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Energy Conservation: Puck on Wood on Frictionless Surface

  1. Jul 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Hey all, another problem. There is a 1 meter wooden panel that is on a frictionless surface. A puck is placed on one end of the panel and is pushed towards the other side. The mass of the panel is 10 times that of the puck. Also, there is a friction coefficient of uk=0.15 between the puck and the board. I have to find the minimal initial velocity that the puck must have to be able to slide off the board.

    2. Relevant equations
    Conservation of Energy: K1 + U1 + Wother = K2 + U2
    Work: W=Fs
    Friction Force: Fk= ukN
    Kinetic Energy: 0.5mv2
    Newton's Second Law: F=ma
    Kinematics: x= xo + vot + 0.5at2

    3. The attempt at a solution

    First thing I've done is set up the conservation of energy equation. The puck's initial kinetic energy is converted to kinetic energy of the wooden panel and some is "lost" as friction (Wother). Using this logic, and adjusting the mass of the wooden panel in terms of the puck, my conservation equation is the following:

    0.5mvo2 = 5mvp2 + ukmgs

    I need to find the unknown vp (the velocity of the panel when the puck has crossed the entire panel). This is where I'm not sure what to do. I decided to go with kinematics, and where I end up is a huge mess and I'm not sure I can isolate vo. Here is my work/methodology:

    The movement of the panel is caused by the reactionary force of the kinetic friction acting on the puck. Thus the force is the same as the kinetic friction force (mass of panel is adjusted for):

    Fk,r = ukmg=10map
    ap = [ukmg] / 10m = 0.1ukg

    This is the acceleration, and I need to find the velocity when the puck is at x=1. Acceleration is a=vt, so I need to find how long it takes for the puck to travel 1 m. I used this as my equation for the motion of the puck on the wooden panel:

    x = vot - 0.5ukgt2 = 1m
    vot - 0.5ukgt2 - 1 = 0

    Using the quadratic formula, I found the following solution:

    t= [-vo + (vo2 - 4ukg)0.5] / ukg

    I plugged this into the equation for the velocity of the panel (vp = at), getting (the 0.1 is from accounting for the mass of the panel vs puck):

    vp = 0.1(-vo + (vo2 - 4ukg)0.5)

    I've plugged this back into my energy conservation equation, and now the equation is a sizable mess and I'm not sure I can isolate vo. Is there any mistake in my math/logic, an easier way to find vb I haven't thought of, or am I just going to have to work through the complicated equation I'm at now?
     
  2. jcsd
  3. Jul 23, 2017 #2
    If it's a minimal initial velocity to get off the plate than what should the velocity be as it just barely gets off the plate?

    However, you have another problem, the board should also be moving with the puck due to the friction.

    Edit: Here's a clue, the answer to many of these kinds of two object problems is to use both conservation of energy AND conservation of momentum
     
    Last edited: Jul 23, 2017
  4. Jul 23, 2017 #3
    The velocity of the puck as it just gets off the panel would be 0, hence why there isn't a term for kinetic energy of the puck on the right side of the conservation of energy equation.

    Wouldn't the movement of the board really not affect the movement of the puck since the puck is moving on the board and we only care about it crossing the board?

    But isn't momentum not conserved because there is friction present?

    Edit: I may see what you mean by considering the velocity of the puck as it barely gets off the plate. Since I know that velocity, I can use the kinematic equation:

    d = 0.5(vo + vf)t

    to solve for time. It makes t=2/vo which is considerably simpler. Can you confirm that this is okay?
     
    Last edited: Jul 23, 2017
  5. Jul 23, 2017 #4

    TSny

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    0 with respect to the earth, or 0 with respect to the board?

    It's true that the movement of the board does not affect the value of the friction force that the board exerts on the puck. So, the movement of the board does not affect the acceleration of the puck relative to the earth.

    Friction between the puck and board is an internal force of the puck-board system. Internal forces do not affect the total momentum of the system.
     
  6. Jul 23, 2017 #5
    0 with respect to the board, that is.

    So would I set up the conservation of momentum the same way as I normally would?

    mvo = mvp where vp is the velocity of the board when the puck has just barely reached the other side of the board? I feel like this doesn't account for the friction...
     
  7. Jul 23, 2017 #6

    haruspex

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    TSny wrote that the momentum of the "puck-board system" is conserved.
    What is the initial momentum of that system in terms of the two masses and vp,initial? What is its final momentum in terms of the two masses and vp,final?
     
