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


by iJamJL
Tags: deals, formula, position, work
iJamJL
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#1
Mar9-12, 01:18 PM
P: 58
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
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Delphi51
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#2
Mar9-12, 01:38 PM
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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*t rather than the cubic formula of this problem.
tiny-tim
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Mar9-12, 01:39 PM
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hi iJamJL!
Quote Quote by iJamJL View Post
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!

(what is worrying you about that? )
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

iJamJL
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#4
Mar9-12, 01:57 PM
P: 58

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


Thanks for the replies!

Quote Quote by tiny-tim View Post
however it might be simpler to use the work-energy theorem, and calculate the difference in kinetic energy
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!
tiny-tim
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Mar9-12, 02:03 PM
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Quote Quote by iJamJL View Post
x = 0.484t3 - 33.6t
dx/dt= 2*(.484)t^2 - 33.
erm 3*t2
iJamJL
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Mar9-12, 02:26 PM
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Quote Quote by tiny-tim View Post
erm 3*t2
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
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Mar9-12, 03:07 PM
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Quote Quote by iJamJL View Post
W=1/2*m*v^2
no, W = 1/2*m*v22 - 1/2*m*v12
gneill
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Mar9-12, 03:18 PM
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Quote Quote by iJamJL View Post
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
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#9
Mar9-12, 03:19 PM
P: 58
Quote Quote by tiny-tim View Post
no, W = 1/2*m*v22 - 1/2*m*v12
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?


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