Find Work of a Force Acted on Mass over Frictionless Surface

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mrsteve
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Homework Statement


A force given by [tex]F(x)=5x^3 (in \frac {N}{m^3})[/tex] acts on a 1-kg mass moving on a frictionless surface. The mass moves from x = 2m to x = 6m.

a) How much work is done by the force?

Homework Equations



[tex]W=\int F_x from x_o to x_f[/tex]

The Attempt at a Solution



[tex]\int 5x^3 from x=2 to x=6 = 1.6*10^3[/tex]

Is this correct?
 
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Also, if the speed of the mass is 2 m/s @ x=2 what is the speed at x=6?

[tex]KE=\frac{mv^2}{2}[/tex]

[tex]1600=\frac{(1)(v^2)}{2} => v=5.6*10^1 = 60 m/s[/tex]

Would that be correct?
 
Ignea_unda said:
Looks correct to me, mrsteve.

The book has 2,000 J. I suppose that's because there's only 1 significant digit in mass = 1kg?

[tex]1.6 *10^3 => 2 *10^3[/tex]