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Homework Help: How to find work from mass, friction given?

  1. Mar 6, 2013 #1
    a Child drags a 25-kg box at a constant speed across a lawn for 14.0 m and along a sidewalk for 36 m; the coefficient of friction is 0.25 for the first part of the trip and 0.55 for the second. If the child always pulls horizontally, how much work does the child do on the box?

    I know that Work= force*distance but how would I find these variables?
    idk if this works but I can do 25kg * 9.81=245 as force
    Last edited: Mar 6, 2013
  2. jcsd
  3. Mar 6, 2013 #2


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    Hint: The child is doing work by pulling against the frictional force on the box. Do you know how to calculate this force?
  4. Mar 6, 2013 #3
    That is a relevant force yes, but it is acting only in y, and is cancelled out by the normal force.

    First. Draw a free body diagram. What is the free body? What forces are acting on this free body?

    Hint: How much force does it take to overcome the friction in each individual part of the trip?
  5. Mar 6, 2013 #4
    245 n *0.25=61.25n
  6. Mar 6, 2013 #5
    Good, now what?
  7. Mar 6, 2013 #6
    we have to find the distance , I dont suppose we add 14 + 36=50m for the distance, do we?
  8. Mar 6, 2013 #7
    is the 55 sec relavent in this problem?
  9. Mar 6, 2013 #8
    Not quite, you have two different surfaces. Each with a different coeffecient of friction.
    I believe that is .55 for the coefficient of friction for the sidewalk.
  10. Mar 6, 2013 #9
    1st part)25*.25*14=87.5J
    2nd part) 25*.55*36=495J
  11. Mar 6, 2013 #10
    You're missing something. What are you multiplying? You found the force needed to move across the grass here:
    Yet, you just threw that value to the wind and multiplied 3 values together.

    You know that W=FDcosθ. And you know the the displacement. So what now?

    *where theta is the angle between the Applied force and the displacement. (In this case 0, cos0=1.)*
  12. Mar 6, 2013 #11
  13. Mar 6, 2013 #12
    I added them both and it was correct, Thank you!
  14. Mar 6, 2013 #13
    Nice! Just keep in mind that Force and Displacement are vectors. Work isn't just F times D. It's F times D times the cos of the angle between them. In nice situations the angle is 0. But in some cases, the there will be an angle between them, and then it can get a little tricky!
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