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theEric
 2 years 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:
theEric
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1You can use this relation

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

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

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

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

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

theEric
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1379925139264:dw

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

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

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