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Quick work question

  1. Sep 25, 2006 #1
    "a force of 104 N is applied to a 26 kg mass at an angle of 32 degrees to the horoizontal. calculate the work done when the mass is moved through a horizontal distance of 5 m."

    my answer is zero because 26kg times 9.8 m/s^2 = 255 N, so a 104N force wouldnt be enough to move it. does that sound right?

    ~Amy
     
  2. jcsd
  3. Sep 25, 2006 #2

    radou

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    Since there is no friction mentioned, the force should move the mass. It is not the same as lifting the mass. In order to lift it, you should have a vertical force which is greater than the weight of the mass. So, the force F = 104 * cos32 is doing work along a distance of 5 m. Unless I'm missing something big here. :biggrin:
     
  4. Sep 25, 2006 #3
    that was my first answer F = 104 * cos32 = 441 J.

    i dunno. it just seems like the 26 kg should be included somehow.

    can anyone else shed some insight?

    ~Amy
     
  5. Sep 25, 2006 #4

    radou

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    Maybe they're included to confuse you. :smile:
     
  6. Sep 25, 2006 #5
    k, looks like youre right. thanks for the help!

    ~Amy
     
  7. Sep 25, 2006 #6
    another question. there's a force displacement graph. (quick vent) i asked the school that im signed up with and they have no idea about the below question :eek:. they have these chat rooms that take for ever to get any help and more than 50% of the time with my questions they are no help at all. today i was lucky and only wasted a half hour waiting (usually it takes an hour or more). but if you send an email they'll give you a good answer, but it takes 5 business days. grrrr...

    part a starts at 40N, 0 displacement in m
    then it goes to 40N at 4 m
    =160J

    then from 40N it slopes at 60N by 8 m
    = 200J

    so the total = 360J

    and the question asks "state two possible outcomes of the work (above) being done on the object". im not sure what "possible outcomes" mean.

    ~Amy
     
  8. Sep 25, 2006 #7

    radou

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    Could you make a sketch of the diagram and post it on a link?
     
  9. Sep 25, 2006 #8
    i dunno how links work. but i'll try to sketch it here just using typing.

    N
    60...................... /
    ..................../
    40------- /

    20

    0......2.....4.... 6....8 (displacement in m)

    imagine the /s are a straight normal slope..

    k, this is the best drawing i can do.

    ~Amy
     
    Last edited: Sep 25, 2006
  10. Sep 25, 2006 #9

    radou

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    Hm, the work should equal the area under the graph, but I don't understand the 'two possible outcomes' part, neither.
     
  11. Sep 25, 2006 #10
    thanks anyways. :smile:

    if anyone else can help, it'd be appreciated!

    ~Amy
     
  12. Apr 2, 2008 #11
    Its because the graph starts from 10N of force instead of 0. which means if you were to count the squares, instead of using the formula W=F*D , you will notice that you are not able to count from 0-10 N because its not included on the graph, thus changing the outcome
     
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