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Force/Kinematics Problems

  1. Jan 10, 2010 #1
    Hey everyone. Struggling a little bit with a force test I have tomorrow. I've tried a few problems, and don't even know where to begin on another. I have the answers, and I would like to see if they are correct. Thank you so much!

    1. An object of mass 70 kg is accelerated by a net force of 20 N. What is its acceleration?

    My answer is .286 N


    2. A speedboat has a mass of 5.0 x 103 kg. It starts from rest and travels 2.0 x 102 m in 6.0 seconds. The boat undergoes uniform acceleration during the 6.0 seconds. What is the net force on the boat?

    My answer is 55,550 N


    3. A 75 kg girl traveling at a constant velocity on ice skates suddenly experiences an acceleration of -4 m/s2. What is the sum of the forces acting on her?

    My answer is (HAH!). I don't even know where to begin. Any tips?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 10, 2010 #2
    This looks correct.

    I don't think this is correct. Calculate the acceleration then use with Newton's second law (F = ma) to solve for the net force.

    HINT: Vf = Vi + at

    Hmmm... Sum of the forces acting on her... I think you've got to use Newtons 2nd law. ;)
     
  4. Jan 10, 2010 #3
    So for #2 if I were to do:

    Vf=Vi+at
    200=0+a6
    200=6a
    a=33.33333

    Then: F=ma
    F=5000(33.333)
    F= 166,667 N


    Right?
     
  5. Jan 10, 2010 #4
    Looks okay to me! Now try number 3!
     
  6. Jan 10, 2010 #5
    And for the latter,

    F=ma
    F=75(-4)
    sumF= -300

    Yay?
     
  7. Jan 10, 2010 #6
    Correct! You don't need to write sumF =, because when you write F = ma you're already implying that:

    [tex]\sum F = ma[/tex]
     
  8. Jan 10, 2010 #7
    Awesome. Thanks so much for the help! One last one possibly?

    A brick has a mass of 1.2 kg. A force of 5.4 N just begins to move the brick along the floor with a constant velocity. What is the coefficient of static friction?

    So I would use f=μn

    5.4=μ*11.76

    μ=0.46

    Would that be right? Thanks so much for your help!
     
  9. Jan 10, 2010 #8
    Correct! Another key observation I want to bring to your attention is that when you have an object moving at constant velocity then,

    [tex]\sum F = 0[/tex]

    Why's that you ask? Well remember, with constant velocity acceleration is simply 0. Applying Newtons second law,

    [tex]\sum F = ma = m(0) = 0[/tex]
     
  10. Jan 10, 2010 #9
    Oh. That makes sense! Thanks so much for your help!
     
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