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Homework Help: Physics question

  1. Sep 15, 2004 #1
    I am having trouble with this physics question, i would appriciate it if some could show me how it can be solved.

    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
    How did you attempt to solve this problem? Where specifically are you having trouble?
     
  4. Sep 15, 2004 #3

    JasonRox

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    Do not post the question everywhere!
     
  5. Sep 15, 2004 #4
    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.
     
  6. Sep 15, 2004 #5

    Pyrrhus

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    You seem to have forgotten
    this force has a vertical component.
     
  7. Sep 15, 2004 #6
    how do i solve it by incoporating the vertical component?
     
  8. Sep 15, 2004 #7

    Pyrrhus

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    1.50 * 10^2*sin(25) = Fy
     
  9. Sep 15, 2004 #8
    i still do not understand.
    after you find fy, then where do you incorporate taht in finding the velocity?
     
  10. Sep 15, 2004 #9

    Pyrrhus

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    Sorry, i was helping someone else, Fy is pointing up as the normal force, so Fy+ n= mg
     
  11. Sep 15, 2004 #10
    k tahnks i got the answer
     
  12. Sep 15, 2004 #11

    Pyrrhus

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    no problem, it's good to be of help.
     
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