Not sure which formula to use Deals with work and position

In summary, the homework statement states that a one-dimensional force acts on a particle of mass 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.
  • #1
iJamJL
58
0

Homework Statement


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.

Homework Equations


F=mg
F=ma
W=F*distance

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)JW=Fs
W=mg*s
W=6.26*(-9.81)*(-48.5)
W=2978.4 J
 
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  • #2
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.
 
  • #3
hi iJamJL! :smile:
iJamJL said:
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:)
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:
 
  • #4
Thanks for the replies!

tiny-tim said:
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:
 
  • #5
iJamJL said:
x = 0.484t3 - 33.6t
dx/dt= 2*(.484)t^2 - 33.

erm :redface:3*t2 :wink:
 
  • #6
tiny-tim said:
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.
 
  • #7
iJamJL said:
W=1/2*m*v^2

no, W = 1/2*m*v22 - 1/2*m*v12 :rolleyes:
 
  • #8
iJamJL said:
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.
 
  • #9
tiny-tim said:
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:
 

1. How do I determine the formula for work and position?

To determine the formula for work and position, you need to identify the specific variables involved in the problem. Then, you can use the following formula: Work = Force x Displacement, where Force is measured in Newtons (N) and Displacement is measured in meters (m).

2. Can I use the same formula for all problems involving work and position?

No, the specific formula you use will depend on the given variables. For example, if you are given the force and displacement but need to find the work, you would use the formula mentioned in question 1. However, if you are given the work and need to find the force, you would rearrange the formula to be Force = Work / Displacement.

3. How do I know which units to use in the formula for work and position?

The units you use for the variables in the formula will depend on the units given in the problem. It is important to convert all units to the correct form before plugging them into the formula. For example, if the displacement is given in centimeters, it should be converted to meters before using the formula.

4. Are there any other formulas that can be used for work and position?

Yes, there are other formulas that can be used for specific situations. For example, if the force is not constant, you can use the formula Work = Force x Distance, where Distance is measured in meters. Additionally, if the force is not in the same direction as the displacement, you can use the formula Work = Force x Displacement x cos(theta), where theta is the angle between the force and displacement vectors.

5. What is the difference between work and position?

Work is a measure of the energy transferred when a force is applied over a displacement. It is a scalar quantity, meaning it only has magnitude. Position, on the other hand, is a vector quantity that describes the location of an object in relation to a reference point. While both involve displacement, work is a measure of the energy involved, whereas position is a measure of the location.

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