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Kinetic Energy

  1. Oct 16, 2005 #1
    Two blocks (with masses 10.1 kg and 65 kg) are connected by a string. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the block and machine is 0.299. The acceleration of gravity is 9.8 m/s^2.

    Find the change in the kinetic energy of the block on the incline (block on the incline is the 10.1 kg one; the 65 kg is hanging down and pulling the 10.1 kg mass up the incline) as it moves a distance of 16.4m up the incline (which is 38.3 degrees) if the system starts from rest. Answer in units of kJ.

    Ok guys, I tried doing this problem for 2 hours now and got 5/6 attempts wrong online when I input my previous answers. So I basically have one more try left at this one. What I did b4 was find the net force on the entire system and multiply it by the distance, 16.4, and I got 9059.8 J. Now since Kf - Ki = W and Ki = 0, wouldn't Kf just be equal to the work? Apparantly not, so I then tried to subtract the work done by the 65 kg block since they are asking for the change in kinetic energy of the 10.1 kg block, but that didn't work either. I'm up for any suggestions because my friend from college has not called me back yet. Thanks guys ...
     
  2. jcsd
  3. Oct 16, 2005 #2

    Chi Meson

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    Double check that net force. In the positive direction is the weight of the hanging mass. In the negative direction is the parallel component of the weight of the block onthe incline and the frictional force. Frictional force is mu times the perpendicular component of the blocks weight.
     
  4. Oct 16, 2005 #3
    Chi, I'm pretty sure that's the correct net work ... The net force I got is 552.429 N. What I was talking about is my net work which I multiplied that force by 16.4 (using the W = Fd formula) to get (which I then got 9059.83). What do I do with this net force? 9059 is not the correct answer so I doubt I can take that approach by just multiplying the net force by the distance. I think the answer is always coming up wrong because I am taking the kinetic energy of the ENTIRE system rather than just that one block which the question is asking for? If that's the case, how do I go about resolving THAT issue?

    Any more suggestions would be appreciated. I think I'm doing this problem in a totally wrong way. Am I?
     
    Last edited: Oct 16, 2005
  5. Oct 16, 2005 #4

    Päällikkö

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    Of course it is not as 9059,8 J rounds to 9060 J + you were supposed to answer in kJ.

    Could you draw a picture and host it on eg. imageshack, as I'm rather confused about the setup of the problem?
     
  6. Oct 16, 2005 #5
    Haha, should have made myself more clear. Of course I entered 9.0598 kJ on the web as an answer; was still wrong though (I'm only speaking im terms of J to be more clear about what I did exactly). Umm well there's no way for me to scan this thing, but I cropped it off my hw online and uploaded it here:

    http://show.imagehosting.us/show/812253/0/nouser_812/T0_-1_812253.jpg

    Thanks in advance for the help ...
     
  7. Oct 16, 2005 #6

    Doc Al

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    If you want to calculate the change in KE of the mass on the incline, then calculate the net force on that mass. (You'll have to solve for the tension.) Or you can just solve for the acceleration and use kinematics.
     
  8. Oct 16, 2005 #7
    The thing is ... that's exactly what I did Doc. I don't know any kinematics but what you said b4, calculate the net force on that mass. That's what I did, here's the equation I used to calc. the net force (should be correct)

    -(10.1)(9.8)(sin38.3) + T (which you have to use the second block to get, which is equal to (65)(9.8) - (0.299)(10.1)(9.8)(cos38.3)

    in simplified terms: -61.3457 + 637 - 0.299(77.6772) = 552.429 yes ?

    So how exactly do I calculate the net force just on that ONE block? Didn't I just do that since you HAVE to use block #2 to calculate the tension force with? Still not getting this ... :/ BTW, my negative / positive signs in finding the force have to do with the coordinate axis I used, I used it parallel to the friction force (x) and parallel to the normal force (y).

    Anyway, I did what you said and solved for tension without the use of the second block and got 84.5712 (-T = -61.3457 -0.299(77.6772)). But what to do after that?

    Wait, would acceleration be zero on both masses? It would, correct? Erm, this prob. is irritating me now ...
     
    Last edited: Oct 16, 2005
  9. Oct 16, 2005 #8

    Doc Al

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    You are assuming that the tension equals the weight of the hanging mass. Nope! (If that were true, the net force on the hanging mass would be zero: no acceleration.) You have to solve for the tension and the acceleration. Write Newton's 2nd law for each mass separately.

    Another way to solve this is to use energy methods.
     
  10. Oct 16, 2005 #9
    Ok so both masses DO have acceleration? I guess I'll try writing newton's law for each mass again; but in that case, I would HAVE to use the value of T that I got for the second mass into the first mass's equation wudnt I? or else I would have two unknowns: tension and acceleration ... Is the energy method of doing this problem easier; if so, please show me how, I would like to learn it ... Seems newton's law for doing this problem takes a lot more work :confused:
     
  11. Oct 16, 2005 #10

    Doc Al

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    You'll have two equations and two unknowns.
    The initial mechanical energy (PE + KE) = final mechanical energy + energy lost to friction.
     
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