1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with simple work / nrg conservation question

  1. Nov 15, 2004 #1
    The question is
    "Two masses comprise an inertial system: the first, a 2 Kg mass which rests on a rough table top; then second, 5 Kg mass which hangs freely 0.4 m above the floor from an ideal pulley at the edge of the table top. The two masses are connected by an inextensible string run over the pulley. The coefficient of kinetic friction between the table top and the first mass is .40."

    first I tried
    [tex]m_{1} = 2 kg[/tex]
    [tex]T-F_{k}=m_{1}a \ where\ F_{k}= \mu N \ and \ N = m_{1}g[/tex]
    so
    [tex]T=m_{1}a+ \mu m{1}g[/tex]


    Then mass 2
    [tex] m_{2} = 5 kg[/tex]

    [tex] F= MA [/tex]
    [tex]m_{2}g -T = m_{2}a[/tex]

    substituting T for what T = in part 1 gives me
    [tex]m_{2}g -m_{1}a - \mu m{1}g = m_{2}a[/tex]

    so a = [tex] (m_{2}g - \mu m_{1}g)/(m_{1} + m_{2}) [/tex]

    pluging in 5 kg for m2 98. m/s^2 for g .4 for mu and 2 kg for m1 gives me an acceleration of 5.88 m/S^2

    then to solve for velocity I would go [tex] W_{nc} = E_{f} - E_{i} [/tex]
    so [tex] -\mu m_{1}gd = 1/2m_{1}v^2 + 1/2 m_{2}v^2-m_{2}gd[/tex]

    then solving using the same numbers above and using .4 for d v comes out to be 2.55 m/s

    could I also solve for T in equation 1 to get a conservative work force of 19.6 J?

    someone also showed me that

    delta K = work total = work gravity + work friction
    which is 1/2(m2 + m1)(v^2 -0) = -m2g(-d -0) + (- mu m1g)d
    then it comes down to v = [tex] \sqrt{\frac{2gd(m_{2}- \mu m_{1}}{m_{2}+m_{1}}} [/tex]

    but that gives me a v = 4.704 m/s which is not 2.55 m/s


    and someone told me the anwser to the velocity component is 2.88 m/s

    I am confused

    moreover, how could i figure out how far it traveled once the free hanging block hit the ground?
     
    Last edited: Nov 15, 2004
  2. jcsd
  3. Nov 15, 2004 #2

    Diane_

    User Avatar
    Homework Helper

    I presume you're trying to figure the acceleration of the system in your first part.

    I get the same acceleration for the system, 5.88 m/s^2. (Note - in your text, you mention m1 as 3 kg, but you seem to have used the proper number in your calculations.)

    In your energy calculations, though, I'm a little confused by your signs. Try it this way: think of the energy in the system as "gozinta's" and "gozouta's". Energy comes into the system due to the change in potential energy of block 1, at least until it hits the ground. Energy goes "out" of the system, appearing as kinetic energy in the blocks and work done by friction. So:

    change in PE for block 1 = change in KE for block 1 + change in KE for block 2 - work by friction

    m1 gd = 1/2m1v^2 + 1/2 m2v^2 - (mu)m1gd

    Note a simplification:

    m1 gd = 1/2(m1 + m2)v^2 - (mu)m1 gd

    In other words, it's the total kinetic energy of block 1 and block 2 that matters - they act as a unit. This should make a lot of sense.

    Solving for v, we get

    v = sqrt(2gd(m1 + (mu)m2)/(m1+m2))

    Note the difference in sign from your version.

    This results in a velocity of 2.55 m/s, which you indicate is the correct answer.

    For the rest of it, find the kinetic energy of the block on the table when the other one hits the floor and figure out how far it would have to slide for friction to do exactly that amount of work on it.
     
  4. Nov 15, 2004 #3
    i was checking over this and noticed that

    [tex] v_{f}^2 = v_{o}^2 + 2a delta x [/tex]
    2.55(m/s)^2 = 0 + 2 (5.88m/s^2) (.4m)
    is not valid

    so does that mean it is wrong? or am i just confused?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Help with simple work / nrg conservation question
Loading...