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Work problem: cannot figure out

  1. Jul 2, 2009 #1
    1. The problem statement, all variables and given/known data
    n Figure 7-32, a constant force of magnitude 82.0 N is applied to a 3.00 kg shoe box at angle f = 53.0°, causing the box to move up a frictionless ramp at constant speed. How much work is done on the box by when the box has moved through vertical distance h = 0.150 m??

    2. Relevant equations
    see link for picture of figure... http://www.lowellphysics.org/beta/T...rk/Problems/c07x7_11.xform_files/nw0314-n.gif

    3. The attempt at a solution
    I think I should be using W=fdcos(theta)...but since the displacement given is verticle instead of horizontal, i don't think i could use h=.150m without manipulating the problem more. but i am also confused because they do not give the angle of the incline, so i cannot figure out what the angle between the force and displacement is. Perhaps I am just missing a simple geometry relationship, but I cannot figure out what. Please help! This problem is really bothering me because it doesn't seem to be a difficult one.
  2. jcsd
  3. Jul 2, 2009 #2


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    Homework Helper

    Welcome to PF.

    Have you covered the relationship with potential energy and work in your coursework?
  4. Jul 2, 2009 #3


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    Assuming the question is asking for the wotk done by the applied force, consider using the conservation of total energy equation. Or note that since thr block is moving at constant speed, the work-energy principle for total work done by all forces should also help.
  5. Jul 2, 2009 #4
    ok now i'm having a sort of concept problem...i got the right answer but i'm not sure why it works....

    i put the Kinetic energy equal to Potential energy...so .5mv^2=mgh and then solved for v, which i then plugged back into .5mv^s to get work... this gave me the correct answer....but why does this give me the work due to applied force when this doesn't account for work due to gravity?

    not sure if this question makes sense....
  6. Jul 2, 2009 #5
    hmmm , as in, wouldn't this answer be total work?
  7. Jul 2, 2009 #6
    We apply a constant force, yet it's moving at constant speed...get the hint yet?
  8. Jul 2, 2009 #7
    okay...i see what u are saying in that a constant force implies that the velocity shouldn't be constant...but it is...soooooo that means, i don't know? im really starting to feel like an idiot now, it is flying right over my head.... :-/

    i still don't get why the net work is equal to the work due to the applied force...
  9. Jul 2, 2009 #8
    Are you sure net work is equal to work done by applied force *only*?

    Constant force implies constant acceleration, yet we have constant speed! Constant speed implies no net force, therefore....?

    Hint: draw a free body diagram (as you always should when doing such problems)
  10. Jul 2, 2009 #9
    ehhhh okay, trying not to be really irritating by the fact that i totally am confused....but okay, so my book asked for work done by applied force, when i solved for .5mv^2, the answer i got was the correct answer....but change in KE should be net work....so i'm confused on what happened to gravitational work....which i wouldn't be confused it gravitational work equaled zero, but that should only be so if it was 90 degrees from the displacement, which it is not...

    i'm trying to figure out what u are hinting at...does it have to do with that the displacement is in the y direction and that in the y direction there is zero net force?...
  11. Jul 3, 2009 #10
    Why are you only trying to solve for KE? Does the object not move up the ramp? It will then gain PE, right? And you were trying to find change in KE? Is there actually a change in KE?

    In both posts I emphasized *constant speed* so there is NO change in KE. I'm trying to hint to you that since there is no net force, there is no work being done to change KE. Thus only PE changes, and using PE = mgh to find the answer is trivial.

    What you did is to somehow say KE = PE (which is incorrect), and then plugging it backwards to get the work done (which is actually the reversal of your first step), then get confused because you didn't realize all you had to do was to find PE. That's all there is to this question.
  12. Jul 3, 2009 #11
    oh dear, thanks so much! it just all clicked, i totally got stuck and my brain just kept going around in circles....so yea, net work=change in KE....so W(gravity)+W(applied force)=0....so then W(Gravity)=mgd costheta....which equals -4.41 so then W(applied force)=4.41 j....it all makes sense thanks for helping!
  13. Jul 3, 2009 #12
    Why are you still insisiting net work = change in KE? There is PE too!@.@ If this is something you have memorized from somewhere, forget it now! It's not always the case!

    Net work = change in energy ==> correct
  14. Jul 3, 2009 #13
    it is what my text book and my lecture notes say
  15. Jul 3, 2009 #14
    perhaps it is because we have not gotten to potential energy yet in the book, it is the next chapter...maybe that is where the confusion is!
  16. Jul 3, 2009 #15
    yea i just skimmed forward to see what the formulas for the next chapter looked like....and it does say that work=change in energy...they should have been more clear that the forumlas they use are shortened based on whatever the chapter is on....kind of misleading and confusing! thanx for pointing it out...
  17. Jul 3, 2009 #16
    Nvm edit
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