Answer:
4E
Explanation:
The elastic potential energy of an elastic material (e.g a spring, a wire), is the energy stored when the material is stretched or compressed. It is given by
U = [tex]\frac{1}{2}kx^2[/tex] --------------------(i)
Where;
U = potential energy stored
k = spring constant of the material
x = elongation (extension or compression of the material).
From the first statement;
when elongation (x) is 4cm, energy stored (U) is E
Substitute these values into equation (i) as follows;
E = [tex]\frac{1}{2}k(4)^2[/tex]
E = 8k
Make k subject of the formula
k = [tex]\frac{E}{8}[/tex] [measured in J/cm]
From the second statement;
It is stretched by 4cm.
This means that total elongation will be 4cm + 4cm = 8cm.
The potential energy stored will be found by substituting the value of x = 8cm and k = [tex]\frac{E}{8}[/tex] into equation (i) as follows;
U = [tex]\frac{1}{2}\frac{E}{8} (8)^2[/tex]
U = [tex]\frac{1}{2}{8E}[/tex]
U = [tex]{4E}[/tex]
Therefore, the potential energy stored will now be 4 times the original one.
