I had a homework problem with this graph of F(Newtons) vs x(meters).

The graph has a straight line (constant slope) from (0,0) to (8,10).

And the slope ((10-0)/(8-0)) came out to be 1.25, making the function of the graph be F = 1.25x.

Then the question was to find the "Work done by force during the displacement".

Which is to simply take integral of and plugging in the numbers.

Integral came out to be 0.625x^2.

To find the area under curve from 0 to 8 meters (Work done by force during displacement from 0 to 8 meters),

I just plugged in 8 meters into the function and the number came out to be 40.

However I'm not quite sure about what the unit is for this problem.

It HAS to be Joules (N*m).

But I'm not sure how I should reformat/move around to make the unit be N*m.

Is there any proof on this?

  • $\begingroup$ My unit came out to be m^2, and I know 40N*m should be coming out if you approached it as finding area of the triangle method. But I want to know how to get the units right by using integration method. $\endgroup$ – hs2345 Jun 21 '15 at 23:42
  • $\begingroup$ What are the units of the slope? What does that make the units of function? $\endgroup$ – dmckee Jun 22 '15 at 1:14
  • $\begingroup$ For this particular problem an integral isnt even necessary since its a triangle, but yeah Leonard takes care of a more general case $\endgroup$ – Triatticus Jun 22 '15 at 2:30

The slope you calculated is $\frac{10 \rm N}{8\rm m}=1.25\,{\rm \frac Nm}$, so the force function becomes $F=1.25\,{\rm \frac Nm}*x$. Hence, the work is $$ W=\int{(1.25\,{\rm \frac Nm}\times x)dx}=1.25\,{\rm \frac Nm}\int{xdx}=0.625\,{\rm \frac Nm}x^2 $$ (If $F(0)=0$). Because the unit of $x^2$ is $\rm m^2$, then the unit of the work is, as expected, $\rm N/m\times m^2=N*m$

  • $\begingroup$ Okay. so I have to keep N/m*x as the unit! Thanks! $\endgroup$ – hs2345 Jun 22 '15 at 0:06
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    $\begingroup$ Don't forget, everytime you integrate, you multiply by a dx. So, you'll always add a power to whatever you integrate against. The opposite is of course true with a derivative. You will always divide (or remove a power) of whatever you are taking the derivative against. $\endgroup$ – Mark Jun 22 '15 at 3:22

Since the scales of the graph are in Newton and meters, your answer is fine.

Think of integration as simply adding small individual work for small displacements.

For a displacement ds, work done will be dw=F.s

Its unit will be in J (or Nm), right. Integrating dw will not change the unit, as adding doesnt change the dimensions.

Hope the explanation is clear.


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