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Force, momentum, work, and power eqzn

  1. Nov 13, 2007 #1

    ~christina~

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    [SOLVED] Force, momentum, work, and power eqzn

    1. The problem statement, all variables and given/known data
    The coordinate of a 2.00kg object in linear motion is described by the function:

    [tex]x(t)= (2.00m/s^3)t^3 -(7.00m/s^2)t^2 + (7.00m/s)t[/tex]

    for the time interval t= 0s to t= 2.00s. The center of mass of this object can be treated as a point particle.

    a) find change in momentum of the mass for the time interval given.

    b) sketch graphs of the momentum of the object and the force on the object as functions of time

    c) How much power is delivered to the particle at any time t?

    d) How much work is done on the object fo the time interval given?

    2. Relevant equations
    F= ma

    P= F* v

    I= [tex]\int \sum F dt[/tex]


    3. The attempt at a solution

    I was okay I think for part a) then I didn't really know what to do. Here's what I did do.

    Does it matter that this is treated as a point particle at it's center mass?

    a) to get change in momentum

    v(t)= x'(t)= (6.00m/s^3) t^2 - (14.00m/s^2) t + 7.00m/s

    ti= 0s

    v(0)= 7.00m/s

    tf= 2.00s

    v(2.00)= (6.00m/s^3)(2)^2 - (14.00m/s^2) (2) + 7.00m/s

    v(2.00)= 2.00m/s

    I= Pf- Pi = m(vf-vi)= 2.00kg (7.00m/s - 3.00m/s)

    I= 8 kg*m/s ==> is this alright?? (not sure about if I did it the right way)


    b) graph

    how would I sketch graphs of the momentum of the object and the force on a object as a function of time?


    Once again I'm not sure but I think this is how I'd do this...

    to get momentum as a function of time...I think I'd multiply the mass into the velocity equation:

    p= mv

    v(t)= (6.00m/s^3) t^2 - 14.00m/s (t) + 7.00m/s

    m= 2.00kg

    p(t)= (12.00kg*m/s ^3) t^2 - 28.00m/s^2 (t) + 7.00m/s


    For the force as a function of time I think I'd have to multiply the mass * acceleration

    a(t)= v'(t)= (12.00m/s^3) t- (14.00m/s^2)

    F= m*a

    m= 2.00kg

    F(t)= (24.00kg*m/s^3) t - (28 kg*m/s^2)

    graph: [​IMG]

    c) How much power is being delivered to the particle at any time t?

    Not sure..

    P= F*V

    would I go and multiply the force equation with the velocity equation?

    F(t)= (24.00kg*m/s^3) t - (28kg*m/s^2)

    v(t)= (6.00m/s^3) t^2 - 14.00m/s (t) + 7.00m/s

    Not quite sure how to multiply those two though...Help?
    d)How much work is done on the object for the time interval given?

    since W= Fd

    would I go and multiply the
    force eqzn that I found with the distance equation given??

    I need help on this too...


    Thanks
     
    Last edited: Nov 13, 2007
  2. jcsd
  3. Nov 13, 2007 #2

    andrevdh

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    a)

    I= Pf- Pi = m(vf-vi)= 2.00kg (7.00m/s - 3.00m/s)

    this should be (3.00 - 7.00)

    b) Seems fine. I can unfortunately not access your image.

    c) Personally I would attempt it by drawing both F(t) and v(t) graphs and generate some points for the P(t) graph by multiplying a few F and v values at the same t values. To generate the formula - just multiply the two equations like you did in maths...

    (at - b)(ct^2 - dt + e) = .....

    d) The work done is given by

    [tex]\int F\ dx[/tex]

    since the force changes with time (displacement).
     
    Last edited: Nov 13, 2007
  4. Nov 13, 2007 #3

    ~christina~

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    Then I get a -8.00kg*m/s

    It works fine for me..I just click on the pic..and it pops up in another window.

    But how would I do that if I just had the equation and didn't want to do the plotting would I multiply them together?

    It seems kinda long...My method may be incorrect though

    After I get the integral of the Force equation (I think that's what you mean)

    I get back to

    W= m[tex]\int a dt [/tex]

    which equals the power vs time equation...
    F(t)= (24.00kg*m/s^3 )t - (28.00m/s^2)

    W= [tex]\int F dt = \int m*a dt = (12.00kg*m/s^3)t^2 - (28.00m/s^2)t + 7.00m/s[/tex]


    Is this alright?
    I still don't know how to do c)
     
    Last edited: Nov 13, 2007
  5. Nov 13, 2007 #4

    ~christina~

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    Hm..I guess nobody is going to help me...even if I say please?

    part c needs most help...
     
