal spring experiments a elongation 'x' with a force 'F' according to its spring constant 'k', as:
[tex]F=kx[/tex]
If a force of 20 N holds the spring, the spring is exerting a force of -20 N (the spring is flat); it is possible to calculate the elongation:
[tex]x=\frac{-20N}{10\frac{N}{m}}=-2m[/tex]
The potential energy stored is the potential work that the spring could do if the external force disappear, so, according to the differential definition of work,
[tex]dw=Fdx[/tex] which means that a differential of work is given by the multiplication of a force F by the displacement dx. This equation can be integrated to give:
[tex]dw=kxdx[/tex]
[tex]W=\frac{1}{2}kx^2[/tex]
So, the potential energy is:
[tex]PE=W=\frac{1}{2}10\frac{N}{m}*(2m)^2=20J[/tex]
