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Mechanics Free Response

  1. Nov 16, 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
    honestly, I have no idea.
     
  2. jcsd
  3. Nov 16, 2008 #2
    ok what i have so far is that the velocity = 1.5t^2+2, which means the speed at 0= 2.
    Then when it asks for the kinetic energy im not sure if it would be 1/2(100)(2^2) or 1/2 (100)(1.5t^2+2).
     
  4. Nov 16, 2008 #3
    anyone
     
  5. Nov 16, 2008 #4

    alphysicist

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    Hi dbb2112,

    Your answer for part a looks right to me.

    For part b, neither one is quite right. The first form 1/2(100)(2^2) is the kinetic energy at t=0.

    The second one:

    1/2 (100)(1.5t^2+2)

    is closer, but you need to square the speed. What do you get for the other answers?
     
  6. Nov 16, 2008 #5
    O ya i just forgot to put the squared. But for the net force F=ma so i would get 100(3t) or 300t for the net force?
     
  7. Nov 16, 2008 #6

    alphysicist

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    That looks right to me.
     
  8. Nov 16, 2008 #7
    Um about the kinetic energy again whenever i have 1/2(100)(1.5t^2+2)^2 would i have to square the 1.5 t part or could i leave it as 50(1.5t^2+2)^2
     
  9. Nov 16, 2008 #8

    alphysicist

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    I believe this last expression is right (I'm not sure what you mean by square the 1.5 t part).
     
  10. Nov 16, 2008 #9
    nevermind. About the power being delivered im not really sure what to do here.
     
  11. Nov 16, 2008 #10

    alphysicist

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    What two ways of calculating power are there; what are their formulas and which one applies to this case?
     
  12. Nov 16, 2008 #11
    Power=work/time?
     
  13. Nov 16, 2008 #12

    alphysicist

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    No, that is the average power (over an interval). What's the other?
     
  14. Nov 16, 2008 #13
    power= force*velocity
     
  15. Nov 16, 2008 #14

    alphysicist

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    That's right.

    How would you find the net work done from t=0 to t=2?
     
  16. Nov 16, 2008 #15
    honestly, i am not sure.
     
  17. Nov 16, 2008 #16

    alphysicist

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    What is the work equal to? In other words, why is it impotant? For example, if I tell you that 10J of work is done on an object, what do you know about that object?
     
  18. Nov 16, 2008 #17

    LowlyPion

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  19. Nov 16, 2008 #18
    How about the change in Kinetic energy is equal to the net work
     
  20. Nov 16, 2008 #19
    I know i but when you got offline i wanted someone else and therefore posted a new one.
     
  21. Nov 16, 2008 #20

    alphysicist

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    That's correct about the net work.

    However, duplicate threads are not allowed on this forum. Please stay with the original thread for this problem.
     
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