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Forces please helpp

  1. Jan 11, 2005 #1
    Forces!!!! please helpp!!!

    1) 2500kg car stops from 14 m/s to full stop in a distance of 25.0 m. what is the coefficient of kinetic friction?q

    2ad= v squared - v(starting) squared
    2a25 = - (14) squared
    50a = -196
    a = - 3.92 m/s

    i got this far and then to find Ffriction = u(coefficient)x normal force....and here i got lost.

    i know that to stop the car a force of - 9800 N is needed. (F = ma = 2500(-3.92) = -9800N

    but how do i get the friction of force? can somebody help me...please?

    do i use Fg at all? Fg=2500(9.8)=24500N???

    and then 2) box of 5 kg is moving at velocity of 3 m/s due to Force of 10N. What is the normal force?

    I dont understand normal force. i know its a weight of an object and any other force added to it.

    so in this case would it be 10N + 5(9.8)N which equals to 59N?
  2. jcsd
  3. Jan 11, 2005 #2
    Is there anymore to number 1? It seems they only give enough information to find the stoping force... Do they give the braking force? if they did then just subtract from the net force which is the 9800N
  4. Jan 11, 2005 #3
    no they dont give any more information.
    is number two right though?
    or should i use pythagorian theorem for that since the speed is constant? so the net force is equal to zero?
  5. Jan 11, 2005 #4
    [tex] f_{k} = \mu N [/tex] , [tex] f_{k}[/tex] is the friction. and what deccelarate the car is friction..... N is normal, which is the force the car applied on the floor...... after you have these two, finding [tex] \mu [/tex] shouldn't be a problem
  6. Jan 11, 2005 #5
    He can get normal, but how does he get friction force?with given info Normal force in this case is = to mg
  7. Jan 11, 2005 #6
    could the force the car needs to stop equal the force of friction since it drives at icy road?
  8. Jan 11, 2005 #7
    for your question, yes....in your daily life.....NO
  9. Jan 11, 2005 #8
    lol, next time copy whole question.
  10. Jan 11, 2005 #9
    well i wasnt sure....but okay
  11. Jan 11, 2005 #10
    ima real sorry guys...but didnt seem important at the beginning...ay ay

  12. Jan 11, 2005 #11
    Normal force is the force that the surface exerts on the box... if the box is on a flat surface, and the force is being applied perfectly horrizantally than it should just be equal to the force that the box is exerting on the floor caused by gravity, in which case it would be mg
  13. Jan 12, 2005 #12
    thanx a lot...man why cant they explain it clearly like this in the book? why cant my physics teacher teach for a change?...and stop making us study by ourselves from the book...*sigh*..........THANX YOU ALL who helped me today.
  14. Jan 12, 2005 #13

    as vincentchan suggested, [tex] friction = \mu R[/tex], you can get the frictional force by using [tex] Friction = m a [/tex] where a is the magnitude of the deceleration. There are no other horizontal forces being applied, so, you can easily evaluate for [tex] \mu [/tex]
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