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Please help - Motion of a hockey puck

  1. Aug 28, 2012 #1
    Please help -- Motion of a hockey puck

    1. The problem statement, all variables and given/known data
    Hi guys, can you please help me answer this problem?


    A hockey puck with mass 0.160 kg is at rest at the origin on the horizontal, frictionless surface of the rink. At time t=0 a player applies a force of 0.250 N to the puck, parallel to the x-axis; he continues to apply this force until t=2.00 s. (a) What are the position and speed of the puck at t=2.00 s? (b) If the same force is again applied at t=5.00 s, what are the position and speed of the puck at t=7.00s


    I got the 1st one, but I couldn't get the second one.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 28, 2012 #2

    jedishrfu

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    Staff: Mentor

    Re: Please help

    welcome to PF!

    You need to show your work before we can help. So why not show us how you solved for (a) and where you got stuck for (b).
     
  4. Aug 28, 2012 #3
    Re: Please help

    Given: F=0.250 N
    m=0.160 kg
    F=ma

    a=F/m

    leads to:

    a=0.250/0.160

    i got a= 1.5625

    then I used the formula a=(2s)/(t^2) to get the distance

    then I got s=3.13 m



    then, couldn't solve for b, please help
     
  5. Aug 28, 2012 #4
    Re: Please help

    Btw, here are the answers

    a) 3.13m, 3.13m/s
    b) 21.9m, 6.25m/s
     
  6. Aug 28, 2012 #5

    jedishrfu

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    Staff: Mentor

    Re: Please help -- Motion of a hockey puck

    extend (a) to 5 secs what is dist and velocity then these become the initial conditions in your eqns.

    so at 2 secs the force ends and the puck glides along at a constant velocity (assume no friction) for 3 secs (so you're now at 5 sec mark)

    vfinal = vinitial + a * t where t =7 - 5

    dfinal= dinitial + vinitial * t + 1/2 a * t
     
  7. Aug 28, 2012 #6
    Re: Please help -- Motion of a hockey puck

    ahm, sir, where would I get my vinitial?
     
  8. Aug 28, 2012 #7
    Re: Please help -- Motion of a hockey puck

    Apply Newton's first law of motion.
     
  9. Aug 28, 2012 #8
    Should V initial at 5s be any different than V at 2s? Can you find V at 2s?
     
  10. Aug 28, 2012 #9
    Re: Please help -- Motion of a hockey puck

    sir, I got the final velocity

    I converted a= 1.563 m/(s^2) to m/s

    so I got v= 3.13 m/s

    then I used it as my vinitial to get my vfinal using the equation you gave me, my vfinal= 6.256 m/s

    but when I tried to substitute that to the 2nd equation you gave me, I didn't got the correct answer.
     
  11. Aug 28, 2012 #10
    Re: Please help -- Motion of a hockey puck

    @Jakeus314: I don't think they are different sir, ahm, already got the answers for 2s sir
     
  12. Aug 28, 2012 #11
    Re: Please help -- Motion of a hockey puck

    I have another question, I got letter a) but why is the unit I got is m/(s^2) the answer should be m/s, but I got the right answer. so I just randomly thought about converting the acceleration to velocity, because the unit of acceleration is m/(s^2) and the velocity is m/s, but if I do that in a) the unit would be correct but the answer would be wrong.
     
  13. Aug 28, 2012 #12
    try not to say you converted an acceleration in to a velocity... You may find a velocity using an acceleration etc.

    V = V0 + a*t
    V from pt A is good for V0 for pt B why?
     
  14. Aug 28, 2012 #13
    Good.. Because no force is acting on the puck from 2s to 5s.
     
  15. Aug 28, 2012 #14
    The position in pt B comes from three distances traveled while moving. Pt A distance contributes some, then some more while moving at a constant speed, then more while accelerating again. Find the position for pt B from those pieces and you'll do fine.
     
  16. Aug 28, 2012 #15
    Re: Please help -- Motion of a hockey puck

    ahm sir, sorry bout the convertion thingy, I don't get is sir.
     
  17. Aug 28, 2012 #16
    Re: Please help -- Motion of a hockey puck

    I tried to find the distance for 5s, so I used the same acceleration


    so my equation is (1.563*(5^2))/2

    then I got 19.5375
     
  18. Aug 28, 2012 #17
    Maybe I'm just being picky. When people say it that way I start thinking they're missing some algebra/units/equation concepts.

    Is English not your first language?
     
  19. Aug 28, 2012 #18
    5^2 is an error here. Has the puck been accelerating for 5 seconds?
     
  20. Aug 29, 2012 #19
    Check my post 14 again. Do pt B by adding three things. Pt A answer, d=r*t during the 3 seconds of no acceleration, and last add the distance traveled while accelerating for 2 seconds with a starting velocity of ptA velocity.
     
  21. Aug 29, 2012 #20
    I can't say for sure why you got acceleration units.

    Do unit analysis to see what it should be, given the values you added or multiplied.
     
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