Calculating Work Done by a Variable Force on a Particle-Like Object

[mbrmbrg]

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 W = \int F(x)dx = \int madx
To find acceleration, I took the second derivative of the position function:
x(t) = 3t-4t^2+t^3
v=x'(t) = 3-8t+3t^2
a=x''(t) = -8+6t

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

 

[quasar987]

"Yes". The change of variable theorem states that,

\int_{x(t=0)}^{x(t=3)}F(x)dx = \int_0^3F(x(t))\dot{x}(t)dt

But as you cleverly noted, F(x(t))=ma(t).

 

[PhanthomJay]

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 W = \int F(x)dx = \int madx
To find acceleration, I took the second derivative of the position function:
x(t) = 3t-4t^2+t^3
v=x'(t) = 3-8t+3t^2
a=x''(t) = -8+6t

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

That looks like it will take away the evil and give you the correct solution!l

 

[mbrmbrg]

Thank you very kindly!