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!

Speed and coefficeint of friction part 2

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Three smooth stones, A, B and C are initially at rest, in contact with each other on the smooth surface of a frozen lake. The masses of the stones are A = 400 g, B = 300 g and C = 120 g and the coefficient of sliding friction between the stones and the ice is μslide = 0.0150. An explosion between the stones causes them to fly apart across the surface of the ice. Stone A flies off due North with an initial speed of 4.50 m s−1. Stone B flies off due east with an initial speed of 6.60 m s−1.

    If 20% of the total energy of the explosion was converted to the kinetic
    energy of the stones, what was the total energy released in the explosion?

    2. Relevant equations

    KE= 1/2 * M * V^2

    PE= mgh

    3. The attempt at a solution

    total energy= ke+ pe

    work done= force*distance

    force= uR

    μslide = 0.0150

    i dont know R. And i dont know the distance travelend to calculate work done.

    Need help please.
  2. jcsd
  3. Apr 8, 2007 #2

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    1. Use cons of momentum to find the momentum and then the KE of stone C.
    2. The KE of A + B + C equals 1/5 of the energy in the explosion.
    You don't use PE, work done, or frliction.
  4. Apr 8, 2007 #3
    why have they given friction in the quesiton at all??
    i didnt use friction in the first part either.
  5. Apr 8, 2007 #4
    I dont see a need for it either. (here)
  6. Apr 10, 2007 #5

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Maybe it is a smokescreen to make you realize that friction plays no part in collisions.
  7. May 10, 2007 #6
    stone C= m1u1 + m2u2 = m1v1 +m2v2

    m1= 0.120 kg
    u1= 0

    m2= would it stone B + stonce c mass
    u2=? i got no idea what value to put here, as i have the speed of stone b and c.

    could i not just calculate the total energy by just calculating KE , or do i have to also calculate conversation of momentum?
  8. May 11, 2007 #7

    Meir Achuz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Your first equation is not relevant here.
    Start with momentum conservation
    [tex]0=m_A{\vec v_A}+m_B{\vec v_B}+m_C{\vec v_C}.[/tex]
    to find v_c. Then use energy.
  9. May 11, 2007 #8

    speed of c is aprox 36meters/sec in the south west direction.

    KE after collison= 0.5mava^2 + 0.5Mbvb^2 + 0.5McVc^2
    = 0.5 ( 0.4 * 4.5^2 + 0.3*6.6^2 + 0.120* 36^2)
    = 0.5 (8.1+ 13.068+ 155.52)
    =0.5* 176.688
    =88.3 J
  10. May 15, 2007 #9
    is this correct to for the calculation to work out KE
  11. May 15, 2007 #10


    User Avatar
    Homework Helper

    Are you sure that's the value you found for the speed of stone C? Looking at the other thread, I think you determined you had made a mistake with that. Double check it.
  12. May 16, 2007 #11
    is the method correct though to work it calculating KE
  13. May 16, 2007 #12


    User Avatar
    Homework Helper

    It's right for finding the total kinetic energy after the explosion.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Speed and coefficeint of friction part 2