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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 $t = 0s$ to $t = 7.0 s$.

My thinking: $dW = Fdr$ where $F = force$, $dr =$ change in position $F = ma$ where $m =$ mass, $a =$ accelleration $a =$ second derivative of position with respect to time $= 6t-8$ $dW=m(6t-8)(3.0t − 4.0t^2 + 1.0t^3)dt$ integrate from $0$ to $7$ seconds and I'm off by a couple orders of magnitude. Where did I go wrong?

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3 Answers 3

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You are correct that the force F is ma, where $a = (6t-8)$. The work is the integral of $Fdx$, with $dx = (3-8t+3t^2)dt$. So, $$dW=m(6t-8)(3-8t+3t^2)dt=\frac{m}{2}d[(3-8t+3t^2)^2]$$

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  • $\begingroup$ Doing this for my new problem where x is the same, m = 2g and t = 0 to 5s gives me 1435 J but the expected answer is 1440 J. $\endgroup$
    – Anazopyreo
    Feb 13, 2018 at 15:11
  • $\begingroup$ I get 1444Jpules $\endgroup$ Feb 13, 2018 at 16:01
  • $\begingroup$ 1444 is for t = 5, at t=0 it's 9, so 1444-9 = 1435. It turns out the answer is 1440 because of significant figures. $\endgroup$
    – Anazopyreo
    Feb 14, 2018 at 22:25
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The work done is equal to the change in Kinetic energy.

$v(0)=3m/s$

$v(7)=94m/s$

That gives: $\text{Work} = \text{KE}_7 - \text{KE}_0 =31kJ$

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  • $\begingroup$ We haven't covered energy yet, but 31 KJ is about what I get when I try integrating acceleration and velocity with respect to time as in my comment to the other answer. But that apparently is not correct. $\endgroup$
    – Anazopyreo
    Feb 13, 2018 at 14:46
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You're integrating Force times Position over Time (the units don 't work). Try integrating Force times Velocity (Power) over Time, since Power times Time is Energy.

Note that:

$$ W = \int{F\cdot dx} \ne \int{ma(t)x(t)dt} = m\int{(6t-8)(t^3-4t^2+3t)dt} $$

because:

$$ dx = \frac{dx}{dt}dt = v(t)dt = (3t^2 - 8t + 3)dt $$.

So:

$$ W = m\int{(6t-8)(3t^2-8t+3)dt} = m[4.5t^4-24t^3+41t^2-24t]^{t=7}_{t=0}$$

Turns out, this is the same as the change in kinetic energy:

$$ \Delta T \equiv \frac 1 2 m v(t)^2|^{t=7}_{t=0}$$

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  • $\begingroup$ I had thought this, (Since Fva gives kg m^2/s^3 and integrating results in kg m^2/s^s or J) but the answer that I was supposed to arrive at was closer to 1kJ. I'm working on a practice problem on a website. The practice problem has changed so I don't have access to the old "correct" answer. The function is the same, but now it's a 2kg particle, and the time is 0-5 sec. The answer I arrived at is 6435 J, the answer I was supposed to get is 1440 J. $\endgroup$
    – Anazopyreo
    Feb 13, 2018 at 14:26
  • $\begingroup$ What really confuses me is that there is a shortcut that says I can subtract the square of the initial velocity from the square of the final velocity and multiply that by half the mass. That gets me close enough to the expected answer that it will accept it, but not the expected answer. I'm confused because that shortcut is found by integrating ma dx. $\endgroup$
    – Anazopyreo
    Feb 13, 2018 at 14:54
  • $\begingroup$ After reading Chester's answer and working out the math for myself I understand the shortcut, but not why my answer is still not right. $\endgroup$
    – Anazopyreo
    Feb 13, 2018 at 15:22
  • $\begingroup$ @Anazopyreo: not sure why this was down voted, since it's right: you asked what you did wrong--and you what did wrong was integrate $\int{ma(t)x(t)dt}$, which doesn't even have the units of energy. I said integrate $\int{ma(t)v(t)dt}$, which has the correct units, is the correct answer, and is equal to $\frac{1}{2}m v(t)^2$. $\endgroup$
    – JEB
    Feb 13, 2018 at 19:37
  • $\begingroup$ I noticed that my original question and some answers were down-voted right from the start. $\endgroup$
    – Anazopyreo
    Feb 13, 2018 at 22:57

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