Explanation:
Given that,
Work done, W = 4.57 J
Distance, [tex]x_1=9.11\ cm[/tex]
Distance, [tex]x_2=9.54\ cm[/tex]
We know that the work done by the spring is given by :
[tex]W_=-kx_1^2[/tex]
[tex]k=\dfrac{W}{x_1^2}[/tex]
[tex]k=\dfrac{4.57}{(9.11\times 10^{-2})^2}[/tex]
k = 550.65 N/m
To find,
Let [tex]W_2[/tex] is the extra work is required to stretch it an additional 9.54 cm. It can be calculated as :
[tex]W_2=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]
[tex]W_2=\dfrac{1}{2}\times 550.65(9.54^2-9.11^2)[/tex]
W = 2207.96 J
So, the extra wok done is 2207.96 J.
