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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.

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  • $\begingroup$ A note: It is called the work energy theorem. $\endgroup$ – Steeven Oct 6 '16 at 14:23
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Your question did not specify the mass of the particle, but you need that, and it sounds like you are taking it to be 3. Normally we'd want units for all these quantities, but let's assume the units are implied. So if the mass is indeed 3, then your F(t) is correct, it's just ma. Your integrated work is wrong because you did the integral over t, but your expression correctly shows the integral needs to be over x. To convert from an integral over x to an integral over t, use dx = dx/dt dt = v(t) dt, and you can figure out v(t) just like you figured out a(t). Then do the integral over t from 0 to 4, it should work (no pun intended).

By the way, if you wonder why we use F(x) dx in the usual definition of the work integral, instead of F(t)v(t) dt, it's because we usually know F(x) not F(t)-- this problem is a bit unusual that way. One could argue that if you start out knowing x(t), then you already know everything you need, so the whole problem is in some sense being done backward, starting from what you want to know, and reasoning steps along the way that you normally would use to get what you already know here!

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  • $\begingroup$ Pun intended. But thank you very much sir. $\endgroup$ – Ian Limarta Oct 6 '16 at 14:56
  • $\begingroup$ I do have a side question, how often do physicists create statistically well-behaved models for specific scenarios as the one shown above? $\endgroup$ – Ian Limarta Oct 7 '16 at 1:11
  • $\begingroup$ In my opinion, this model exists for what we might call pedagogical purposes only. In other words, it is a model designed to teach us something, or familiarize us with some of the mathematics behind the work-energy theorem. $\endgroup$ – Ken G Oct 7 '16 at 1:37
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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
$\int_0^4F(x)\,dx = \int_0^4 6(3t-4)(3-8t+3t^2)dt=528$.

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