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Energy, work, 2 masses on incline

  1. Jul 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Two blocks with masses m1=4.0kg and m2=5.0kg are connected by a light rope and slide down a frictionless wedge. The angle of the side on which m2 is moving is 53 degrees, and the angle of the side on which m1 is moving is 37 degrees. Basically the system looks like a triangle with the masses sliding down the two sides. If the system starts from rest, what is the speed of m2 after it has moved 40.0cm along the incline?
    GIVEN: m1=4.0kg m2=5.0kg d2(displacement of mass 2)=40.0cm




    2. Relevant equations
    Ek= 1/2mv^2 Eg=mgh




    3. The attempt at a solution
    I know that in this situation, the acceleration is going to be towards m2 so m2 is going to lose gravitational potential energy and gain kinetic. I know that m1 is going to gain gravitational potential energy and lose kinetic energy. I also know that the energy is conserved. I could calculate Eg for m2 but im not sure if i would use 40.0cm for the height.
     
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  3. Jul 15, 2008 #2

    alphysicist

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    That's not correct for [itex]m_1[/itex]; it will gain potential energy, but if it starts from rest, how can it lose kinetic energy?

    The [itex]h[/itex] in the gravitational potential energy formula is the position in the vertical direction; so how much it has moved in the horizontal direction does not appear in [itex]h[/itex] at all. Do you see how to set it up now? What do you get?
     
  4. Jul 15, 2008 #3
    I figured the kinetic energy for m1 is going to decrease and be converted to potential energy.
    For Eg of m2 i could use the formula Eg=mgh. The block slides down 40.0cm on an angle of 53degrees so the horizontal height would be 0.4m sin53= 0.32m. I substituted that into Eg=mgh
    = 5 x 9.8 x 0.32
    = 15.68J
     
  5. Jul 15, 2008 #4

    alphysicist

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    Since the smallest value for the kinetic energy is zero, there is no way for it to decrease. Both objects are speeding up, and so both of their kinetic energies are increasing.

    That's looks like the right magnitude, but you'll need to be careful with the sign. Most people would say that the original height of m2 was h=0, and so this height is h=-0.32 m (since it's sliding downwards).

    But I've seen other people assemble their equation a bit differently, so what do you get for the entire energy equation?
     
  6. Jul 15, 2008 #5
    h=0 is my reference point.
    for the kinetic energy of m2 i got 15.68 J
    Eg=mgh
    = 5 x 9.8 x 0.32
    = 15.68J
    but i have no clue where to go from there:( i need to somehow involve m1 into this and the law of conservation of energy.
     
  7. Jul 15, 2008 #6

    alphysicist

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    No, that's not right for the kinetic energy of m2. You have to deal with both masses together because they affect each other.



    Since you said energy is conserved, that means that the total energy of both masses at the starting point will be equal to the total energy of both masses at the final point, [itex]E_i = E_f[/itex].

    So at the beginning, when both masses are at rest, what is the total energy of the system? To figure this out, you've already listed the types of energies possible, so what is the kinetic energy and potential energy of m1? what is the kinetic energy and potential eneryg of m2? Remember that you already have the masses, so all you need is the speed and height of each one.


    Then do the same for the final point: what is the kinetic and potential energy of each mass at the final point? The speed at the final point is the unknown, so all you need is the height of each mass at each of their final points.


    Altogether you'll have eight energy terms, but four of them should be zero.
     
  8. Jul 15, 2008 #7
    okay but arent the speed and the height of both masses zero since they start at rest (v=0) and my reference point is 0 (h=0)? substituting zero to the equation will get me zero...that doesnt seem right.
     
  9. Jul 15, 2008 #8

    alphysicist

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    No, that's perfect! Remember you're not setting a whole equation to zero (yet), its just the initial energy. So the initial energy is zero, or:

    [itex]E_i[/itex] = (initial KE of [itex]m_1[/itex] ) + (initial PE of [itex]m_1[/itex]) + (initial KE of [itex]m_2[/itex]) + (initial PE of [itex]m_2[/itex]) = 0

    So now calculate the four energy terms for [itex]E_f[/itex] (just like you did for [itex]E_i[/itex]; and then to get your equation, you'll set

    [tex]
    E_i = E_ f
    [/tex]
     
  10. Jul 15, 2008 #9
    Oh...okay. Yea I tried it and surely enough I had 8 terms and 4 were zero.
    I got the height using 0.4sin53= 0.32 and the and speed of mass 1 using: Eki+Egi=Ekf+Egf
    vi=2.5m/s
    i substituted that into the 8 term equation below:
    1/2m1v1+1/2m2v2 +m1gh +m2gh= 1/2m1v1' +1/2m2v2' +m1gh' +m2gh'
    0 +0 +0 +15.68= 12.5 +2.5v2^2' + (-12.544)
    v2'=2.5m/s
    Does make sense? would it be correct? I hope i didn't forget to to accoutn for anything else.
     