  8. Jul 23, 2017 #7
    So initially the only thing moving would be the puck. Would its momentum be as simple as pi=mvo?

    The final momentum is definitely a bit more confusing. As the puck finishes traversing the panel, the panel is still moving forward. That means relative to the panel, the puck is moving backwards with a velocity of -vp, right? I'm not sure how to incorporate the masses though....

    On another note, I was able to use the kinematic equation d = 0.5(vo + vf) to find the time it takes for the puck to cross the panel. I then used this time and multiplied it by the acceleration of the panel (found from newton's third law and the reactionary force from friction acting on the panel) to find the velocity of the panel just as the block barely reaches the end. I then managed to solve for vo, min after a good bit of algebra. Is there anything wrong with this approach?
     
  9. Jul 23, 2017 #8

    TSny

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    Yes, in order for the puck to barely fall off the far end of the board, it should reach the far end with essentially zero velocity with respect to the board. What does this tell you about the relation between the final velocity of the puck relative to the earth and the final velocity of the board relative to the earth?

    Yes. That would be the initial total momentum of the system relative to the earth.

    No. As you correctly stated above, the puck has zero velocity relative to the panel when it reaches the end of the panel.

    We would need to see the details. But, I think for now it would be best to stick with energy and momentum. Your energy equation in your first post was almost correct. But it does not include the final kinetic energy of the puck relative to the earth.
     
  10. Jul 23, 2017 #9

    haruspex

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    I do not see the difficulty. You are told the mass of the panel is 10 times that of the puck.
     
  11. Jul 24, 2017 #10
    If the puck has (almost) zero velocity with respect to the board, then its velocity with respect to the earth would be the velocity of the board, right?

    I just want to make sure I'm thinking of this correctly. When the puck moves on the board in the positive x-direction, the friction force is acting in the opposite direction. This friction force is applied to both the board and the puck. Since the board doesn't have an initial velocity, it would just move backwards due to the friction force, right?

    I'm trying to find the momentum with respect to the Earth, since both the puck and the board are moving in this problem. Their momenta are conserved between each other, but only with respect to the earth?

    The initial momentum (with respect to the earth) is: mpuckvo
    The final momentum (w/ respect to the earth) is: mpanelvpanel, f + mpuckvpuck, f

    Given that the puck's velocity with respect to the earth near the end of the movement is equal to the board's velocity:

    Final Momentum: mpanelvf + mpuckvf

    Adjusting the mass of the panel in terms of the mass of the puck (m):

    Final Momentum: 10mvf + mvf = 11mvf

    Conservation of Momentum: mvo = 11mvf

    It seems that my energy conservation is also slightly off, as I wasn't thinking relatively to earth.

    Initially, the only energy is kinetic energy: 0.5mvo2

    At the end, the initial kinetic energy is converted to kinetic energy of the board and puck (moving together), and some is lost as friction:

    Final energy: 5mvf2 + 0.5mvf2 + ukmgs = 5.5mvf2 + ukmgs

    Energy Conservation: 0.5mvo2 = 5.5mvf2 + ukmgs

    If everything above looks good, then I'm confident in the rest. I would use the conservation of momentum to solve for vf:

    vf = vo / 11

    I can then insert this value of vf into my conservation of energy and solve for vo.

    I just wanted to thank you guys for hanging with me through this problem. I feel like I've got a better understanding of this problem and I'm getting close to a solution!
     
  12. Jul 24, 2017 #11

    haruspex

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    Looks good.
    One question, just to be sure - what value are you using for s?
     
  13. Jul 24, 2017 #12

    TSny

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    The board will not move backwards relative to the earth. Think about the direction of the friction force on the board. Both the board and the puck will move in the same direction relative to the earth (but they will have accelerations in the opposite direction). I agree with haruspex that your energy and momentum equations now look correct.
     
  14. Jul 24, 2017 #13
    I'm using "s" as displacement. In this case it would be 1m since that's how far the puck would travel.
     
  15. Jul 24, 2017 #14

    haruspex

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    Well, it travels 1m relative to the panel, but rather further than that relative to the Earth. But you have chosen the right one. Can you explain why it's right?
     
  16. Jul 24, 2017 #15
    The displacement term is part of the equation of work done by the friction force. This force acts between the puck and the board. Since the puck crosses the entire length of the board, which is 1m, the friction force acts over 1 m.

    Thank you guys so much for all of your help!
     
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