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 iJamJL Mar9-12 01:18 PM

Not sure which formula to use! Deals with work and position

1. The problem statement, all variables and given/known data
A one-dimensional force acts on a particle of mass m = 6.26 kg in such a way that its position is given by:

x = 0.484t3 - 33.6t

Find W, the work done by this force during the first 1.49 s.

2. Relevant equations
F=mg
F=ma
W=F*distance

3. The attempt at a solution
To find the distance, I plugged in 1.49 into the formula they gave, and I got (-48.5). I know that to find work, the formula is W=F*s, so in order to find the force, I'm not sure which formula to use.

I think I should be using F=ma, and to find the acceleration, I can take the second derivative of the given formula. However, I forget whether that is allowed in one-dimension. If not, then do I use F=mg? If I use F=mg, then the answer for work will come out positive. If I use F=ma, then the work comes out negative.

Here are my results:

W=Fs
W=ma*s
W=6.26*18.18*(-48.5)
W=(-5519.6)J

W=Fs
W=mg*s
W=6.26*(-9.81)*(-48.5)
W=2978.4 J

 Delphi51 Mar9-12 01:38 PM

Re: Not sure which formula to use! Deals with work and position

Quote:
 to find the acceleration, I can take the second derivative of the given formula.
Yes! Do that and you will find that the acceleration varies with time so you can't use W=Fs; instead you have to integrate F*ds.

Gravity is not involved in this. It would cause a constant acceleration and a distance formula of s = Vi*t + .5*a*t² rather than the cubic formula of this problem.

 tiny-tim Mar9-12 01:39 PM

hi iJamJL! :smile:
Quote:
 Quote by iJamJL (Post 3806716) I think I should be using F=ma, and to find the acceleration, I can take the second derivative of the given formula. However, I forget whether that is allowed in one-dimension. If not, then do I use F=mg?
if you need to find the force, then yes you use F = ma

and yes the acceleration is the second derviative, even in three dimensions! :smile:

(what is worrying you about that? :confused:)
Quote:
 To find the distance, I plugged in 1.49 into the formula they gave, and I got (-48.5). I know that to find work, the formula is W=F*s, so in order to find the force, I'm not sure which formula to use.
yes, that does give you the total distance

but W = F*s only works if F is constant

here, F isn't constant, so you would have to use W = ∫ F.ds
however it might be simpler to use the work-energy theorem, and calculate the difference in kinetic energy :wink:

 iJamJL Mar9-12 01:57 PM

Re: Not sure which formula to use! Deals with work and position

Thanks for the replies!

Quote:
 Quote by tiny-tim (Post 3806738) however it might be simpler to use the work-energy theorem, and calculate the difference in kinetic energy :wink:
I didn't realize that until you pointed it out! I tried it out, but I think I made a mistake somewhere. Here's what I did:

To find the final velocity, which for this case is at 1.49s, we take the first derivative at 1.49:
x = 0.484t3 - 33.6t
dx/dt= 2*(.484)t^2 - 33.6
v= (-30.38)m/s

W=ΔKE
W= 1/2*m(v-final^2) - 1/2*m(v-initial^2)
**v-initial is 0 because t=0, x=0**
W=1/2*6.26*(-30.38)^2
W=2888.8 J

I tried entering that into my online homework system and it came out wrong. The system doesn't really care for the form that the answer is, as long as there are 3 significant digits. That means that I solved for this incorrectly! :cry:

 tiny-tim Mar9-12 02:03 PM

Quote:
 Quote by iJamJL (Post 3806770) x = 0.484t3 - 33.6t dx/dt= 2*(.484)t^2 - 33.
erm :redface: … 3*t2 :wink:

 iJamJL Mar9-12 02:26 PM

Re: Not sure which formula to use! Deals with work and position

Quote:
 Quote by tiny-tim (Post 3806784) erm :redface: … 3*t2 :wink:
lol I wrote that on my scrap paper, but typed and calculated it incorrectly!

My end result came to:

v=(.484)*3*(1.49^2) - 33.6
=(-30.4)m/s

W=1/2*m*v^2
W=1/2*6.26*(-30.4^2)
W=2892.6 J

That's still coming out as incorrect.

 tiny-tim Mar9-12 03:07 PM

Quote:
 Quote by iJamJL (Post 3806820) W=1/2*m*v^2
no, W = 1/2*m*v22 - 1/2*m*v12 :rolleyes:

 gneill Mar9-12 03:18 PM

Re: Not sure which formula to use! Deals with work and position

Quote:
 Quote by iJamJL (Post 3806770) Thanks for the replies! I didn't realize that until you pointed it out! I tried it out, but I think I made a mistake somewhere. Here's what I did: To find the final velocity, which for this case is at 1.49s, we take the first derivative at 1.49: x = 0.484t3 - 33.6t dx/dt= 2*(.484)t^2 - 33.6 v= (-30.38)m/s W=ΔKE W= 1/2*m(v-final^2) - 1/2*m(v-initial^2) **v-initial is 0 because t=0, x=0**
Just because t=0 and x=0, that does not imply that v(0) = 0. Plug t=0 into your velocity formula.

 iJamJL Mar9-12 03:19 PM

Re: Not sure which formula to use! Deals with work and position

Quote:
 Quote by tiny-tim (Post 3806888) no, W = 1/2*m*v22 - 1/2*m*v12 :rolleyes:
Wow, I'm bad at this. I initially thought that v-initial would be 0, but then I looked back and realized it's not. *facepalm*

v(0)=(-33.6)m/s
v(1.49)=(-30.4)m/s

W=1/2*m*(v2^2) - 1/2*m*(v1^2)
W=1/2*6.26(-30.4^2) - 1/2*6.26*(-33.6^2)
W=2892.6 - 3533.6
W=(-641) J

Did I finally get the correct answer? :approve:

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