1. Limited time only! Sign up for a free 30min personal 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!

Law of Conservation of Mechanical Energy Exercise

  1. Feb 8, 2016 #1
    1. The problem statement, all variables and given/known data
    A mass ##m_1 = 5.0 kg## is hanging from the end of a thread, of negligible mass, that slides on a pulley, of negligible mass too and without friction. At the other end of the thread, at the same height of ##m_1##, there is another hanging mass ##m_2 = 3.5 kg##. Using the Law of Conservation of Mechanical Energy, determine the velocity of ##m_1## when it goes down, starting still, of a height of ##2.5 m##.

    2. Relevant equations
    ##E = K + U##
    ##K_f + U_f = K_0 + U_0 \Rightarrow E = const##

    3. The attempt at a solution
    So, since we have two masses that move at the same time but have different masses, we can start writing these:
    ##K_0 = \frac{1}{2} * m_1 * v_0^2 + \frac{1}{2} * m_2 * v_0^2##
    But since we know it starts still, ##v_0 = 0## and so ##K_0 = 0##.
    ##U_0 = 0## because it hasn't moved yet.
    ##K_f = \frac{1}{2} * m_1 * v_f^2 + \frac{1}{2} * m_2 * v_f^2##
    with ##v_f## being the velocity we are searching for.
    ##U_f = m_2 * g * h - m_1 * g * h##
    So, using the second relevant equation we have:
    ##K_f = - U_f##
    ##\frac{1}{2} * m_1 * v_f^2 + \frac{1}{2} * m_2 * v_f^2 = m_1 * g * h - m_2 * g * h##
    ##\frac{v_f^2}{2}(m_1 + m_2) = g * h(m_1 - m_2)##
    ##v_f =\sqrt{ \frac{2 * g * h(m_1 - m_2)}{(m_1 + m_2)}} = 3 \frac{m}{s}##
    Is this method I used to find the velocity with the law good? Or I could have done better and faster?
     
  2. jcsd
  3. Feb 8, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    This is about it, can't do much better than ##\Delta K + \Delta U = 0## :smile:
     
  4. Feb 8, 2016 #3
    Okay! Thank you for replying! :D
     
  5. Feb 8, 2016 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're welcome.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Law of Conservation of Mechanical Energy Exercise
Loading...