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Homework Help: Big Test Prep-Questions

  1. Nov 30, 2004 #1
    Big Test Prep--Questions

    Alright we are having a review test and I needed to go back to the stuff we did earlier this year and I am having some trouble remembering what to do....
    :shy: how can i forget this stuff you ask... I don't know


    7.
    A 52.3 kg trunk is pushed 5.95 m at a constant speed up a 28.0 degree incline by a constant horizontal force. The coefficent of kinetic friction beeween the trunk and the incline is .19 Calculate the work done by (a) the applied force (i got b)


    Answer:2160 J
     
  2. jcsd
  3. Nov 30, 2004 #2

    James R

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    The work done by a force is

    [tex]W = \vec{F}.\vec{s} = |\vec{F}||\vec{s}|cos \theta[/tex]

    where [itex]\theta[/itex] is the angle between the force vector and the displacement vector.
     
  4. Nov 30, 2004 #3
    Just [tex]W=Fd\cos{\theta}[/tex]. Can you work with the force vectors acting on the block and figure it out? Start with a FBD.
     
  5. Nov 30, 2004 #4
    Indded that is the forumula, however it is in calculating F, that I am having difficulty
    I did just realize that it is horizontal so the [tex] cos \alpha [/tex] will apply as opposed to if they were pushing directly at the 28 degree angle,
     
  6. Nov 30, 2004 #5
    I am being really slow with friction, its been awhile I tried setting it up like follows




    [tex]\sum{x=0}P-f cos\alpha=0 [/tex]
    [tex]\sum{y=0}N-mp-fsin\alpha=0[/tex]

    however I have a strong feeling this is incorrect, I have also tried 2 other ways but they are more involved and require a lot more work to show, and I know they don't work
     
  7. Nov 30, 2004 #6
    P, being the force
    N- Normal
    f= uk
    there should also be spaces betwen 0 and p and 0 and N
     
  8. Nov 30, 2004 #7
    The applied force is horizontal. So you have four forces: The normal force (points directly up), friction (pointing back), gravity (pointing downwards) and your applied force, pointing horizontally. These are the equations i get:

    Fy = Fn - Asin28 - mgcos28 = 0
    Fx = Acos28 - Ff - mgsin28 = 0

    You might want to double check that. I may have made a mistake somewhere. Solve for normal force in Fy and then plug that into Fx. Then you can solve for your applied force. When I did it, i got a slightly different answer. That may have been an algebraic mistake (hopefully). Either way, do it for yourself and see what you come up with.
     
  9. Nov 30, 2004 #8
    Yea, i made a small mistake. It does work.
     
  10. Dec 1, 2004 #9
    thanks for you help it worked out when I did the work
     
  11. Dec 1, 2004 #10
    [tex]Fy=N-Psin28-mgcos28 = 0[/tex]

    [tex]Fx =Pcos28 - f - mgsin28 = 0[/tex]

    [tex] N=Psin28-mgcos28 [/tex]

    [tex] Pcos28-ukN-mgsin28 = 0 [/tex]

    [tex] Pcos28-uk*(Psin28-mgcos28)-mgsin28=0 [/tex]

    [tex] Pcos28-ukPsin28-ukmgcos28-mgsin28=0 [/tex]

    [tex] P(cos28-uksin28)=ukmgcos28+mgsin28 [/tex]

    [tex] P=\frac{ukmgcos28+mgsin28}{cos28-uksin28} [/tex]

    [tex]P=\frac{.19*52.3*9.8*cos28+52.3*9.8*sin28}{cos28*.19sin28} [/tex]

    [tex]P=411.474[/tex]

    [tex] W=P(dot)d [/tex]

    [tex] W=P*d*cos28 [/tex]

    [tex] W=411.474*5.95*cos28=2161.69 [/tex]

    thanks for the help, just wanted to show you that I actually used your advice
     
  12. Dec 1, 2004 #11
    The thing I need help is with comming up with the initial values the
    [tex]Fy=N-Psin28-mgcos28 = 0[/tex]
    [tex]Fy=N-Psin28-mgcos28 = 0[/tex]
     
  13. Dec 1, 2004 #12
    Well the only suggestion i can make is that you draw yourself a free-body diagram and then start labelling your forces. Resolve them into your x and y components. Then just add them up.
     
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