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Force of Friction

  1. Oct 23, 2005 #1
    This is my problem

    A 15 kg child slides, from rest, down a playground slide tht is 4.0 m long, as shown in the figure. The slide makes a 40 degrees angle with the horizontal. The child's speed at the bottom is 3.2 m/s. What was teh force of friction that the sldie was exerting on the child?

    Ok. The answer in the texbook says 75N.


    Here is my WROng attempt

    work=Kinetic Energy 2 - Kinetic Energy 1
    work = 1/2 m (V2squared - v1squared)
    w=1/2 (15 kg) (3.2 m/s squared - 0)
    w=7.5 (10.24 m/s)
    w=76.8 J

    then I..

    Work=force (distance) (cos40)
    f= w/d (cos 40)
    f=76.8 J / 4.0 (cos 40)
    f=14.7 N

    Where is my error.. HELP PLEASE
     
  2. jcsd
  3. Oct 23, 2005 #2
    I've bolded the problem. Kinetic energy is not conserved, total energy is. That means that you must take potential energy into account.
     
  4. Oct 23, 2005 #3
    still no correct answer

    I took what you said into effect. I still dont get 75 N as a result.

    Given

    m=15 kg
    g=9.81m/s squared
    final velocity=7.2 m/s
    initial velocity = 0m/s
    d=4.0 m

    w= gravitational potential energy 1 - gravitational potential energy 2
    w= [(m) (g) (h2)] - [(m) (g) (h1)]
    w= mg (h2-h1)
    w= 15 kg (9.81 m/s squared) (4.0m - 0)
    w= 588.6 J


    then

    work = force (distance) (cos 40)
    f=w/d cos 40
    f= 588.6 J / 4.0 m (0.7660)
    f= 588.6 J / 3.0641
    f=192.1

    therefore the answer is still wrong according to the textbook

    How does Velocity relate to this problem?
     
    Last edited: Oct 23, 2005
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