Physics Opponauts! I request assistance!

"Haimatox" (haimatox) 11/16/2014 at 21:22 • Filed to: None

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I'm studying for a AP Physics test tomorrow on work/energy. There is a question about work done by a general variable force, and I don't understand this part of things. Here's the question. "A single force acts on a 3.0kg particle-like object in such a way that the position of the object as a function of time is given by x = 3.0 t - 4.0 t ^2 + 1.0 t ^3, with x in meters and t in seconds. Find the work done on the object by the force from t = 0 to t = 4s." Help me out, please?

I'm not sure if I am allowed to post this, because it's so ridiculously off-topic. Here's a Blipshift design I did last night, that way this post is still car-related.

JQJ213- Now With An Extra Cylinder! > Haimatox 11/16/2014 at 21:27

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This is Oppo! Virtually anything is allowed here! We have some really smart folks here who can help.

Sadly I am not one of them.... good luck

Arch Duke Maxyenko, Shit Talk Extraordinaire > Haimatox 11/16/2014 at 21:27

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w=fd and f=3, so just plug in 4 into your equation and it'll give you the d. Easy shit, man.

Haimatox > Arch Duke Maxyenko, Shit Talk Extraordinaire 11/16/2014 at 21:29

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I presume this was a sarcastic comment and not an actual explanation? haha.

BrownMiataDieselWagon > Arch Duke Maxyenko, Shit Talk Extraordinaire 11/16/2014 at 21:32

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Bro, where's your integrals? Do you even calculus?

ly2v8-Brian > Haimatox 11/16/2014 at 21:33

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No it's not Work=Force*Distance. So, Work/Force=Distance.

unclevanos (Ovaltine Jenkins) > Haimatox 11/16/2014 at 21:34

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Since W= F*d and F= ma, plug in for t to get the distance first. Then derive the equation two time as this yields in the acceleration equation. The first derivative will yield dx/dt which give you velocity, then dv/dt yields the acceleration. Once you get the acceleration, plug in for t again. So now you can find F and you now also have d to solve for work.

thedevilinside > Haimatox 11/16/2014 at 21:34

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Work is the integral of Force times velocity right? You can get velocity as derivative of x(t) and force as mass time derivative of velocity?

Arch Duke Maxyenko, Shit Talk Extraordinaire > Haimatox 11/16/2014 at 21:35

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Actual explanation.

promoted by the color red > Haimatox 11/16/2014 at 21:35

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I'm a little rusty, but you can get acceleration and distance based on that position equation (diff 1x for velocity, diff again for accel). Then f=ma, so a little this, a little that, and you can figure out w=f*d. :)

BrownMiataDieselWagon > Haimatox 11/16/2014 at 21:38

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EDIT: nvm, read question wrong.

 V8 Rustler > unclevanos (Ovaltine Jenkins) 11/16/2014 at 21:40

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Displacement is a better term.

Axial > unclevanos (Ovaltine Jenkins) 11/16/2014 at 21:45

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Was about to post this, but had to do maths first.

yamahog > Haimatox 11/16/2014 at 21:54

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You may want to keep studying ;)

Haimatox > yamahog 11/16/2014 at 22:00

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... Not very comforting.

unclevanos (Ovaltine Jenkins) >  V8 Rustler 11/16/2014 at 22:03

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I was a distance runner, I use distance a lot.

wallaby13 > Haimatox 11/16/2014 at 22:45

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Work = F*d. Take the second derivative of x(t) to find a(t). Then multiply by m to get m*a(t), then multiply by x(t) again and you have an expression W(t)= m*a(t)*x(t). You can simply plug in 0 and 4 to get W(4) -W(0) = Total Work. However your initial position is zero so you don't need the W(0) term. However it's always good practice to make sure you start at 0.

GTO62 > Haimatox 11/17/2014 at 01:35

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Easiest way is to use the work-energy theorem: W=DeltaK, where K is kinetic energy K(t)=1/2mv(t)^2, so W=K(4)-K(0)=1/2m[v(4)^2-v(0)^2]. To find v as a function of time, take the first derivative of the position v=dx/dt=3.0-8.0t+3.0t^2. At t=0 s: v=3 m/s. At t=4 s: v=3.0 -8*4+3*16=19 m/s. So, W=1/2*3.0*[19^2-3^2]=528 J.