  11. Jul 15, 2008 #10
    account*
     
  12. Jul 15, 2008 #11

    alphysicist

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    It's almost there, but there are a few problems.

    Your energy equation and the equation with values is:

    1/2m1v1+1/2m2v2 +m1gh +m2gh= 1/2m1v1' +1/2m2v2' +m1gh' +m2gh'

    0 +0 +0 +15.68= 12.5 +2.5v2^2' + (-12.544)

    The left hand side is okay; in particular, you have set the original height of m1 to be zero, and the final height of m2 to be zero. That's fine, we just have to keep track of it.

    On the right side, there are a couple of problems. You don't have the kinetic energy term for m1. Also, you have two potential energy terms for m1, one with a positive sign and one with a negative sign. To decide which is which, since m1 started at h=0, does it end up higher or lower than it began? If it's higher, the final h will be positive (and so will the final potential energy of m1); if it's lower, h will be negative.

    (Also, I don't think the number 12.5 is correct; I think you might have used the wrong angle for m1.)

    So you still need to find the following terms:

    [tex]
    \begin{align}
    \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 +m_1gh_1 +m_2gh_2= \frac{1}{2}m_1v_1'^2 +\frac{1}{2}m_2v_2'^2 +m_1gh_1' +m_2gh_2'\nonumber\\
    0+0+0+15.68=(\mbox{kinetic energy term for m1?})+2.5v_2'^2+(\mbox{potential energy term for m1?}) + 0\nonumber
    \end{align}
    [/tex]

    Once you get those last two terms in with the right variables and values, you'll be able to find the speed.
     
  13. Jul 15, 2008 #12
    im getting 2.5m/s for my kinetic energy for m1final. For the Eg for m1final I'm getting -12.544. Im not sure if these numbers are right though since in mgh i made g= -9.8.
     
  14. Jul 15, 2008 #13
    oops mistake...for my kinetic energy for m1final I'm getting 12.5 and for the Eg of m1 final Im getting zero
     
  15. Jul 15, 2008 #14

    alphysicist

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    The g always stands for the magnitude 9.8 m/s^2; you don't make it negative. So then the final PE for m1 is positive (since the final h for m1 is positive).

    But how did you get the number 12.544? Did you use the wrong angle? If you're still getting 12.54 J, please post the actual numbers that you used. (And remember, m1 is on a 37 degree incline.)

    For the kinetic energy of m1, you can't find the actual number yet; you don't know how fast m1 and m2 are going. That's what you are trying to find in this problem. So the KE term for m1 should have the unknown variable v in it.
     
  16. Jul 15, 2008 #15
    okay I think Im doing something incredibly wrong in my angles. When i try to get the height of the triangle I am doing 0.40m x sin 53 which gives me 0.32. The height occurs several times in the 8 term equation which is why my answer keeps coming wrong. Im not sure what to do about the height.
     
  17. Jul 15, 2008 #16
    I wish i could upload an image. This is what the triangle looks like. 0.40 is the hypotenuse of the triangle with the angle 53degrees when you split this triangle into two. That is why I am getting 0.32m as my height. Something's wrong...
     
  18. Jul 15, 2008 #17

    alphysicist

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    The mass m2 is on the 53 degree angle incline; so it moves downards a vertical distance of 0.32 meters.

    The mass m1 is on the 37 degree angle incline; so how far vertically does it move upwards? You would do it the same way, just replacing the angle.
     
  19. Jul 15, 2008 #18

    alphysicist

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  20. Jul 15, 2008 #19
    well the angles are switched. 53degrees is on the left with m2 and 37degrees is on the right with m1
     
  21. Jul 15, 2008 #20
    to get the height im doing 0.40sin53. So when i split the triangle that you've drawn, into 2 the height of both is the same therefore its 0.32.
     
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