Respuesta :
Answer:
[tex]29.75J[/tex]
Explanation:
Fist we must find the constant of the spring. The equation for work in a spring is:
[tex]W=\frac{1}{2}kx^2[/tex]
where [tex]W [/tex], is the work, in this case [tex]W=5J[/tex], [tex]k[/tex] is the constant of the spring, and [tex]x[/tex] is the distance from the equilibrium length: [tex]x=2.2cm=0.022m[/tex]
clearing for [tex]k[/tex] and substituting known values:
[tex]2W=kx^2\\\frac{2W}{x^2}=k[/tex]
[tex]\frac{2(5J)}{(0.022m)^2}=\frac{10J}{4.84x10^{-4}m^2} =20,661.16J/m^2[/tex]
So work to stretch the spring, an additional 3.6cm is:
[tex]W=\frac{1}{2}kx^2[/tex]
where [tex]x=2.2cm+3.6cm=5.8cm=0.058m[/tex]
[tex]W=\frac{1}{2}(20,661.16J/m^2)(0.058m)^2[/tex]
[tex]W=\frac{1}{2}(20,661.16J/m^2)(3.364x10^{-3}m^2)[/tex]
[tex]W=\frac{1}{2}(69.5J)[/tex]
[tex]W=34.75J[/tex]
And since we already have 5J, the difference, or the additional work is:
[tex]34.75J-5J=29.75J[/tex]
