Find the work done by the force F=10i+15j in moving an object from P(0,0) to Q(2,3).
What is the distance between P(0,0) and Q(2,3)?
Yeah I think so...(2,3)?
Sorry, \[W = F*d\]
My son is playing some silly video game in the background and it is destroying my brain :-)
Distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is given by \[d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\]
Haha no problem (:
easy, because one of the points is (0,0)!
Yeah that wasnt so bad... I got square root of 1
;-( how?
wait was that formula supposed to be x1-y1 and x2-y2?
No, I wrote it like I meant it. It's the Pythagorean theorem, in disguise
Okay so I did square root of (0-0)^2 + (2-3)^2 is that right?
Nope. \((x_1,y_1) = (0,0)\) and \((x_2,y_2) = (2,3)\) Plug into formula\[d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}\]
\[d = \sqrt{(0-2)^2 + (0-3)^2} = \]
Alright now I got square root 13 haha
Woah sorry about that huge space hahaha my computers crazy
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just Pythagorean theorem, as I said
\[d=\sqrt{13}\] is correct
Now I think we find the magnitude of the force vector, and multiply it by \(d\) to get our answer
Okay Ill do that now
Alright I got 18
and then when I multiply them together would it just be 18 square root 13
How did you get 18?
square root of 10^2 +15^2?
oh or would it just be square root of 18?
\[\sqrt{10^2+15^2} = \sqrt{100+225} = \sqrt{325} = \sqrt{25*13} = 5\sqrt{13}\]
Now we multiply \(d = \sqrt{13}\) by \(||F|| = 5\sqrt{13}\) and get \(W = \) drum roll please....a hush falls over the crowd...will she get it?
Umm 65? :D
the crowd goes nuts!!!
Yayyyyy thank you! I think my brain is fried but I understand everything a lot better now! (:
Come on back, let's finish off the cube root problem...
Okay haha Ill try to do it quick because I have to go to bed :P
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