# Work done by a Variable Force

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?

Last edited:

Homework Helper
Gold Member
"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).

Homework Helper
Gold Member
mbrmbrg said:
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!