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Spring mechanics with 2 different masses

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Two masses m and 3m are joined together by a compressed massless spring of constant k = 750 N/m. When spring is released it pushes masses apart and they slide on a horizontal tabletop as shown below. Mass 3m slides to the right and its track is frictionless. Mass m slides to the left and part of its track of length of .75 m has a friction where mu = .15. Compression of the spring is .25 m. For the mass 3m there is a ramp at the end of the tabletop. The ramp makes an angle of 20 degrees with the horizontal. Mass m is .5 kg. A length of the table is 2 m. Spring regains its initial length before mass m starts to slide with friction and mass 3m reaches the ramp. Using the data given above answer the followings:
    1. calculate the potential energy of the spring.
    2. Calculate the speed of the mass m before it starts to slide with friction.
    3. Calculate speed of the mass 3m before it enters the ramp.
    4. Calculate the work of the friction force for the mass m.
    5. Calculate the speed of the mass m when it leaves the table.
    6. Calculate the speed of the mass 3m when it leaves the ramp.
    7. For the mass m find the time that is required to reach the floor (from the edge of the tabletop to the floor).
    2. Relevant equations and attempt at a solution
    for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J

    for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s

    I believe problem 3 is solved the same way but that we change the mass so V = 5.59 m/s

    problem 4 I believe would use 2 formulas W = fd and F(f) = mu* mg, where F(f) = force friction so it would be mu *mg *d = .552 J

    problem 5 if I am not mistaken would use the equation V = (square root 2g delta h) = square root 2 *9.81 * .9 = 4.2

    Problem 6 is where I'm a little hazy. Also I am not sure If I'm using the right equations. I have not done mechanics since November so if some one could check my work I would very much appreciate it. Also please excuse the English my teacher is Russian and I wrote the problem word for word.
     
  2. jcsd
  3. Apr 12, 2008 #2
    this picture should help
     

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  4. Apr 12, 2008 #3
    Is anyone there?
     
  5. Apr 12, 2008 #4
    I believe I made a mistake on # 5 so what I think I was suppose to do was take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.
     
  6. Apr 12, 2008 #5

    Astronuc

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    The energy of the spring becomes the kinetic energy of both mass, m and 3m. Conservation of energy. The same force is applied to both block, F = kx. Also, the total (net) momentum before the blocks move is zero, and then think about conservation of momentum.
     
  7. Apr 12, 2008 #6
    I'm confused how does that help me find the speed of mass 3m when it leaves the ramp? Your answer seems to be the perspective I need to look at this problem, is their something wrong with how I have been solving the problems? If so where did I mess up and if not using work/energy formulas how do I find the velocity of mass 3m going off the 20 degree ramp?
     
  8. Apr 12, 2008 #7

    Kurdt

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    Take this one step at a time. What is the potential energy of a compressed spring?
     
  9. Apr 12, 2008 #8
    I've already solved that, I'm on step 6 but the answer is 22.44, all the info you need is in the first, second, and fourth post.
     
  10. Apr 12, 2008 #9
    The way I see it I found the x direction velocity in problem 3 so I need to find V that accommodates two dimensions (the x as well as the y). so i am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s. So I'm wondering if my method is correct?
     

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  11. Apr 12, 2008 #10
    so T should = .43 because y=.5at^2, so t = square root y/(.5a)
     
  12. Apr 12, 2008 #11
    final answers:
    for problem 1: PE =KE, where PE = potential energy and KE = kinetic energy formula two equals .5K delta X^2. this means PE = (750*.25^2)/2 = 23.44 J

    for problem 2 I believe you use V = square root (2KE)/m = square root 2(23.44)/.5 = 9.68m/s

    I believe problem 3 is solved the same way but that we change the mass so V = 5.59 m/s

    problem 4 I believe would use 2 formulas W = fd and F(f) = mu* mg, where F(f) = force friction so it would be mu *mg *d = .552 J

    problem 5 take the work of the spring - the work of friction = (mv^2)/2 now multiply the new work by 2 divide by m and square root both sides and you get velocity. so the answer should be 9.568 round to 9.57.

    problem 6 I am using the formula v0cos 20 = vx so vx/cos20 = v0 which gives me 5.55m/s.

    problem 7 T should = .43 because y=.5at^2, so t = square root y/(.5a)

    I just want someone to tell me if 6 and 7 are correct
     
  13. Apr 13, 2008 #12

    Kurdt

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    For parts two and three you haven't applied the conservation of energy and momentum as Astronuc suggested.
     
  14. Apr 13, 2008 #13
    why doesn't my answer take into account the conservation of energy and momentum/how do It the right way?
     
  15. Apr 13, 2008 #14

    Kurdt

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    The kinetic energy of the two masses must add up to the potential stored in the spring. You have calculated the kinetic energy for each mass as if they both gain all the potential energy.
     
  16. Apr 13, 2008 #15
    OK I'm not quite sure how to do that, But this is my attempt. Because of conservation of momentum I know PE = (.5V1^2)/2 + (1.5V2^2)/2 but that leaves me with two variables, so what do I do? M =.5 and 3M = 1.5, which means I am using the formula KE = (mv^2)/2
     
    Last edited: Apr 13, 2008
  17. Apr 13, 2008 #16

    Kurdt

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    one must also conserve momentum.
     
  18. Apr 13, 2008 #17
    Can you give me a formula or a procedure that I can use, cause I know I'm not doing this right, but I want to know the right way.
     
  19. Apr 13, 2008 #18

    Kurdt

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  20. Apr 13, 2008 #19
    I'm sorry I can't figure this out.
     
  21. Apr 13, 2008 #20
    I'm running out of time can't you just tell me what I need to do? I can do the rest and I have shown an attempt at the problem. Also I know that 1,4,5,and 7 are correct. and their were actually 11 problems and I can solve those If I can find the answer to 2 and 3.
     
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