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Homework Help: Integration and work

  1. Oct 30, 2004 #1
    hey guys i'm in first year university

    ok these problems should seem really easy but our teacher hasn't done crap in the class so i have a problem gettin it
    if someone can help me these two i feel like i could do the rest by myself
    thanks alot

    A reluctant 5-year old, whose mass is 26.2 kg, is pushed at constant speed into a washroom for a bath. His father pushes him 9.53 m along a level floor with a force directed 28.1° below the horizontal. If the co-efficient of kinetic friction between the floor and the child's socks is 0.186, how much work is done by his father in pushing him into the washroom?

    How much energy was lost through the frictional force?
  2. jcsd
  3. Oct 30, 2004 #2
    Consider the forces acting on the child and consider the sum of these forces if the child is moving at a constant speed. Recall that [tex]F_k = \mu_kF_n[/tex] where [tex]F_n[/tex] is the normal force from the surface of the level floor and [tex]\mu_k[/tex] is the coefficient of kinetic friction. Also recall that [tex]W = Fdcos(\theta)[/tex]

    PS Welcome to PF!
    Last edited by a moderator: Oct 30, 2004
  4. Oct 30, 2004 #3
    hey thanks alot
    i really appreciate it
    i'll go and try to do the problems now

    vsage, u go to the univeristy of florida

    GATORS rock baby
    lol bad loss today against the bulldogs
  5. Oct 30, 2004 #4
    Yeah I was at the game. Go gators! It was painful though :(
  6. Oct 30, 2004 #5
    man i dunno why i can't get it
    i know the force exerted by the father is the same as the force exerted by friction

    so F= 9.81*26.2*.186=47.81N


    w= 47.81 *(cos 118.1) *9.53 = -214.6 J

    i also tried 214.6 J

    this program gives me 10 tries to get it right

    the angle i got was 118.1 degrees becasu he was pushing below he horizontal so the that's the difference between the force and the displacement in degrees

    any other suggestions
  7. Oct 30, 2004 #6
    He was pushing at an angle of 28.1 degrees. That means the force was applied towards the object (child). 118.1 degrees would be correct if he was pulling the child but then he'd go the wrong way :P.

    Hope that helps.
  8. Oct 30, 2004 #7
    that gives me 402 j.. which is also wrong lol
  9. Oct 30, 2004 #8
    maybe the trick is in the ''5 year old" because when you are 5 i think your muscle structure starts to grow so maybe i should triple the force normal


    oh well it was worth a try lol
  10. Oct 30, 2004 #9
    You have to remember that the father is pushing at 28.2 degress below the horizontal. He is using cos(28.2) of his force to push the child in the horizontal direction, and using sin(28.2) of his force to push the child down, adding to the weight of the child, and consequently to both the normal force and the kinetic friction opposing the horizontal force of the father.

    Think of it this way, slid your physics book accross the table, it is very easy, now, try pushing down hard on the book while sliding it, it is harder to push now, not only because you are using force to push it down, but it also takes more force to push it horizontally because you are making the atoms in the book interact more strongly with the table and thus it is harder to overcome friction and move the book.

    Hope this helps,

  11. Nov 1, 2004 #10
    ok i see what you are saying

    however i'm still not sure for one part of the problem


    we have d (9.53 m) so i need to find the force of friction


    We have Mk (.186) so i need to find the force normal

    the parts i'm gonna put "" is the part i'm not sure

    N= ( (9.81)(26.2) + (sin28.1) " (9.81)(26.2) "))

    see you were talking about the horizontal component to his pushing, which would add to the force normal. however, i'm not sure what to multiply the sin28.1 by. Like for example, if it said "the father was pushing with 60.2 N, i would multiply it by sin28.1 *60.2 . however, i don't know the force he is exerting so i can't find the horizontal compontent of that force.

    any tips
  12. Nov 1, 2004 #11

    Doc Al

    User Avatar

    Staff: Mentor

    You need to solve for that force! Call it "F". Set up equations for the vertical and horizontal components. (Both vertical and horizontal forces add to zero.) You'll have two equations and two unknowns, F and N.

    Solve for F. (or N) Then use it to calculate the work.
  13. Nov 3, 2004 #12
    Hey, that's like the exact problem I am having trouble with, except my numbers are a bit different.

    So far this is what I have

    Fxnet = F(father)X - fk = 0
    Fynet = N - (W+(F(father)Y) = 0

    I drew a graph, I used the angle from this problem just so that I don't confuse anyone.

    Edit: Meh!, seems the image link wont work :grumpy:

    So basically

    Now for the force of the father I think it's equal to

    cos 61.9 degrees = (9.8 * 26.2)/F(Father)X

    F(Father)x = 256.76N/cos 61.9

    So am I doing this right, or have I misunderstood the question somehow?
    Last edited: Nov 3, 2004
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