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Physics problem

  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

    ahrkron

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    Gold Member

    What have you done so far?
     
  4. Sep 15, 2004 #3
    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.
     
  5. Sep 16, 2004 #4
    You have not accounted for the angle anywhere.
    Resolve the applied force into its horizontal and vertical component.

    -- AI
     
  6. Sep 16, 2004 #5

    cronxeh

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    Gold Member

    Ffriction=mu * Fn * cos (angle) -- this is the force due to friction, acting in opposite direction to pull

    ( pull <-- object --> Ffriction)
     
    Last edited: Sep 16, 2004
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