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Find the final speed of the box

  1. Feb 4, 2007 #1
    there are 2 part for this problem I got the part one but i didnt get another one so helpppp please.......

    1)A 44.4 kg box initially at rest is pushed 5.3m along a rough, horizontal floor with a constant applied horizonta force of 181.646N.
    The acceleration of gravity is 9.8 m/s2.
    If the coefficient of friction between box and floor is 0.38, find the work done by the friction. Answer in units of J.
    The answer for this question is -876.33

    2) find the final speed of the box. Answer UNits of m/s. Please help me in this problem....post the reply as soon as possible if u can.......
     
  2. jcsd
  3. Feb 4, 2007 #2

    Doc Al

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    Staff: Mentor

    Hint: What's the net work done on the box? And its resulting kinetic energy?
     
  4. Feb 4, 2007 #3
    the fnet= ma that is 44.4 kg times 9.8 and i got 435.12 is that correct
     
  5. Feb 4, 2007 #4
    to find the final speed...do i use v=vf-vi/t
     
  6. Feb 4, 2007 #5

    Doc Al

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    Staff: Mentor

    No. The acceleration is 9.8 only if it were falling freely. What you've calculated is the weight of the box.

    You can either find the net force and calculate the acceleration, or find the net work and calculate the kinetic energy.
     
  7. Feb 4, 2007 #6
    do u subtract the applied horizontal force and the work done by friction to find the net force
     
  8. Feb 4, 2007 #7
    i have a problem very similar to this..do u subtract the applied horizontal force and the work done by friction to find the net force?
     
  9. Feb 4, 2007 #8
    You either find the net force or the net work. If you want the net force, subtract the friction force from the horizontal force. Then you can find acceleration and use a kinematic.

    Or you could find the net work (subtract friction work from horizontal work) and find the change in kinetic energy. Then you can find the final velocity
     
  10. Feb 4, 2007 #9
    so it would be -876.44 + 181.646 = -694.684
    Fnet = ma
    -694.684 = 44.4a
    a = -15.65 ?
     
  11. Feb 4, 2007 #10
    Youve got FORCES and WORKS. 1 of them is multiplied by the distance 5.3m. You cannot add them together until you have all FORCES or all WORKS
     
  12. Feb 4, 2007 #11
    so the net work would be -876.44 + 962.7 (which is 181.646 x 5.3) = 86.28
     
  13. Feb 4, 2007 #12
    Yes, and this is the change in kinetic energy
     
  14. Feb 4, 2007 #13
    ok, now that ive got the change in kinetic energy:
    do i do:
    KE = 1/2mv^2
    86.28 = 1/2(44.4) v^2
    divide both sides by 22.2 and get 3.886 and then find the square route of that to get the speed = 1.97 m/s
     
  15. Feb 4, 2007 #14
    by the way, thank you very much!
     
  16. Feb 4, 2007 #15
    Looks good
     
  17. Feb 6, 2007 #16
    thank u aiken fan...no actually mrs aiken..lol...
     
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