Question

I need help with this problem

Answer 1

\[W = Fx\]Can you figure out the work done in each of the intervals? One interval is from x = 3 to x = 6, another is from x = 6 to x = 9 (I'm guessing, it's hard to read), and the final one is from x = 9 to x = 10.

Answer 2

But how do I find Fx?
All of your X= are correct btw

Answer 3

read them off the graph! \[F = \text{ force (the y-axis)}\]\[x = \text{ distance (the x-axis)}\]

So, as the thing goes from \(x = 3\) to \(x = 6\), that's \(x = 6 -3 = 3\) . What is the force applied in that range of \(x\)?

Answer 4

Is the Force in the X direction 12N? There is 12N marked on the graph plus there is something with 1.5m, is that your doodling or is it part of the graph?

Answer 5

Just doodling. I understand that whpalmer. But I'm guessing that I was guessing that I have to find something like the integral min=3 max=10 with a function

Answer 6

Line integral.

Answer 7

No, you're making it too complicated. From x = 3 to x = 6, the force is -8 N. W = F * x = -8N * 3 m = -24 J

From x = 9 to x = 10, the force is 4 N. W = F*x = 4N * 1m = 4 J

The only moderately tricky part is from x = 6 to x = 9. Hint: find the average force there.

Answer 8

Ok thanks I got it -24-6+4 = -26 J

Answer 9

If you like complicated, you could write the the force as a piecewise function:

\[Fx(x) = -8,\, 0\le x \le 6\]\[Fx(x) = -8+4(x-6),\, 6\le x\le 9\]\[Fx(x) = 4,\,9\le x\le 10\]

and then compute \[W = \int_3^{10} F\,dx\]

Answer 10

Yep, that's what I got as well.

Answer 11

Which is what I said all along.