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Homework Help: What is the speed of the box

  1. Sep 15, 2004 #1
    A 1.50 * 10^2 N force is pulling a 50.0 kg box along a horizontal surface. The force acts at an angle of 25.0°. If this force acts through a displacement of 12.0 m, and the coefficient of friction in 0.250, what is the speed of the box, assuming it started from rest?
     
  2. jcsd
  3. Sep 15, 2004 #2
    this is how i attempted to solve the question:

    Fg=mg
    =(50kg)(-9.81m/s2)
    = -491 N

    Fnet=Fn-Fg
    Fn= 491 N

    Ff=uFn
    =(0.250)(491N)
    =123 N

    W=Ek
    W=Fd
    W=(123 N)(12.0 m)
    W=1472 J

    Ek=1/2mv2
    V=squareroot 2Ek/m
    V=squareroot 2(1472 J)/(50.0 kg)
    V= 7.7 m/s

    The actual answer is 2.53 m/s. Can you tell me what i am doing wrong, thanks.
     
  4. Sep 15, 2004 #3
    im going to call the 1.5*10^2 N Fp

    okay. the normal force is the weight minus the vertical component of the force, becaues the force is at an angle of 25 degrees, right? so its weight-Fpsin25

    use that to calculate friction, mu*normal

    horizontal force is then Fpcos25-Ffriction

    divide that by 50 kg to get acceleration, reduces to a kinematics problem
    you get?
     
  5. Sep 15, 2004 #4
    thanks i got the answer
     
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