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A single force acts on a 7.0 kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0t - 4.0t^2 + 1.0t^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 3.0 s.

I want to use the equation [tex]W = \int F(x)dx = \int madx[/tex]

To find acceleration, I took the second derivative of the position function:

[tex]x(t) = 3t-4t^2+t^3[/tex]

[tex]v=x'(t) = 3-8t+3t^2[/tex]

[tex]a=x''(t) = -8+6t[/tex]

When I go to plug values into my work integral, I get [tex]W=\int m(6t-8)dx[/tex]

I would then take mass out of the integral, but having t and dx in the integral is evil. Do I say that dx=x'(t) and so substitute my velocity function (with dt tacked onto the end) for dx?

I want to use the equation [tex]W = \int F(x)dx = \int madx[/tex]

To find acceleration, I took the second derivative of the position function:

[tex]x(t) = 3t-4t^2+t^3[/tex]

[tex]v=x'(t) = 3-8t+3t^2[/tex]

[tex]a=x''(t) = -8+6t[/tex]

When I go to plug values into my work integral, I get [tex]W=\int m(6t-8)dx[/tex]

I would then take mass out of the integral, but having t and dx in the integral is evil. Do I say that dx=x'(t) and so substitute my velocity function (with dt tacked onto the end) for dx?

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