Calculating work done vs. calculating final energy [closed]

Question

I'm trying to solve for work after 2 seconds given $v(t)=3t^2$ and mass$=1kg$. There are 2 approaches:

• Just calculate kinetic energy after 2 seconds: $E_k=.5*mv^2 = .5 *1 * (3*2^2)^2 = .5* 144 = 72J$
• And calculate the amount of work applied: $W=Fx$. Which is where I'm having troubles - can't get the same number.

Here's what I'm doing:

1. I know that total distance after 2 seconds: $x(t)=\int3t^2dt=t^3$, which gives $x(2)=8m$.
2. Acceleration (to calculate force): $a=v' = (3t^2)' = 6t$
3. Since $x(t)=t^3$ this gives us: $t=x^{1/3}$. Thus acceleration as a function of distance (needed for the integral below): $a(x) = 6x^{1/3}$
4. And since acceleration (and thus force) are not constant: $W=\int_0^{8m} Fdx = \int_0^{8m} ma \, dx = m\int_0^{8m} 6x^{1/3} \, dx = 6m\frac{4}{3}8^{4/3}= 8*8^{4/3} = 128J$

Which is a completely different result. Where do I make a mistake?

Update: I didn't integrate correctly as was pointed out by @DavidWhite. Should've been: $W=\int_0^{8m} Fdx = \int_0^{8m} madx = m\int_0^{8m} 6x^{1/3}dx = 6m\frac{3}{4}8^{4/3}= \frac{9}{2}8^{4/3} = 72J$

Thanks!

Answer 1

Or you can express F and dx in terms of t and then evaluate work.

$F = 6mt$ and $dx = 3t^2 dt$

$F = \int_{0}^{2}6mt3t^2 dt = 72$

Answer 2

We can also write $\vec a =v \times dv/dx$ .

$dv/dt= 6t$ ,

$dx/dt= 3t^2$ , $\implies $ $dx=3t^2dt$

$v\times dv/dx = 6t$

$Fdx = madx = m \times v\times dv/dx\times dx = (6t)\times3t^2 =18t^3$

integrate from 0 to 2 and i get 1 st method is correct.