Respuesta :
The elastic potential energy (Ep) is given by [tex]Ep = \frac{1}{2}*k*x^2 [/tex]
Data:
Ep = 5184 J
k = 16200 N/m
x (displacement) = ?
Solving:
[tex]Ep = \frac{1}{2}*k*x^2 [/tex]
[tex]5184 = \frac{1}{2}*16200*x^2[/tex]
[tex]5184*2 = 16200x^2[/tex]
[tex]10368 = 16200x^2[/tex]
[tex]16200x^2 = 10368[/tex]
[tex]x^2 = \frac{10638}{16200} [/tex]
[tex]x^2 = 0.64[/tex]
[tex]x = \sqrt{0.64} [/tex]
[tex]\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\quad\checkmark[/tex]
Data:
Ep = 5184 J
k = 16200 N/m
x (displacement) = ?
Solving:
[tex]Ep = \frac{1}{2}*k*x^2 [/tex]
[tex]5184 = \frac{1}{2}*16200*x^2[/tex]
[tex]5184*2 = 16200x^2[/tex]
[tex]10368 = 16200x^2[/tex]
[tex]16200x^2 = 10368[/tex]
[tex]x^2 = \frac{10638}{16200} [/tex]
[tex]x^2 = 0.64[/tex]
[tex]x = \sqrt{0.64} [/tex]
[tex]\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\quad\checkmark[/tex]