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Physics question - Forces on a mass on horizontal surface

  1. Feb 24, 2014 #1
    Physics question -- Forces on a mass on horizontal surface

    1. The problem statement, all variables and given/known data
    A 2.5 kg mass at rest on a horizontal surface is acted upon by an applied horizontal force of 150 N. A frictional force of 100 N acts on the mass. How much work is done on the mass?


    2. Relevant equations
    I'm not too sure if any equation would be used.


    3. The attempt at a solution
    Again, I am not sure how to answer this question. I think that the work would be 0 but I am not sure.
     
  2. jcsd
  3. Feb 24, 2014 #2
    Hi! Which direction is the horizontal force moving this mass. Can you draw the diagram (if any)?
     
  4. Feb 24, 2014 #3

    collinsmark

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    Hi suzy123,

    Welcome to Physics Forums!

    In addition to the previous comment, there seems to be information missing in the problem statement. Does the original problem statement tell you how far the mass moves, or the distance to which the force is applied? Or maybe the time period that the force is applied? (Anything along those lines?)

    Do you know if it is acceptable to give your answer in terms of some other variable like distance traveled or time?
     
  5. Feb 24, 2014 #4
    Hi there!

    This is all of the information i was given for this question and no diagram was included.
     
  6. Feb 24, 2014 #5

    BvU

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    Different tack: what equations would you have at hand in case we manage to convince you that, in order to come up with a quantitative answer, you do need to make use of some equations nevertheless ?
     
  7. Feb 24, 2014 #6

    berkeman

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    You need to show some effort of your own, or this thread will be deleted. Our Homework Help rules here on the PF require that you show some effort on your question, and that you do the bulk of the work.
     
  8. Feb 24, 2014 #7
    I made another attempt at the question.
    so I used the f=ma formula. I substituted (150 N-100 N) for f, 2.5 kg for m and I solved for a. I got 20 m/s^2 for a. would I use one of the accelerated motion formulas to possibly solve for d? then substitute it into the w=fd formula?
     
  9. Feb 25, 2014 #8
    Well I'm not sure how to find the work honestly. Typically the equation w=F*d, however you aren't given distance or time. So I guess you could use (not sure though) the equation v=at. so it seems you have right answer for a, then just equate d=v and you should have a work eq. with t.
     
  10. Feb 25, 2014 #9

    adjacent

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    The information given are mass,and force.
    Even though mass and force gives acceleration,
    none of the four kinematics equations will be useful here.
    ##v=u+at## -
    ##S=\frac{1}{2}(u+v)t##
    ##S=ut+\frac{1}{2}at^2##
    ##v^2=u^2+2as##

    The equations require at least time to be given(to find distance) or distance to find work done.
    So this question lacks enough information.
    It's like saying what is the work done if a force of 1N acts on a mass of 1kg.We don't know the distance the forces acted.It can be 100m,1000m or 10^100m.We don't know.
     
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