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Mechanics free response

  1. Nov 14, 2008 #1
    1. The problem statement, all variables and given/known data

    A 100 kg box is being pulled along the x axis by a student. The box slides along the a rough surface and its position x varies accoring to the equation x=.5tcubed +2t, where x is in meters and t is in seconds.

    a. Determine the speed of the box at t=0
    b. Determine the following as functions of time t.
    i. The kinetic energy of the box
    ii. the net force acting on the box
    iii. The power being delivered to the box.
    c. Calculate the net workdone on the box in interval t=0 t t=2
    d. Indicate below wheter the work done on the box by the student in the interval t=0 to t=2 wold e greater than, less than or eequal to the answer in part c.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 14, 2008 #2

    LowlyPion

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    How would you think to approach the problem?

    You have a displacement/time graph ... so how do you read velocity?
     
  4. Nov 16, 2008 #3
    I dont really know how to approach the probelm or how i read velocity
     
  5. Nov 16, 2008 #4

    LowlyPion

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  6. Nov 16, 2008 #5
    before i go there this is just guessing but would i take the derivitive of the equation to find the velocity?
     
  7. Nov 16, 2008 #6

    LowlyPion

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    Since that's what velocity is - the rate of change of position wrt time - then yes that's what you will need to be doing.
     
  8. Nov 16, 2008 #7
    ok, so at t=0 the velocity would be 2m/s.
    It then asks to Determine the following as functions of time t. I do not understand what that means.
     
  9. Nov 16, 2008 #8

    LowlyPion

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    You have the velocity function.
    You will need the acceleration function too.

    KE - 1/2*m*v2 = 1/2*m*vt2

    F = m*a

    P = w*/t = F*d/t = Ft*vt

    Basically you are using these functions in place of the usual functions you might employ in determining more traditional uniform gravitational kinematics.
     
  10. Nov 16, 2008 #9
    i do not really understand what finding those as functions of time means still. But i am assuming that i should just plug in the velocity i found for kinetic energy and mass they already give you to obtain the answer of 200J. And then derive the velocity again to get the acceleration of 3t? that is where I kind of get lost.
     
  11. Nov 16, 2008 #10

    LowlyPion

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    In b) they are asking you for these things as functions of time. (Remember a is the derivative of v and equals 3*t, which is what you apparently have determined.)

    In c) you need to integrate the product of F and X from t=0 to t=2.
     
  12. Nov 16, 2008 #11
    so with the acceleration being 3t would i sub anything in there or just multiply 3 by the mass.
     
  13. Nov 16, 2008 #12

    LowlyPion

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    No. Wherever you need to know a is where you would sub in 3*t, just as for velocity wherever you have v you sub in (3*t2/2 +2) wherever you have v.

    These variables are functions of time. And the equations in b they are asking for are in terms of these.
     
  14. Nov 16, 2008 #13
    O ok so the force woudl be 300t.
     
  15. Nov 16, 2008 #14
    and for the next part, i would integrate 300t *.5tcubed+ 300t*2t?
     
  16. Nov 16, 2008 #15

    LowlyPion

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    That looks correct. It would be the definite integral over the time given.
     
  17. Nov 16, 2008 #16
    the only problem now is im not really sure how to do that.
     
  18. Nov 16, 2008 #17
    would it be 30t^5+300t^3? and then just plug in the times they give me?
     
  19. Nov 16, 2008 #18

    LowlyPion

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    It's 1/3 of 600 isn't it?
     
  20. Nov 16, 2008 #19
    o yes so it would be 30t^5+200t^3 but after that im not too sure what to do would i plug in 0 and then plug in 2
     
  21. Nov 16, 2008 #20
    could you tell me why we integrated there also?
     
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