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I have this problem :

A 2.19-kg cart on a long, level, low-friction track is heading for a small electric fan at 0.21m/s . The fan, which was initially off, is turned on. As the fan speeds up, the magnitude of the force it exerts on the cart is given by $at^2 $ , where a = 0.0200 $N/s^2$.

I need to find the speed of the cart after 3.5 s of the fan being on and after how many seconds the velocity of the car will equal 0.

I'm not entirely sure what I should do to solve, I've tried drawing a free body diagram because that's the section this problem is from but that doesn't feel right. I've tried

$$\Sigma F=F(fan)=ma \space \\ \Sigma F=mg-N=ma=0 \\ v=v(initial)+at\\v=0.53$$

Which I am pretty sure is all wrong.

How is this suppose to be diagramed? Is this actually a free body diagram problem? If it is, how do I account for velocity since it isn't coming from the object. If it is not a free body diagram problem, do I just use general velocity equations to solve? I know conservation of mass and energy do not apply so I am not sure what else to do.

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closed as off-topic by ACuriousMind, BMS, Brandon Enright, JamalS, Rob Jeffries Nov 4 '14 at 13:23

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  • $\begingroup$ You can draw a FBD for this but it wont help much. it will only have one force, the one exerted by the fan. $\endgroup$ – stackErr Nov 3 '14 at 21:14
  • $\begingroup$ Okay, that's pretty much what I have been seeing. What should I do with the force then? Is the force even relevant to a problem if it's the only force acting? $\endgroup$ – K P Nov 3 '14 at 21:16
  • $\begingroup$ Have you learned integration yet? The way the question is phrased is that the force of the fan is varying with time, so it looks like some kind of integration is needed. $\endgroup$ – Yandle Nov 3 '14 at 21:18
  • $\begingroup$ @Yandle I dont think that he would require integration...it would help though $\endgroup$ – stackErr Nov 3 '14 at 21:18
  • $\begingroup$ I have learned it in a general sense, I'm in calc 3 right now. Physics professor has not given any problems or done examples of problems using integration in the past though, so I don't know why he would do a problem requiring it now. $\endgroup$ – K P Nov 3 '14 at 21:20
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The rate of change of velocity (momentum) of an object is proportional to the force:

$$m\Delta v = F \Delta t$$

In your case you have force $F(t)=at^2$, so you can now write the equation of motion:

$$m \frac{dv}{dt} = at^2$$

See if you can get velocity as a function of time from here... note that you need to pay attention to the direction of the force and the initial velocity (they are in opposite directions...)

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  • $\begingroup$ So if I do this I get $$dv/dt=at^2/m \\ v=at^3/3m \\ v= 0.626$$ Which isn't the right answer. So I am assuming I missed a step? Am I supposed to do something with the initial velocity? But this minus initial still wouldn't be right, I don't think. I'm not sure what else you would do besides that. $\endgroup$ – K P Nov 3 '14 at 21:42
  • $\begingroup$ You need to think about the initial velocity, yes. At t=0, velocity is -0.21 m/s . But when I evaluate the expression you get for $v$, I don't get your answer. Not even close. Check your calculation. I get $\Delta v = 0.02*3.5*3.5*3.5 / (3 * 2.19) = 0.131 m/s $ $\endgroup$ – Floris Nov 3 '14 at 21:51
  • $\begingroup$ @KP when you perform the integral there is an arbitrary constant of integration. The integral is really $v=at^3/3m+C$. You need to plug in v(t = 0) = -0.21m/s $\endgroup$ – Yandle Nov 3 '14 at 21:54
  • $\begingroup$ Ahh I seee. Thank you so much! I forgot to put the entire denominator in parenthesis on my calculator so it was dividing by 3 instead of 3*2.19. Okay. I got the answer now. Do you know if I need this found velocity to solve for when t=0? My professor in class today said that all problems should be solved using what was originally given. $\endgroup$ – K P Nov 3 '14 at 21:56
  • $\begingroup$ Calculators are only as good as the person driving it... Just eyeballing the expression tells me "first three terms multiply out to 0.2 since 3.5*3.5 is about 10, then I have two terms 3.5/3 which roughly cancel, and then another divide by 2. Answer should be about 0.1." - and then you check with your calculator... You don't need this intermediate velocity (remember to subtract 0.21 m/s!) to get the answer to the second question (which is "solve for t when v = 0", not what you wrote in your comment...) $\endgroup$ – Floris Nov 3 '14 at 22:01

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