# Comparing two approaches to work

When a particle's position can be described by $x(t)=3t-4t^2+t^3$, the work done from $t_0=0$ to $t_1=4$ sec is $528\space{J}$ using the work-kinetic theorem.

Now if we were to find the force function, we know that the acceleration function is $a(t)= 6t-8$ Should the force function them be $$F(x)=18t-24$$

Should the work then be the integral of force on the x-axis? In this case I come up with a work of $$\int_0^4F(x)\,dx = 48\space{J}$$ Which is clearly incorrect (I always had more confidence in the work-kinetic theorem than the integral )

What is the discrepancy?

Note: I would like to note that I did consider the force function to be completely "in" the x component only.

• A note: It is called the work energy theorem. – Steeven Oct 6 '16 at 14:23

The discrepancy is your treatment of $dx$.
Work done is the integral of $Fdx$. Here $F=6(3t-4)$ but after differentiating the expression for $x$ we get $dx=(3-8t+3t^2)dt$. Therefore
$\int_0^4F(x)\,dx = \int_0^4 6(3t-4)(3-8t+3t^2)dt=528$.