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Work and Force

  1. Oct 20, 2007 #1
    Is More work is done to lift a 4.00-kg object a distance of 1.00 m at a constant velocity than to push a 1.00-kg block a distance of 1.00 m with a force of 39.2 N?

    I can find the work done in the 2nd case using W=fd= 39.2 J.
     
  2. jcsd
  3. Oct 20, 2007 #2

    Astronuc

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    Staff: Mentor

    What is the weight of a 4.00 kg object? Remember weight is a force (due to gravity).
     
  4. Oct 20, 2007 #3
    oh yes, W=mgd= 4*9.8*1= 39.2J
    So the work done is the same!
     
  5. Oct 20, 2007 #4
    Total Work

    You raise a 2.00-kg pail of water 1.00 m and then lower it 1.00 m. What is the total work done?
    W=mgd
    W=2*9.8*1=19.6J
    Thats the work done when the pail is lifted.
    The work done when the pail is lowered is
    W= mgcos(theta)d
    and theta is 180
    so that would make w= -19.6J

    so total work will be zero. yes..?
     
  6. Oct 20, 2007 #5
    opps..the question is what is the total work done by gravity
     
  7. Oct 20, 2007 #6
    Work done (in a gravitational field, such as that of the earth) depends only on the initial and final positions, and its path independent.
    So if the initial and final positions as in this case are the same, the work will be the same. You don't need to calculate anything.
     
  8. Oct 20, 2007 #7
    Work done by gravity is just numerically the negative of the work done by you. [tex]W_{you on object} = -W_{gravity on object}[/tex]
     
  9. Oct 20, 2007 #8
    so the total work done will be 19.46J..?
     
  10. Oct 20, 2007 #9
    oh..i get it..so its -19.6J..thankz..!
     
  11. Oct 20, 2007 #10
    Over the entire trip. i.e up and then back down, it is zero. As I said any work, be it by you or by gravity is dependent only on its initial and final positions, which are same in both cases, so the work is zero.
     
  12. Oct 20, 2007 #11
    Another doubt, more work is done on an object by lifting it straight up than lifting it diagonally yes?
     
  13. Oct 22, 2007 #12
    That isn't the right thing to say. Work is a scalar quantity given by the dot product(scalar product of two vectors) [tex] \vec{W}= \vec{F} . \vec{s} [/tex]. So it depends on two parameters(viz. the component of the force responsible for the work and the displacement in that direction) and not just one. You'd have to have a condition for both the parameters. What you wanted to say I guess is that knowing the magnitude of a constant force, F and a constant displacement, s ; you are merely changing the direction of the force so that only one of its component is responsible for the work. In that case, yes, you are right. But you have to take care about what you assume.
     
    Last edited: Oct 22, 2007
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