When a force of F=26 N is applied to the spring, the spring is stretched with respect to its rest position by
[tex]\Delta x=16 cm - 10 cm=6 cm = 0.06 m[/tex]
This means that the constant of the spring is
[tex]k= \frac{F}{\Delta x}= \frac{26 N}{0.06 m}=433.33 N/m [/tex]
The work required to stretch the spring from 10 cm to 13 cm, which corresponds to an elongation of
[tex]\Delta x_2 = 13 cm - 10 cm=3 cm =0.03 m[/tex]
is equal to the variation of elastic potential energy of the spring, so:
[tex]W=\Delta U= \frac{1}{2}k(\Delta x_2)^2= \frac{1}{2}(433.3 N)(0.03 m)^2=0.19 J [/tex]
