MathGeek99
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Please Help!!! It's for a quiz!!

Work-energy theorem: A 7.0-kg rock is subject to a variable force given by the equation
F(x) = 6.0 N - (2.0 N/m)x + (6.0 N/m2)x2
If the rock initially is at rest at the origin, find its speed when it has moved 13.9 m.

Respuesta :

Answer:

38.8 m/s

Explanation:

Force F(x) = 6 - 2x + 6x²

work

[tex]W=\displaystyle\int_{0}^{13.9}F(x)dx=\displaystyle\int_{0}^{13.9}(6-2x+6x^2)dx[/tex]

[tex]=6x-x^2+2x^3|_{0}^{13.9}\\=5261 J[/tex]

W = mv²/2=7v²/2 = 3.5v² = 5261

v = 38.8 m/s

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