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Generated Heat Quantity

  1. Apr 17, 2006 #1
    Let the mechanical equivalent of heat be 4.2[J/cal]. 70 percent of lost mechanical energy was converted into heat. Find the generated heat quantity given an object with mass 10 kg moving at speed 49 m/s with coefficient of kinematic friction 0.5 and gravitational acceleration of 9.8 [tex]\frac{m}{s^2}[/tex].
     
  2. jcsd
  3. Apr 17, 2006 #2

    Hootenanny

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    Joules per calorie?
     
  4. Apr 17, 2006 #3
    Yes Hootenanny! I really cannot get it because the givens are quite unique and confusing.
     
  5. Apr 18, 2006 #4

    Hootenanny

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    I think theres a typo in the question because the units of J/cal are nonsensical. Heat is 'energy in transit' therefore the SI units are Joules.
    Regards,
    ~Hoot
     
    Last edited: Apr 18, 2006
  6. Apr 18, 2006 #5
    I am trying to solve the problem but all I can get is the stopping distance. I am still researching for the formula for the time respected to the problem.
     
  7. Apr 18, 2006 #6

    Hootenanny

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    Nevermind, I have seen my mistake. If you have calculated the stopping distance you have all the information you need.

    HINT: What is work done?

    Regards,
    ~Hoot
     
  8. Apr 18, 2006 #7
    Can I use the formula [tex]x-x_o=v_ot + \frac{1}{2}at^2[/tex] to get the stopping time?Or is there any other formula?
     
  9. Apr 18, 2006 #8

    Hootenanny

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    If you have calculated the stopping distance, you have all the information you need. Work done = force times distance. In this case work done = energy 'lost'.

    Regards,
    ~Hoot
     
  10. Apr 18, 2006 #9
    If I get the work with this= 0.5 * 10 kg * 9.8 [tex]\frac{m}{s^2}[/tex] * 255 m, then what is the purpose of the given "mechanical equivalent of heat be 4.2[J/cal]. 70 percent of lost mechanical energy was converted into heat"?
     
  11. Apr 18, 2006 #10

    Hootenanny

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    You should get 245m, but 255 is close enough. You have calculated the energy lost. However, only 70% of this 'lost' energy is converted into heat...

    Regards,
    ~Hoot
     
  12. Apr 18, 2006 #11
    Yeah you're really right. That was really close enough for the stopping distance. But how can I compute for the generated heat quantity given the mechanical equivalent of heat be 4.2[J/cal]. The final answer is suppose to be in calorie unit.
     
  13. Apr 18, 2006 #12

    Hootenanny

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    Ahh, I think what they question meant was 1 calorie = 4.2 Joules.

    Regards,
    ~Hoot
     
  14. Apr 18, 2006 #13
    Now I get it buddy. 12495 Joules is the work done or the energy lost and I will multiply it by 0.70 to get the 70 percent lost. I get 8746.5 Joules. Dividing it by 4.2 j/cal gives me 2082.5 cal. Is 2082.5 right buddy?
     
  15. Apr 18, 2006 #14

    Hootenanny

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    Yep, that looks good to me.

    Regards,
    ~Hoot
     
  16. Apr 18, 2006 #15
    Thanks Hoot. You were a great help to me. Till next time.
     
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