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Homework Help: 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

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    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

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    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

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    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


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    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


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    It's right for finding the total kinetic energy after the explosion.
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