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Net work

  1. Oct 15, 2007 #1

    dnt

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    1. The problem statement, all variables and given/known data

    a man pulls a 10 kg box up an incline at a constant speed to a vertical height of 5 m. what is the net work being done?

    2. Relevant equations

    w= fd

    3. The attempt at a solution

    i know in the plane perpendicular to the surface there is no net work done because all the forward forces (the pull) cancel out the backward forces (friction and gravity). however, isnt there a net force being done by gravity because he rose the box up 5 meters?

    would the answer be (5m)(9.8)(10kg) = 490 J

    or would it be 0 J because its not accelerating at all? whihc one is correct? thanks.
     
  2. jcsd
  3. Oct 15, 2007 #2

    Doc Al

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    Staff: Mentor

    No. Gravity is not the only force acting on the box. The net force is zero.

    That's the work done by the man, not the net work.

    Exactly.
     
  4. Oct 15, 2007 #3

    dnt

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    thanks. what if the quesiton did say he was accelerating up the slope with a = 5 m/s2? would that change the answer then?
     
  5. Oct 15, 2007 #4

    Doc Al

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    Staff: Mentor

    Absolutely. In that case there would be net work done on the box.
     
  6. Oct 15, 2007 #5

    dnt

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    would it be:

    (10kg)(5 m/s2)(5m) = 250 J

    is that right?
     
  7. Oct 15, 2007 #6

    Doc Al

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    Staff: Mentor

    That implies that the box accelerates straight up! The net work done depends on the angle of the incline and thus how far the force needs to act. For example, if the angle were 30 degrees you'd have to drag the box up 10 m to raise it 5 m vertically. So, the net work would be: (10 kg)(5 m/s^2)(10) = 500 J.
     
  8. Oct 15, 2007 #7

    dnt

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    so basically you can take the net force (ma) in the same direction as its moving and then just calculate that distance to multiply to get the net work?

    so gravity has no real part here?
     
  9. Oct 15, 2007 #8
    Theres not enough information to solve the problem. Net work done on the object is equal to its change in kinetic energy

    [tex]W_{net}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 [/tex]

    to know the net work performed you need to know the initial velocity (did the box start from rest) and the angle of the ramp
     
  10. Oct 15, 2007 #9

    dnt

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    but if i know the distance the box traveled i could also multiply that by ma (net force) to get work, right?
     
  11. Oct 15, 2007 #10
    Make sure it is net distance (not just vertical), so if you are using W = F*D*Cos(theta) which is W = M*A*D*Cos(theta) those are all scalar values where theta is the angle between the force and distance vectors (distance vector should always be along the ramp). If the man is pulling along the ramp then theta = 0 and cos(theta) = 1. So if you are talking about distance along the ramp and the man is not pulling at an off angle then yes that should work
     
  12. Oct 15, 2007 #11

    Doc Al

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    Staff: Mentor

    Yep.

    Gravity is already included in the net force.

    Sure. You don't need the initial velocity to find the net work done. (You would need the initial velocity to find the final kinetic energy.)
     
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