Integrating Work from Force(time)

[ecophysicist]

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?

 

[phyzmatix]

Could it be:

$$W=\int_{x_i}^{x_f}F(x)dx$$
$$=\int_{x_i}^{x_f}Ma_x dx$$
$$=\int_{x_i}^{x_f}M \frac{dv}{dt}dx$$

But, by the chain rule

$$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}$$

so

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

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

so, finally

$$W=(19m/s)M$$

?

NB: I'm not 100% sure of myself, so don't take this as a definite!

 

[Redbelly98]

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.

 

[ecophysicist]

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.

 

[phyzmatix]

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.

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?

I'd love to see those formulas as I have to admit I can't really visualise your solution

 

[Redbelly98]

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.

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:

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

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

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. basically, 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.

 

[phyzmatix]

You were okay up to here:

[tex]=M{[\frac{1}{2}v^2]}\right|_{v_i}^{v_f} $$<br /> After that, the v² became simply v (the "²" was omitted).$$[/tex]

[tex]$$<br /> Smacks forehead while muttering "<span style="font-size: 9px">stupid idiot" to self. <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/redface.gif" class="smilie" loading="lazy" alt=":redface:" title="Red Face :redface:" data-shortname=":redface:" /><br /> <br /> Thanks Red!</span>$$[/tex]