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Applications of Newton's Law

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-3-30_0-16-10.png

    upload_2016-3-30_0-15-53.png

    2. Relevant equations
    d = v1(t) + ½ a (t)^2
    Newtons second Law
    3. The attempt at a solution
    ok, so i know that once the puck leaves the stick, the only force acting on it would be the frictional force (1.0 N). From reading other forums, i guess i need to calculate the acceleration first, then use the acceleration to calculate the distance travelled in 3.0 s.
    i just dont know what numbers and formulas to use, to solve for the distance, im really confused :P
    anyhelp would be appreciated. thanks.
     
  2. jcsd
  3. Mar 29, 2016 #2
    i do not know ,whether you know newtons laws or not but a force applied on a body produces acceleration equal to force divided by mass of the body- if the force applied is in opposite direction to already moving mass ,this will produce acceleration in the opposite direction called deceleration.
    so in your equation a=frictional force/mass of the puck ; but it will be negative.
     
  4. Mar 30, 2016 #3
    so the frictional force is 1.0 N and the mass of the puck is 164 g or 0.164 kg.
    a = 1.0 N / 0.164 kg = 6.1 m/s^2 [forward]
    ??
     
  5. Mar 30, 2016 #4

    PeroK

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    That looks right so far, although I'm not sure what [forward] means. There are really two aspects to a problem like this:

    1) To understand what is happening physically.

    2) Using the equations and numbers to get a solution.

    Do you understand the problem? After the puck has been hit, does it:

    a) Race across the ice getting faster and faster
    b) Slow down due to friction?
     
  6. Mar 30, 2016 #5
    no-forward as the friction opposes the motion.
    so you know the acceleration produced by frictional force on the puck - care should be taken that it will act resisting the motion so you can use your equation to find the distance traversed by the puck which has the initial velocity given ... and the time taken .
    i think you can calculate the distance .
     
  7. Mar 30, 2016 #6
    "a) Race across the ice getting faster and faster
    b) Slow down due to friction?"

    so after the puck has been hit, it will slow down due to the force of friction (which is 1.0 N [backwards], because friction opposes the puck in the opposite direction?)
    i wrote [forward] to refer to the puck accelerating in that direction. is it supposed to be backwards since friction opposes the motion of the puck??

    so to solve for the distance the puck travels in 3.0 s, i substituted all the known information into the following equation:
    d = v1(t) + ½ a (t)^2
    d = 0 + 1/2 x(6.1 m/s^2) x (3.0s)^2
    d = 27.45 m
    but im assuming this is incorrect, im not sure if the initial velocity is 0 or 45 m/s, it was mentioned that the puck was initially at rest meaning it had a velocity of 0 initially?? i dont know if im correct here.

    so here is my second answer using 45 m/s as the initial velocity
    d = (45 m/s) (3.0 s) + 1/2 (6.1 m/s^2) (3.0s)^2
    d = 135 + 27.45
    d = 162.45
    so the puck will travel about 162 m in 3.0 s??
     
  8. Mar 30, 2016 #7

    haruspex

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    I'm afraid both those attempts are wrong.
    In the first one, you took the initial velocity as zero. The SUVAT equations (s=vit+at2/2 etc.) are only valid for constant acceleration. The puck is first accelerated from 0 to 45m/s, then decelerates, so acceleration is not constant throughout. It is only constant having reached 45 m/s, so that must be your initial velocity.
    In the second, you need to get the signs right. Take the positive direction as being the direction in which the puck was hit. So the initial velocity is +45m/s, but what is the acceleration now? Remember, it will slow down, not speed up.
     
  9. Mar 31, 2016 #8
    so would the acceleration be -6.1 m/s^2 since the puck slows down.
     
  10. Mar 31, 2016 #9

    PeroK

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    Hallelujah!
     
  11. Mar 31, 2016 #10
    d = (+45 m/s) (3.0 s) + 1/2 (-6.1 m/s^2) (3.0s)^2
    d = 135 - 27.45
    d = 107.55
    so the puck will travel about 108 m in 3.0 s
    ??
     
  12. Mar 31, 2016 #11
    seems good
     
  13. Mar 31, 2016 #12
    ok thanks everyone!
     
  14. Mar 31, 2016 #13
    wait one more question
    so if i were to re-write the acceleration part, would i make 1.0 N negative
    a = -1.0 N / 0.164 kg = -6.1 m/s^2 [forward]
    and would it still be [forward]?
     
  15. Mar 31, 2016 #14
    actually the 'forces' are vectors having 'magnitude' means value in numbers and direction say x,y,z, direction or in common parlance 'forward' i.e. +x or backward -x .
    the friction force has a nature-if the body is moving +x direction i.e. forward then it will act in -x direction; if the body is moving backward or -ve x direction then friction will act opposite -forward direction.
    so , in your case the puck was moving forward so it will act backwards- so you can not write forward -you wish to say that you have put in a negative sign so you compensate it by calling forward, but it can create confusion as some people may err in looking at -ve sign and raise issues, actually -ve sign is saying it is acting backwards and actively reducing the speed-as your computation for distance traversed in +x direction.
    i think i have'nt confused you further and its clear now.
     
  16. Mar 31, 2016 #15
    a = -1.0 N / 0.164 kg = -6.1 m/s^2 [backwards]
    is this correct?
     
  17. Mar 31, 2016 #16
    yes this fits in your equation for distance calculation.
    by the way the same situation happens when you throw a stone up and -g acts as acceleration opposite to motion
    or a body sliding down a slope and friction acting opposite to motion.
     
  18. Mar 31, 2016 #17

    haruspex

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    My answer is different from @drvrm 's. It depends what you mean by "[forward]". If you just mean it as a comment then I agree with drvrm, but it would be clearer to write [i.e. backward]. But as it stands I read it as defining the reference direction, saying "accelerating at -6.1 m/s^2 in the forward direction", i.e. 6.1 m/s backward.
     
  19. Mar 31, 2016 #18
    The lesson that im doing uses 'backwards' and 'forwards' to identify the direction of the motion. When i say the frictional force is acting backwards, i mean it is opposing the applied force. But since the acceleration is negative in this case, i dont know whether to make the direction forward or backward. Would it also make sense to say:
    -6.1 m/s^2 [forward] , the acceleration is negative because the puck is slowing down, but it is also slowing down in the same direction 'forward'
    here is what the lesson shows:
    upload_2016-3-31_17-4-45.png
     
  20. Mar 31, 2016 #19

    haruspex

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    Based on that example, the bracketed word is not a comment, so your answer would be either "6.1 m/s [backwards]" or "-6.1 m/s [forwards]".
     
  21. Mar 31, 2016 #20
    so i can chose any and they'll both be correct?
     
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