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 one year ago
Hi! I have to find the velocity as a function of displacement for some given force functions.
Here is the question:
Find the velocity \(\dot x\) as a function of the displacement \(x\) for a particle of mass \(m\), which starts from rest at \(x=0\), subject to the following force functions:
 one year ago
Hi! I have to find the velocity as a function of displacement for some given force functions. Here is the question: Find the velocity \(\dot x\) as a function of the displacement \(x\) for a particle of mass \(m\), which starts from rest at \(x=0\), subject to the following force functions:

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theEric
 one year ago
Best ResponseYou've already chosen the best response.0(a) \(F_x=F_0+cx\) (c) \(F_x=F_0\cos(cx)\) If I see the general technique, I might be able to get it! Thanks!

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1You can use this relation

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1dw:1379923967464:dw

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Ah, thank you very much Saeeddiscover!

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1Hold on a moment!

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Does that work out? Is the \(x\) in \(F_x=F_0+c\large x\) a function of \(t\)?

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1For the second one, you may consider this way as well: dw:1379924739346:dw

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1Please wait a moment. I got a mistake!

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Okay! If it's in that last part, I can't find it because I don't know that material yet!

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1dw:1379925139264:dw

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1dw:1379925286166:dw

theEric
 one year ago
Best ResponseYou've already chosen the best response.0That all makes sense, thank you! I'll try to apply those to my problems soon!

theEric
 one year ago
Best ResponseYou've already chosen the best response.0I think I need to head to bed, though...

Saeeddiscover
 one year ago
Best ResponseYou've already chosen the best response.1Both 1 and 2 you are given simply can be solved using the last equation I gave you. It is true and works properly.

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Great! I will use that, then! Have a good day!
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