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Conceptual Understanding: 2nd Law and Coefficient of Friction

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A 1 kg object is moving along at speed of 5 m/s right before colliding with a 4 kg stationary object, in a perfectly inelastic collision.

    Suppose that the objects are moving on a surface that has a coefficient of friction of mu=0.2. (Note you may still assume that the first object moves at 5 m/s immediately before collision. After impact, the objects move off in the velocity you calculated (which is 1m/s) and are then decelerated by friction until they come to a full stop


    2. Relevant equations

    I'm trying to understand how I can apply Newton's 2nd Law to this. We can solve this by using one of the Kinematics Equations, but we need to calculate acceleration. We know that Vi = 1m/s and that Vf = 0m/s (comes to a stop). But how can we calculate acceleration? I know it has to do with how to calculate the Force of Friction which is equal to mu*N (N is the normal force which is equal to -mg)

    3. The attempt at a solution
    So here is what the steps of the equation look like

    (1) Calculate acceleration
    (2) use Vf , Vi, and a to find t through the equation Vf = Vi + at

    Then the solution has this

    ma = F of friction = mu*m*g

    therefore a = mu*g

    (mu is Greek Letter meaning coefficient of friction)...

    But how is this possible? I tried to think of it in terms of Fnet , but F-net :

    Fnet = ma + mu*m*g (Find a from Newton's Second law[i/]

    I don't see how "Find a from Newton's Second Law" works here...

    But to do what the solutions did we would have to make Fnet = 0 ...

    I don't see why we can do this? Is it because we are solving for the distance when it comes to a stop (which is 0 net force?)
     
  2. jcsd
  3. Nov 16, 2007 #2

    Doc Al

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    Staff: Mentor


    I don't understand the question. There's only one (horizontal) force acting on the body, the friction force. So Fnet = mu*m*g. Newton's 2nd law tells us that Fnet = ma, so mu*m*g = ma, and thus a = mu*g (as you had previously noted).

    Can you restate your question?
     
  4. Nov 16, 2007 #3

    Dick

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    Fnet=mu*m*g=ma. Fnet is not zero. The force on the object is constant while it is moving. As you have said, a=mu*g. I'm not sure what it really confusing you. Fnet is only zero if an object is not accelerating.
     
  5. Nov 16, 2007 #4
    I think my confusion came in because I thought there were 2 forces action on the body - as I saw it in my mind's eye - there was a force from 1 the 1kg object that impacted the second object and "pushed" (put a force on it) in the positive x-direction, while it deaccelerated from the Force of Friction acting in the opposite direction.

    Maybe I'm just not good at visualizing/seeing these things. Our teacher had told us that, for example, 2 dimension motion, there is never a force acting in the X-direction, so if I apply that same logic here, I see what you are saying - but I just didn't see why there was no force in the direction opposing friction.
     
  6. Nov 16, 2007 #5

    Doc Al

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    During the collision the objects certainly exert forces on each other! Otherwise the mass at rest would never get moving to begin with. But after the inelastic collision, it's more useful to think of them as one combined object (of mass m1+m2) that is acted upon by friction.
     
  7. Nov 16, 2007 #6

    Dick

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    Not an unusual problem. Many students have the impression that if an object is moving in some direction then there must be a force 'pushing' it in that direction. You just have to listen to Newton and get over it.
     
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