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An unusual spring has a restoring force of magnitude F = (2.00 N/m)x + (1.00 N/m2)x2, where x is the stretch of the spring from its equilibrium length. A 3.00-kg object is attached to this spring and released from rest after stretching the spring 1.50 m. If the object slides over a frictionless horizontal surface, how fast is it moving when the spring returns to its equilibrium length?

Respuesta :

Answer

given,

mass = 3 kg

x = 1.50 m

Restoring Force

F = (2.00 N/m)x + (1.00 N/m2)x²

work done = [tex]\int F.dx[/tex]

   = [tex]\int( (2.00)x + (1.00)x^2)dx[/tex]

   = [tex]x^2 +\dfrac{1}{3}x^3[/tex]

now work done at x = 1.5

   = [tex]x^2 +\dfrac{1}{3}x^3[/tex]

   = [tex]1.5^2 +\dfrac{1}{3}1.5^3[/tex]

  W= 3.375 J

we know

[tex]W = \dfrac{1}{2}mv^2[/tex]

[tex]3.375 = \dfrac{1}{2}\times 3 \times v^2[/tex]

v = 1.5 m/s

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