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KE and PE Work

  1. Oct 18, 2007 #1
    #1
    A 10kg crate is pulled up a rough incline with an initial speed of 1.5 m/s. The pulling force is 100Nparallel to the incline, which makes an angle of 20* with the horizontal. If the coefficient of kinetic friction is .40 and the crate is pulled a distance of 5 m.
    a)HOw much work is done by the 100N
    ----------------
    Fg = 98 N
    Fn = 92.1 N
    Fpar = 33.5 N
    Ffricton = 36.84 N
    Fa = 100 N
    Fnet = 100 - 33.5 - 36.84 = 29.66 N
    W=F.d
    d (in the vertical) = 4.7 m
    a) W=29.66 (4.7) 139.402 J

    #2
    In a circus performance, a monkey on a sled is given an initial speed of 4 m/s up a 20* incline. The combined mass of monkey and sled is 20 kg, and the coefficient of kinetic friction between sled and incline is .20. How far up the incline does the sled move ?
    --------------
    Fn = 184.2 N
    Fpar = 67.03 N
    Ff = 36.84 N
    Fnet = Fpar - Ff = 30.19
    KE = .5(20)(4)^2 = 160 J
    KE = W
    W = F.d
    160 = 30.19 d
    d = 5.3 m

    What i did wrong on those 2 problem ?
     
  2. jcsd
  3. Oct 18, 2007 #2

    learningphysics

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    For #1

    Have you studied dot product?

    The work done by the 100N force is 100N times the component of displacement parallel to the 100N force... what is the component of displacement parallel to the 100N force?

    For #2

    Fpar and Ff are acting in the same direction. be consistent with your signs... down the plane negative... up the plane positive...

    Net work done = change in kinetic energy

    Net work done = -160
     
  4. Oct 18, 2007 #3

    learningphysics

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    I'm wondering, are there more parts to problem #1? Because there's a lot of information that's not necessary for part a).
     
  5. Oct 18, 2007 #4
    #1 No, I haven't studied the dot product yet
    A) HOw much work is done by the gravitational force ?
    b) Ho much work is done by 100 N force ?
    c) What is the change in kinetic energy of the crate ?
    d) What is the speed of the crate after it is pulled 5 m?

    Sorry about that
     
    Last edited: Oct 18, 2007
  6. Oct 18, 2007 #5

    learningphysics

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    Ah... I see. Ok... for all these work problems... to get the work done by a force... you need the force * (component of displacement parallel to the force)

    starting with a)... you need:

    the gravitational force * displacement parallel to the gravitational force

    so you need displacement in the up/down direction (since that is the direction of the force)... take signs into account for both the force and displacement... take up positive and down negative...

    so what is the work done by the gravitational force?
     
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