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Let's say there is a 1kg lump of rock floating in space.

Through some means I apply an acceleration of 3 m/s² to the rock for one hour.

How much energy did I spend, assuming none was wasted?


Here is what I'm trying:

Force = mass * acceleration. So F = 1kg * 3m/s² = 3 newtons.

This must mean I'm applying 3N for 1 hour.

Joules = newtons * metres.

What? This doesn't make sense. I thought we'd already accounted for all derivatives of position. Suppose I stopped for half an hour in the middle of the exercise. I would have spent the same amount of total energy, but the rock would have travelled further.

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2 Answers 2

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So, work, which the energy expended is given by

$ \textbf{W} = \int \textbf{F} \cdot d\textbf{s} $

Where $\textbf{r}$ is the distance over which the force acts. Let's say that the force is applied directly in one direction and we have no other forces going on (i.e. no gravity). We can then drop the integral and the vector notation

$W = Fs$

We now use newton's second law to write

$W = mas$

We turn to our equations of motion for constant acceleration (commonly called SUVATS), and we can write

$ s= ut + \frac{1}{2}at^2 $

Where $u$ is the initial velocity, lets say it is $0$.

So, our expression for work, the energy used, is

$ W = \frac{1}{2}ma^2t^2 = \frac{1}{2}(1)(3)^2(3600)^2 = 58,320,000 J$

Where we pushed the mass for $3m/s^2$ and for 1 hour. It would have travelled 19,440,000m. For 30 minutes the work is

$ W =\frac{1}{2}(1)(3)^2(1800)^2 = 14,580,000 J$

and it would have travelled 4,860,000m.

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  • $\begingroup$ Thanks! If there is an equal, external acceleration happening exactly to counteract my own (for example gravity if it were weaker), does the amount of energy I expend to keep the object in one place change from about 14 million J? If I've understood properly, no it wouldn't change. $\endgroup$
    – matt_rule
    Commented Mar 9, 2017 at 23:40
  • $\begingroup$ (By the way I will study all of your equations in detail.) $\endgroup$
    – matt_rule
    Commented Mar 9, 2017 at 23:43
  • $\begingroup$ You should ask another question if you want to think about that. It's all to do with your frame of reference (i.e. what we consider in the equation). The amount of energy required (work) to move the object is 0 because we didn't move it at all. $\endgroup$
    – Tomi
    Commented Mar 9, 2017 at 23:46
  • $\begingroup$ My mind is blown right now, which is probably a good thing. Asking another question probably won't help - I won't use the right words to describe what I mean - instead I'll have to revise my understanding of acceleration, forces and/or energy. $\endgroup$
    – matt_rule
    Commented Mar 9, 2017 at 23:50
  • $\begingroup$ BBC Bitesize is quite a good High School level resource - bbc.co.uk/schools/gcsebitesize/science/add_aqa_pre_2011/forces/… $\endgroup$
    – Tomi
    Commented Mar 9, 2017 at 23:54
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Let $v$ be the final velocity, $a$ the acceleration and $t$ the time.
Assuming the initial velocity is zero then the distance travelled $x =\frac 12 at^2=\frac 32 t^2$.
The work done is force times distance $=3x=\frac 92t^2$.

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  • $\begingroup$ If I stop pushing on the object, it continues to move through space. Is work still being done? Am I doing it? If the kinetic energy of the object doesn't change from then on, surely the total energy we're measuring, measured in joules, is going to stop increasing even though the object keeps moving. I know I'm wrong, somehow, but I don't know why. $\endgroup$
    – matt_rule
    Commented Mar 9, 2017 at 23:57
  • $\begingroup$ No work is being done. $W=Fs$, your force is 0, so work is 0, and so it doesn't change. $\endgroup$
    – Tomi
    Commented Mar 10, 2017 at 0:07

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