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  1. Oct 8, 2003 #1
    Ok I have been workoing on this problem for at least a couple of hours with no luck


    A 1.0-kg block is released from rest at the top of a curved fric-tionless track as shown in the figure below. (a) What are the speeds of the block at points A and B? (b) If the block goes on a level surface at point C with a coefficient of kinetic friction of 0.50, how far from point C will the block come to rest?

    Attached is a graphical representation

    Here is what I did so far


    KE =1/2mv^2
    M= 1.0 kg
    V= ?
    Wsubg= mgh

    For point A
    Final Kinetic energy = KEInitial - W subg
    W sub g = mgh 1.0 X 9.8 X 2.0
    Kfinal = 20

    20-1/2(1.0)v^2
    40= v^2
    v= 6.3 m/s that is v of point a


    for point b
    Initial kenetic = 20

    Final Kinetic energy = KEInitial - W subg
    W sub g = mgh 1.0 X 9.8 X 6.0
    final kentic = 60-20=40

    40=1/2(1.0)v^2
    80=v^2
    v 8.9 m/s for point b


    I then calculated the maximum static firction
    using the coefficient of kinetic friction of 0.50
    mg X coefficient friction

    1.0X9.8X.50=4.9N

    that is the maximum force that can be exerted wihtout the objectmoving

    but I do not know how to claculate how far from point C will the block come to rest

    I do not know if I did the first three parts correct either

    I have been at this for a while now over 2 hours

    I desperatley need help any advice would be much appreciated
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2003 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Since the object is at rest initially, V= 0 and the potential energy is 1(9.8)(9)= 88.2 Joules

    Why? I assume the "20" is an approximation to 19.8 which is 9.8*2. That's the potential energy not the kinetic energy.

    At point A, the object's height is 2 so its potential energy is
    1(9.8)(2)= 19.6 Joules. That's a decrease of 88.2- 19.6= 68.6 Joules which must have gone into kinetic energy: (1/2)(1)v2= 68.6 so v= 11.7 m/s.

    You're going the right way but without friction the total energy is still the potential energy at the start: 88.2 Joules. At point B the height is 6 meters so its potential energy is 1(9.8)6= 58.8 Joules (you round off to 60). That's a change (from the INITIAL point) of 88.2- 58.8= 29.4 Joules. We must have (1/2)(1)v2= 29.4 or
    v= 7.7 m/s at point B.

    Finally, at C, the height is 0 so potential energy (relative to 0 height of course) is 0. The 88.2 Joules of energy is now all kinetic energy: (1/2)(1)v2= 88.2 so v= 13.3 m/s.

    No, your calculation is correct but 0.5 is the coefficient of KINETIC friction, not static friction! Since friction is imposing a force of 4.9 N, we have f= ma or -4.9= (1.0)a. The acceleration is
    -4.9 m/s2. At that acceleration the object slows from 13.3 m/s to 0 m/s in (v= at) 13.3/4.9= 2.7 seconds. Of course,
    x= v0 t- (a/2) t2 so x= (13.3)(2.7)- (4.9/2)(2.7)2= 18 meters.
     
  4. Oct 8, 2003 #3
    Clarification

    One quick question before I get started



    why is it times 9 and not times 10 if initial height is 10?
     
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