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Masses and pulley problem - stuck

  1. Oct 23, 2003 #1
    Hi. I am trying to solve for two things here. I need the speed of mass 1 when mass 2 hits the ground, and I need to find out how high mass 1 will climb before returning to rest.

    Mass 1 is 3kg, mass 2 is 5kg, and they are connected by a light pulley and string. mass 2 is elevated 4meters, and mass 1 is at rest at height 0. When mass 2 is "let go", what is the speed of mass 1 when mass 2 hits height 0, and how high will mass 1 reach before returning to rest?

    Can someone clue me in as to how to solve this problem? I am trying to help my husband, the student, and don't know how to go about doing so. Any beginnings would help, and if there's a formula out there, I would love to know about it.

    Thanks so much.
     
  2. jcsd
  3. Oct 23, 2003 #2

    jamesrc

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    Hi,

    There are two main ways you can go about solving this. One is drawing free body diagrams and finding the acceleration of the masses and working through the kinematics. The other way is to use the conservation of energy. For the conservation of energy method, all you have to consider is gravitational potential energy and kinetic energy in this problem. There are no losses in the system (no friction) and we assume there are no elastic elements (the strings don't stretch) and there is no energy stored in the pulley (it has no inertia). So if I use T* to indicate kinetic (co)energy and U to indicate potential energy and the subscripts i and f to indicate initial and final, respectively, the conservation of energy can be written:
    T*i + Ui = T*f + Uf

    Gravitational potential energy is found from mgh, where h is the distance above some datum you define (the initial height of m1 is a good choice here). Kinetic energy is given by .5*m*v^2

    This will allow you to solve for the velocity of m1 when m2 is at h = 0. Remember to account for the energy terms of both masses at the initial and final state. Also keep in mind that the geometric constraint of the string requires that the magnitude of the velocity of m1 will equal the magnitude of the velocity of m2.

    For the second part, I'm not clear on what is being asked. Wouldn't it depend on the nature of the collision between the mass and the ground and whether or not you allow the string to go slack? I mean, if you assume m2 sticks to the ground when it hits it and m1 acts as a simple projectile with voy given by the above calculation and initial height of 4 m, you can find out how high it goes after that until its y-velocity is decelerated to 0 (using, for example, v^2-vo^2 = 2*a*d). No matter what your assumptions are there, eventually m1 will come to rest at a height of 4 m, simply because m1<m2. Anyway, I hope that's somewhat useful...
     
  4. Oct 23, 2003 #3
    Assuming that the cord flexes to allow the rising mass to freely lift, is it correct to say that when the rising mass has the same potential energy as the original raised mass had, before it fell, that that is when it will stop? Assuming that the rising mass will act as a projectile once the falling mass hits the ground, is it correct then to say that it will achieve its maximum height when its potential energy is equal to the potential energy the falling mass had to begin with?

    We have the book's answer in the key of the book, but it is not the answer we get on paper when numbers are plugged into the formulae.
     
  5. Oct 23, 2003 #4

    jamesrc

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    I don't think so. When the heavier mass hits the ground, one of two things can happen: it sticks to the ground in which case energy is lost, or it bounces and energy is still tied up in the moving mass. I think that, either way, you would want to find out where the kinetic energy the rising mass has goes to 0.

    So, from the first part, you should get that v ~ 7.2 m/s
    Here, m1 has a kinetic energy of ~78 J and a potential energy of ~118 J. At its maximum height, this will all be potential energy, so it should be ~6.67 m.
     
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