  6. Nov 13, 2007 #5
    yes u are right in part c and D.. u need to multiply those.. gj :)
    by the way.. im not sure how accurate your graph has to be.. but your p(t) should be
    p(t)= (12.00kg*m/s ^3) t^2 - 28.00m/s^2 (t) + 14.00 Kg*m/s and your f(t) graph has the x intercept at the wrong t value.. it should intersect where p(t) reaches a min (aka d(pt)/dt =0)
     
  7. Nov 13, 2007 #6

    ~christina~

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    darn...I've forgotten how to do that..I only remember foil...

    either blame me or the teachers in hs...(they actually never asked us questions on that or maybe I just blocked out remembering since I probably got the answer wrong)

    you wouldn't just happen to know how to do that?

    Actually that was for momentum which is p= mv

    while power is P= F*V so....I'm still not sure how to multiply both

    F(t)= (24.00kg*m/s^3) t - (28kg*m/s^2)

    v(t)= (6.00m/s^3) t^2 - 14.00m/s (t) + 7.00m/s

    together I think...so I'm not sure where you got the equation for power but If that's what you get from multipying both those together..I'll try it...
     
  8. Nov 13, 2007 #7
    sorry for the late response, haha i am still in hs. I have major hw.. but i can still help :)
    look v(t)= (6.00m/s^3) t^2 - 14.00m/s (t) + 7.00m/s so p(t) <-- momentum = m*v(t)=p(t)= what i said earlier... anyway, like i said, I am not sure HOW accurate the graph has to be. as for the slope, its good.. but it is still not accurate.. anyway, moving on..

    multiply f(t)*v(t) = (24t-28)(6t^2-14t+7)= P(t)<-- power not momentum (honestly, depending on your teacher, if you just write that, that is TECHNICALLY acceptable) but lol i'm not sure HOW i can show you to multiply it out other than actually DOING it.. but its very doable..

    Work is f(t)*x(t)= (24t-28)(2t^3-7t^2 +7t) but see, since u know t is 2, u can just plug it in, and its much easier.
     
    Last edited: Nov 13, 2007
  9. Nov 13, 2007 #8
    look carefully, thats impulse.. work is "dx" not dt.. work is the area under a force x DISTANCE graph.
     
  10. Nov 13, 2007 #9

    ~christina~

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    Forgot to say that since it says "sketch" my instructor would mean not to plot since I kinda got told with exclamation points last time I plotted instead of sketched this that plotting was NOT sketching. A picture depicting what the graph looks like would suffice.

    well I looked it up online and I think I know how to multiply it now. It was kinda interesting how they state on that very page that usually teachers don't teach students to multipy anything beyond the foil method with polynomials which doesn't help.:rolleyes:

    hm...I was wondering however that...
    since

    c) How much power is being delivered to the particle at any time t?

    It would entail me just stating the equation of power as a function of time I guess, right?.

    d) how much work is done on the object for the time interval given..

    with integration from t= 0 to t= 2.00m/s

    since you said

    I guess to get work I would multiply force eqzn times the distance equation and then go and plug in t= 0 into that equation, find work for that time, then find work for t= 2.00s and then add or subtract ?

    I'd suspect that since it doesn't say change in work it would be add.


    Thanks for your help!
     
  11. Nov 13, 2007 #10
    u would subtract.. as usual final - initial.. but initial is zero..
    also, just fyi: work is a scaler quantitiy
     
  12. Nov 13, 2007 #11

    ~christina~

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    if
    is that just a side note? I'd think so b/c I do know that even thought force and distance have x and y components the product is a scalar.

    hey but why would it be subtract??
     
  13. Nov 14, 2007 #12
    ya thats a side note, work is technically W=Fdcos(theta) (its a dot product, I am not sure if u know/remember dot products or cross products, but logically it also makes sense.. anyway, lets not go off at a tangent here)

    u subtract, well I can think of two obvious reasons.. first.. because it is fdcos(theta) (where d is the distance traveled, now.. lets say you were initially at 5 meters and u go to 7 meters, the distance traveled isn't 12, it is 2 --> final-inital ) second.. is along the same lines but harder to explain, but i hope u got the point.
     
  14. Nov 14, 2007 #13

    ~christina~

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    Oh..yep I get it..

    Thanks for your help aq1q :smile:
     
  15. Nov 14, 2007 #14
    your welcome, now i have to get back to my hw
     
  16. Nov 14, 2007 #15

    ~christina~

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    I really appreciate you spending your time to help me and you should get back to your homework :tongue:
     
  17. Nov 14, 2007 #16

    andrevdh

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    On second thought the work can also be evaluated via the work-energy principle.
     
  18. Nov 14, 2007 #17

    ~christina~

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    Oh..I shall try that.

    THANKS Andrevdh
     
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