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Thermal energy problem. How was this answer reached?

  1. Oct 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Two railroad cars, each of mass 7650 kg and traveling 95 km/h in opposite directions, collide head-on and come to rest. How much thermal energy is produced in this collision?


    2. Relevant equations
    I'm assuming:
    KE=1/2mv^2
    HE=mc(delta)T


    3. The attempt at a solution
    The answer is 5.3x10^6 Joules, or 5300000 Joules.
    I don't see enough information in the question to find thermal energy, yet somehow an answer was reached. I'm guessing that there is some mix of equations or there is another way of solving that I'm not aware of... A variation of the question is on my homework, so any help in figuring out how this was done is greatly appreciated.

    Thank you!
     
  2. jcsd
  3. Oct 2, 2009 #2

    Doc Al

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    Where do you think the thermal energy comes from?
     
  4. Oct 2, 2009 #3
    Hey Doc,

    I would assume the energy is coming from work the two trains are performing on each other, which should be the same. It could also come from the friction between the trains and the track, but I think that is irrelevant.

    In my notes, I have an equation involving momentum ((delta)p=mv). Would this be the equation used? I understand that momentum is another word for collision (in my class, at least), correct?

    Also, KEi =/= KEf because the two trains should stick together on impact, correct?
     
  5. Oct 2, 2009 #4

    Doc Al

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    It's simpler than all that. Think in terms of energy being transformed from one form to another. What kind of energy do the trains start with?
     
  6. Oct 2, 2009 #5
    Potential energy, no?
     
  7. Oct 2, 2009 #6

    Doc Al

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    No. Hint: The trains are moving.
     
  8. Oct 2, 2009 #7
    Then kinetic energy.
     
  9. Oct 2, 2009 #8
    That is correct.
    Now think about what that energy turns into, and apply the appropriate conservation laws to find the answer you're looking for.
     
  10. Oct 2, 2009 #9
    This is where I keep getting messed up. I'm not sure how to convert the energy.
     
  11. Oct 2, 2009 #10

    Doc Al

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    You don't need to "convert" anything. One form of energy (mechanical KE) just transforms into another (thermal energy). How much KE do the trains have to start with? How much do they end up with?
     
  12. Oct 2, 2009 #11
    I'm getting the KEi as 2,663,628.472 and I'm fairly sure KEf is zero. It's tackling the problem from there that I have no idea about.
     
  13. Oct 2, 2009 #12

    Doc Al

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    That's the KE of each car. What's the total KE? (And what are the units of energy?)
    Good. So how much of the initial KE was "transformed" into thermal energy?
     
  14. Oct 2, 2009 #13
    So you know [tex]E_{k_i}[/tex] and [tex]E_{k_f}[/tex] (Taking Doc Al's corrections into account)

    What can you say about the total energy prior to, and after the collision? In what form is the energy prior to the collision, and in what form is it after the collision? And most importantly, what can you say quantitatively?
     
  15. Oct 2, 2009 #14
    Wow. I feel dumb. All I've had to do this whole time is double the KE. :/ Okay thank you guys!

    Oh, and if I have an angle at which something is pushed and a vertical height that the object gained, how can I find the speed?
     
  16. Oct 2, 2009 #15
    The concept in your new question is exactly the same as that in this one. It is conservation of energy.

    For a closed system, energy can not be created or destroyed, it can only be transformed from one form to another.

    What we did in this question was look at the initial and final states (Prior to, and after the collision).
    We calculated the total energy in each of them, said that they were equal (Conservation of energy!) and isolated for our unknown.

    Writing this out would be:

    [tex]E_{tot_i}=E_{tot_f}[/tex]

    [tex]E_{tot_i}=E_{k_i}[/tex]
    [tex]E_{tot_f}=E_{k_f}+Q[/tex]

    Since we know the initial energy, and the kinetic energy in the final state, we were able to isolate our unknown, the heat, [tex]Q[/tex]

    Though for this question that was especially easy since all the kinetic energy was transformed into heat.

    Try applying the same principle to your new question. Look at the total energy in the initial state, and the total energy in the final state. Use the conservation of energy to say that these two energies are equal and isolate for your unknown, [tex]V_f[/tex]

    It would be wise to use potential energy to solve this new problem.
     
  17. Oct 2, 2009 #16
    Here is the actual question:

    "A sled is initially given a shove up a frictionless 22.0° incline. It reaches a maximum vertical height 1.15 m higher than where it started. What was its initial speed?"

    I'm trying to solve for Vi, and Vf is equal to zero.

    I got the distance traveled by the sled by doing Xsin22=1.15 and solving for X.
    X=3.069887237

    How can I use the angle of 22 degrees along with the other information to solve for Vi?
     
  18. Oct 2, 2009 #17

    Doc Al

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    Two ways to go:
    (1) Conservation of energy. Don't forget the work done by the friction force.
    (2) Dynamics. Find the acceleration (using Newton's 2nd law) and then use kinematics.
     
  19. Oct 2, 2009 #18
    Okay thank you guys again! I got that one too.

    I used PE=KE, or mgh=1/2mv^2.

    The m is canceled out on both sides, so the equation becomes gh=1/2v^2
    Put in the information and you get:
    (9.8)(1.15)=1/2v^2
    11.27=1/2v^2
    divide both sides by 1/2 and you get:
    22.54=v^2
    take the square root of both sides to get v=4.747630988

    This is the correct way to do this, right? I'm pretty sure it is.
     
  20. Oct 2, 2009 #19

    Doc Al

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    Perfect! Just round off that answer to a reasonable number of significant figures and attach the proper units.
     
  21. Oct 2, 2009 #20
    I'm confused on another one... I have the first half of it, but finding the height is proving to be a bit more difficult. Here she is:

    "A vertical spring (ignore its mass), whose spring stiffness constant is 870 N/m, is attached to a table and is compressed down 0.180 m.

    (a) What upward speed can it give to a 0.400 kg ball when released?
    8.226748325 m/s
    (b) How high above its original position (spring compressed) will the ball fly?
    ? m"

    What I know:
    m= .4 kg
    v= 8.226748325 m/s
    k= 870 N/m
    X= it is the compressed distance, correct? then .180 m.

    I have tried to solve for h by using PE=KE, but no luck so far. I don't know another way to do it, so any hints would be helpful.

    Never mind I got it. Simple math error.
     
    Last edited: Oct 2, 2009
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