Integrating Work from Force(time)

In summary, the conversation discussed a challenging physics problem involving integrating force to find work. The problem involved finding the acceleration, work, power, and average power of a rabbit moving according to a given equation. Solutions were provided, but not completely accurate until the Latex code was corrected. The conversation also addressed the issue of giving detailed solutions to homework problems and the use of Latex in posting equations.
  • #1
ecophysicist
2
0
Hi folks:
Question for you guys. I've generalized a question that keeps coming up in class...but that keeps going unexplained. This is a somewhat involved problem which requires integrating force to find work. Thing is, force is always integrated as a function of position NOT time. This type of problem keeps throwing me for a loop because you end up finding a force equation that's basically F(t)...and end up having to integrate it in terms of dx to find work. I know I'm missing something here, but I can't figure out what. It's stage B where things start getting tricky.

Here's the challenge problem of the day then...

Imagine you have a rabbit of mass M whose position is given as a function of time is given by x = (3t - 4t^2 + t^3)i, where x is in meters and t is in seconds.
(A) What is the acceleration of the rabbit for any time t?
(B) What is the work done by the rabbit for the first five seconds of its journey?
(C) What is the power of the rabbit at t = 3.0 seconds?
(D) What is the average power of the rabbit during the first five seconds of its journey?
 
Physics news on Phys.org
  • #2
Could it be:

[tex]W=\int_{x_i}^{x_f}F(x)dx[/tex]
[tex]=\int_{x_i}^{x_f}Ma_x dx[/tex]
[tex]=\int_{x_i}^{x_f}M \frac{dv}{dt}dx[/tex]

But, by the chain rule

[tex]\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/tex]

so

[tex]W=\int_{v_i}^{v_f}Mv dv[/tex]
[tex]=M{[\frac{1}{2}v^2]}\right|_{v_i}^{v_f}[/tex]
[tex]=\frac{1}{2}M[v_f-v_i][/tex]

we know that [tex]v_f=\frac{dx}{dt}]_{t=5}=38m/s[/tex] and [tex]v_i=\frac{dx}{dt}]_{t=0}=0m/s[/tex]

so, finally

[tex]W=(19m/s)M[/tex]

?

NB: I'm not 100% sure of myself, so don't take this as a definite!
 
  • #3
In the future, please don't give detailed solutions to homework problems. It is our policy to help by giving hints, not answers, and only when somebody has shown an attempt at solving the problem themself.
 
  • #4
Thanks for the help physzmatix. The solution, as it turns out, isn't quite that though. It involves integrating work, taking the derivative of x to be dx in terms of dt, and plugging that dx in for the one in the work integral to make things in terms of t. I've got no clue how to type the formulas out, but given the mass 1.5, the answer comes out to

definite integral [0,5] of (27t^3 - 108t^2 + 123t - 36) dt

-> which ends up being around 1000 watts.And RedBelly, respectfully, there's no need for the wiseguy attitude. This wasn't a homework problem, as I explicitly stated from the beginning. Maybe most people write out their homework in part As and part Bs, but that's how I take my notes. Don't jump to conclusions.
 
  • #5
Redbelly98 said:
In the future, please don't give detailed solutions to homework problems. It is our policy to help by giving hints, not answers, and only when somebody has shown an attempt at solving the problem themself.

Hi Redbelly98. I'm aware of that, but since I found the problem immensely challenging myself, I wasn't sure if I was aiding or hindering ecophysicist. I thought of my contribution more as an extension to eco's question (for my own sake) than an answer. Perhaps you can give some input as to where my strategy failed, since ecophysicist says the answer turns out to be something quite different?

Thanks.

ecophysicist said:
Thanks for the help physzmatix. The solution, as it turns out, isn't quite that though. It involves integrating work, taking the derivative of x to be dx in terms of dt, and plugging that dx in for the one in the work integral to make things in terms of t. I've got no clue how to type the formulas out, but given the mass 1.5, the answer comes out to

What are the odds of you brushing up on your Latex skills and showing us? :biggrin:

I'd love to see those formulas as I have to admit I can't really visualise your solution :redface:
 
  • #6
ecophysicist said:
This wasn't a homework problem, as I explicitly stated from the beginning.

I'm sorry but it did look like a homework problem to me, the "challenge problem of the day" part of it at least. I didn't see where you explicity said that was not homework.

We do get people just giving out homework answers. I wasn't trying to be a wiseguy, just trying to educate people about how things are done at PF.

phyzmatix said:
Perhaps you can give some input as to where my strategy failed, since ecophysicist says the answer turns out to be something quite different?

You were okay up to here:

[tex]
=M{[\frac{1}{2}v^2]}\right|_{v_i}^{v_f}
[/tex]

After that, the v2 became simply v (the "2" was omitted).

ecophysicist said:
I've got no clue how to type the formulas out...

Yeah, that was something I resisted when I first started posting here. But it's not so hard once you get the hang of it. Basicly, LaTex equations get typed in between [ tex ] and [ /tex ], but you omit the spaces. For examples, click on some of phyzmatix's equations and the Latex code will show up.

More information is here:
https://www.physicsforums.com/misc/howtolatex.pdf
https://www.physicsforums.com/showthread.php?t=8997

Also, if you click the ∑ button, a menu of LaTex symbols will pop up.

Hope that helps.
 
  • #7
Redbelly98 said:
You were okay up to here:

[tex]=M{[\frac{1}{2}v^2]}\right|_{v_i}^{v_f} [tex]

After that, the v2 became simply v (the "2" was omitted).

Smacks forehead while muttering "stupid idiot" to self. :redface:

Thanks Red!
 

Related to Integrating Work from Force(time)

1. What is "Integrating Work from Force(time)"?

"Integrating Work from Force(time)" is a scientific method used to quantify the amount of work done on an object by a varying force over a period of time. It involves calculating the area under a force vs. time graph.

2. How is "Integrating Work from Force(time)" different from calculating work using W = Fd?

While both methods calculate work, "Integrating Work from Force(time)" takes into account varying forces over a period of time, while W = Fd only considers a constant force over a specific distance.

3. What is the significance of "Integrating Work from Force(time)" in scientific research?

"Integrating Work from Force(time)" allows scientists to accurately measure the work done on an object in situations where the force applied is not constant. This is important in fields such as physics, engineering, and biomechanics.

4. What are the units for work when using "Integrating Work from Force(time)"?

The units for work when using "Integrating Work from Force(time)" are joules (J), which is the same as the units for work in the equation W = Fd.

5. Can "Integrating Work from Force(time)" be used for non-constant forces?

Yes, "Integrating Work from Force(time)" can be used for any force that varies over a period of time. It is a more accurate method of calculating work compared to using W = Fd for non-constant forces.

Similar threads

  • Introductory Physics Homework Help
Replies
17
Views
834
  • Introductory Physics Homework Help
Replies
2
Views
256
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
37
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
657
  • Introductory Physics Homework Help
Replies
8
Views
257
  • Introductory Physics Homework Help
Replies
2
Views
815
  • Introductory Physics Homework Help
Replies
13
Views
827
  • Introductory Physics Homework Help
Replies
8
Views
292
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top