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A single force acts on a $3.4~\text{kg}$ particle-like object in such a way that the position of the object as a function of time is given by $x = 4.2t - 2.1t^2 + 2.5t^3$, with $x$ in meters & $t$ in seconds. Find the work done on the object by the force from $t = 0 \to t = 6.5\: \mathrm{s}$.

I first took the second derivative to find the acceleration. I substituted this in Newtonian equation $$F = m \cdot a$$.

then I used the second derivative and the newly obtained equation, which I integrated with limits of integration $0$ to $6.5$, I am confused why is my answer incorrect. Have I done something wrong? this is what I have tried.

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I would use the following: $$ dW= F dx = m \frac{d^2 x}{dt^2} \frac{dx}{dt} dt $$ where I used $F=ma=m\frac{d^2 x}{dx^2}$ and $dx=\frac{dx}{dt}dt$. Now, integrate this from $t=0$ to $t=6.5s$. You just need to calculate the derivatives of the function that is given for $x(t)$. I get around $147\,KJ$, but I might be wrong.

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