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Work Problem!

  1. Oct 23, 2007 #1
    1. A boy bushes a 10 kg box 3 m slowly up a 30 degree slope. There is a constant 30 N frictional force directed down the slope. What is the minimum work required for the boy to move the box.



    2. W(friction)=fR, Wg=mg3sin(theta)



    3. The work of friction is -90J. the Wg is -147 J, so the minimum work to move the object should be 147+90=237J
     
  2. jcsd
  3. Oct 23, 2007 #2
    i think i did this correct right?
     
    Last edited: Oct 23, 2007
  4. Oct 23, 2007 #3
    midterm soon.... can anyone pllllleasssee help? :)
     
  5. Oct 23, 2007 #4
    I'm not that well versed in this yet, but I think the idea is that if you were talking about forces, you would need to consider the friction force when figuring out the force needed to push the box up the incline (and you know this).

    But, when talking about work, we're just talking about the Force over a distance, so friction doesn't need to be considered for that calculation. So the work done pushing the box up is the 147J, the work done by friction is -90J. If you wanted the 'net' work done on the box, you would use 147J - 90J for a net work of 57J.

    I hope this is a little helpful.
     
  6. Oct 23, 2007 #5
    can anyone else confirm this?
     
  7. Oct 23, 2007 #6
    The work of friction and gravity are not entirely needed (all you really need to know is that you are on earth). The equation for work is W=F*d. Because the force that the boy will exert on the block is in the direction of the distance, then d=3m.

    What you need to do is find the minimum force required to push the box because the boy is pushing the box *slowly*. This would be F=friction+gravity.

    The force of friction is straightforward at 30N. The force of gravity acting on the box down the slope is also straighforward. Because the slope is inclined at thirty degrees above horizontal, the force of gravity is equal to g*sin(30).

    This means that the force that the boy is pushing on the box is F=30+9.8*sin(30). This is the force part of the equation.

    Adding the values, you obtain a final value of W=3*(30+9.8*sin(30))=104.7 N*m.


    Basically the does not have to do with the work that friction and gravity exerts on the box, those are not part of the equation. You need the minimum force that the boy needs to exert on the box and multiply that by the distance.

    Good luck on the midterm!
     
  8. Oct 23, 2007 #7
    thanks but wait... the force of gravity is equal to mgsin30 right?
     
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