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Force given mass, velocity and fricton

  1. Nov 12, 2016 #1
    1. The problem statement, all variables and given/known data
    A 2 kg box is given a brief push so that it slides across the floor, but the applied force is only applied for an instant. The coefficient of friction is 0.20 and the initial speed of the box is 4 m/s.
    Find acceleration of the box after the push
    Find how far the box goes before coming to rest.

    2. Relevant equations
    Net force= mass*accel
    Friction=u*normal

    3. The attempt at a solution
    I created an FBD, found the weight is 19.6N(9.8*2), and because it is being pushed horizontally, the Normal force is also 19.6 N. With the normal force I can find the force of friction. F=un, which means F=.2(19.6), and F=3.92 N. This leaves me missing the force I would need to calculate the Net f and find the accleration. I could only get this far, as I do not know how to find a force with only the velocity, mass, and friction.
     
  2. jcsd
  3. Nov 12, 2016 #2

    TSny

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    Hello. Welcome to PF!
    The push was only used to give the object an initial speed of 4 m/s. After that, the pushing force is zero.
     
  4. Nov 12, 2016 #3
    Oh, I suppose it was easier then I thought! Thank you!
     
  5. Nov 13, 2016 #4
    So, the box's acceleration of the box is slowing down by 9.81 m/s^2? Because it's initial speed is 4 m/s and there are not other forces acting on it?
     
  6. Nov 13, 2016 #5

    TSny

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    No, the box is not decelerating at 9.8 m/s2. You need to find the net force acting on the box and use Newton's second law of motion to find the acceleration.
     
  7. Nov 13, 2016 #6
    Thank you. I got 1.96 m/s2 by dividing the Net Force by the mass. The Net Force, because this is all on a horizontal plane is the normal force multiplied by the coefficient of friction (3.92 N).

    What formula would I need to use to find how far the box goes?

    Thank you for your help.
     
  8. Nov 13, 2016 #7

    TSny

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    Yes.

    You can use the standard kinematic formulas for constant acceleration.
     
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