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How much force is used to pull the block?

  1. Jan 30, 2016 #1
    1. The problem statement, all variables and given/known data

    A 7 kg block on a flat surface moves from 2 m/s to 4 m/s over 5 m. The coefficient of friction between the block and the surface is 1. How much force was used to move the block?

    2. Relevant equations

    Honestly, my physics instructor gave us so many worksheets with so many worksheets, I have no clue which one to use.

    3. The attempt at a solution

    vf^2 = vi^2 + 2a(change in x)

    4^2 = 2^2 + 2a(5)
    a=1.2
    F=ma

    F = 7*1.2
    F=8.4

    I don't know how friction comes into play though....


    Thank you!
     
  2. jcsd
  3. Jan 30, 2016 #2

    haruspex

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    In that equation, what exactly does F represent? Be precise.
     
  4. Jan 30, 2016 #3
    Net force..?
     
  5. Jan 31, 2016 #4

    haruspex

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    Yes. What actual horizontal forces are present here?
     
  6. Jan 31, 2016 #5
    Dillion,think for a minute would you
    The net acceleration is 1.2 m/s^2
    This acceleration can be yielded by applying a force F to the block which acts in the forward direction,opposing the frictional force(acting in the opposite direction)
     
  7. Jan 31, 2016 #6
    i am sorry for posting a complete solution
    It's just that you know I couldn't help myself
     
  8. Jan 31, 2016 #7

    haruspex

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    Please delete it now or I will have to report it.
     
  9. Jan 31, 2016 #8
    Ummm okay
    I will
     
  10. Jan 31, 2016 #9
    i edited it
    Is it okay?
     
  11. Jan 31, 2016 #10

    haruspex

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    Yes , that's fine, thanks.
     
  12. Jan 31, 2016 #11
    You're welcome
     
  13. Jan 31, 2016 #12
    Force of friction = coefficient of friction * normal force
    Force of friction = 1 * 7

    And the pulling force
     
  14. Jan 31, 2016 #13

    haruspex

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    Yes.
    No. You should show units. A force is not just a number. If you fill in the units you might realise you have forgotten something.
    Yes.

    So how do these three relate: the frictional force, the pulling force, the net force that you calculated?
     
  15. Jan 31, 2016 #14
    Ohhh it should be 1 *68.6 (7 * 9.8 = normal force). So the force of friction is 68.6.

    "So how do these three relate: the frictional force, the pulling force, the net force you calculated?"

    The net force is equal to 0 when the block is not moving. So the pulling force has to be more than 68.6, correct?
     
  16. Jan 31, 2016 #15

    haruspex

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    Yes, but you still need to mention the units. How much more?
     
  17. Jan 31, 2016 #16

    68.6 Newtons. I don't know how to calculate how much more....
     
  18. Jan 31, 2016 #17

    haruspex

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    You calculated the frictional force as 68.6N. Let the pulling force be FP. Do those act in the same direction or in opposite directions? What is the net of those two forces?
     
  19. Jan 31, 2016 #18
    The Frictional Force is -68.6N and the pulling force is positive N. The net force is 0?
     
  20. Jan 31, 2016 #19

    haruspex

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    You calculated the net force in your first post in this thread. It wasn't zero. If the net force were zero there'd be no acceleration.
     
  21. Jan 31, 2016 #20
    The net force was equal to 8.4 N.

    Frictional force (-68.6N) plus pulling force = 8.4 N.

    Is the pulling force 77N?!
     
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