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Work-friction Problem

  1. Jan 18, 2007 #1
    1. The problem statement, all variables and given/known data
    How much work is needed to push a 112- kg packing crate a distance of 2.6 m up a frictionless inclined plane that makes an angle of 34 o with the horizontal?
    Answer: 1595.80
    How much work would be required to move the crate the same distance if the coefficient of friction were 0.30?

    2. Relevant equations

    3. The attempt at a solution
    I don't understand the second part is it saying that with friction what would it be? and if so would you not just subtract 0.30*2.6=0.78 and then add that to the first answer? thanks for any help.
    Last edited: Jan 18, 2007
  2. jcsd
  3. Jan 18, 2007 #2
    okay, first of all for the first part, knowing that work is equal to a change in energy (any type) how much energy do you think you would need to get the block up this ramp?
    For the second part, you sort of have the right idea. Knowing the amount of energy needed to get it up without friction, it should make sense that you just have to add to that the work that friction does on the block (or really the opposite of that). Now, you said that W=F*D, so you might be able to go from there?
  4. Jan 18, 2007 #3
    so would this be right 1595.80+0.78= 1596.580234J ?
  5. Jan 18, 2007 #4


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    Cole07; it might be better if you show your working, that is, the equations you used, what values you put in for each term. It's hard to see where you're going if we cant see what you've done. And, of course, simply putting down numbers implies that you expect us to work it out ourselves in order to check your work!
  6. Jan 18, 2007 #5
    yeah, you dont even need to use nubers if you dont want to, just put it in some way so that we can look at your technique. Thanks!
  7. Jan 18, 2007 #6
    to get the 1st answer i drew a diagram of a triangle with the crate on the hypotnuese pointed NE the angle is 34 degrees so i used sin to find the opposite of the angle and then i found the weight of the crate in newtons which is 112kg*9.8= 1097.6 then i used sin to find the y side of the triangle 1097.6sin(34)=613.7701308N then i used the equation W=F*D
    W-613.7701308*2.6 so W=1595.80234J (the first answer) . then for the second question is where i'm stuck but what i think you do is take 0.30*2.6 which is friction force and then add this to the first answer.
  8. Jan 18, 2007 #7
    okay--good job on the first part! (you could have more easily just used W=change in energy=Umax=mgh=mgdsin@ but whatev)
    Now, you are not doing the friction correctly. Frictional force is normal force times the coefficient of kinetic friction. So find that and then multiply it by 2.6 to get the work.
  9. Jan 18, 2007 #8
  10. Jan 18, 2007 #9
    normal force is the force exerted by the ground on the object (newtons third law) if an object sits on a flat surface, the force is simply the weight of the object, but because it is inclined, you need to take a component of its weight in the direction perpendicular to the incline.
  11. Jan 18, 2007 #10
    ok i understand what you are talking about but how do i take a component of its weight don't get that.
  12. Jan 18, 2007 #11
    do you mean what the hypotnuese is?
  13. Jan 18, 2007 #12
    okay Theta=34 degrees. The weight is mg so try to go from there.
  14. Jan 18, 2007 #13
    i'm sorry but i'm not getting this my teacher told me we should be able to do all the questions from this assignment, but i just don't get it.
  15. Jan 18, 2007 #14
    ok i do understand that theta is 34 degrees and that the weight would be 1097.6N.
  16. Jan 18, 2007 #15
    okay, Ill help you out (if it's not too late) the component of weight towards the surface is mgcos34. If you draw some triangles, you should be able to see this. so next, Friction is equal to .3*normal force. with that you have your force so multiply it by the distance to get the work done. does that make sense? By the way--if you are having trouble seeing why I use cosine, imagine that the angle was instead of 34 degrees only 1 degree. you should see that the normal force would be very close to the actual weight and cos1 is extremely close to one (probably like .9999 something). tell me if that helps or not.
  17. Jan 19, 2007 #16
    would this be what Ja4Coltrane has told me to do?
    Q:is the normal force the same as mgcos34?
  18. Jan 19, 2007 #17
    could someone please tell me if i'm on the right track.
  19. Jan 19, 2007 #18


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    Sorry, I didn't read the question properly before replying the first time!

    Your method is correct, so if your arithmetic is right then you have the correct answer. Note, however, that when setting it out, its clearer to write what each line corresponds to! (see the red amendments)
  20. Jan 19, 2007 #19
    Now I think that the most fundemental thing that you need to have learned from this problem is how to calculate frictional force. Remember always that kinetic frictional force (friction when an object is sliding) is always equal to the coefficient of kinetic friction times the force that a surface pushes against the object (aka normal force). If you sit on a chair, the chair must push up on you with your own weight to keep you from moving. It is very important also that you understand why In this case you used cosine. The surface only pushes the object perpendicularly to the surface itself. This means that normal force is only equal to the weight of the object in the direction perpendicular to the surface.
  21. Jan 19, 2007 #20
    My answer still seems to be wrong and i've check my math a few times could someone possibly tell me what i have done wrong please?
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