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Puck sliding accross ice W/ friction and incline

  1. Jun 7, 2013 #1
    A puck, weighing 75KG(yes, 75kg) is launched from 0-20m/s in 4.4s. Determine the unbalanced force and acceleration.



    Fnet=ma A=v/t



    This PART was easy, A=20/4.4, A=4.55m/s^2. Fnet=75(4.55) Fnet=341.25N

    Its the next part that gets tricky...

    The puck, traveling at a constant 20m/s approaches a 15deg incline hill with a frictional mu of 0.100. Determine the force of friction, how far up the hill it will travel before coming to a complete stop, acceleration and the applied force.

    So what I did was first find the Fg which is 735N
    Then I split it up into X and Y components, for the Y component I ended up with 710N because COS15(735N)=710N. This should also be the normal force (I'm just not sure if I found the right component/value for Fg) so then Ff=Fn(mu) Ff=710(0.100) Ff=71N
    Since the Ff is 71N, and there is NO applied force pushing it UP the incline, I did 71/75=-0.95M/s^2 and have that as my acceleration value. Then I subbed it into the V2^2=V1^2+2ad, I ended up subbing in 0=(20)^2+2(-0.95)D 400=1.9D D=210m. Therefore distance before stopping is 210M. It seems off to me so any confirmation/help would be greatly appreciated.
     
  2. jcsd
  3. Jun 7, 2013 #2

    PhanthomJay

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    You forgot another force acting on the puck along the incline besides friction.
     
  4. Jun 7, 2013 #3
    Ah you're right, the force of gravity in the X direction. Would you mind checking if the value of 710N for the Y direction is 100% correct? And you would use Pyth theorem to solve for the other, so it would be 710^2+b^2=735^2 right?
     
  5. Jun 7, 2013 #4

    PhanthomJay

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    Yes, using the cos function is correct in finding the component of the gravity force perpendicular to the incline . You can use Pythagoras to find the comp of the gravity force parallel to the incline, but it might be easier to use the sin function.
     
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