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Homework Help: Help Please! Energy Conservation!

  1. Oct 19, 2005 #1
    Please help me, this is due tonight at 11 and I can't figure out what to do!

    " A 17 kg child descends a slide 3.5 m high and reaches the bottom with a speed of 2.5 m/s. How much thermal energy due to friction was generated in this process?"

    The equation should be:

    kinetic energy intiial + potential energy initial = ke (final) + pe (final) + friction X distance travelled


    1/2mv^2 (initial) + mgy (initial) = 1/2mv^2 (final) + mgy(final) + Friction X d

    The problem is I don't know d!

    0 (no initial KE) + (9.8)(17)(3.5) = 1/2(17)(2.5)^2 + 0 (no final potential energy) + friction X d!

    What is D!

    I tried making the slide a right triangle with 45 degree angles for the other two and using the pythagorean theorem...and got 5 for the distance..but I got a wrong answer..

    huh, could someone PLEASE help me!
  2. jcsd
  3. Oct 19, 2005 #2
    You dont need D to find out the answer, one second ill post my toughts.

    ok lets see... forgive me if im wrong...

    Change E system = Change E mechanical + Change E thermal

    E thermal is the friction and its equal to friction force times delta S

    Change in mechanical energy is Eo-Ef... so 0 = Eo-Ef + E thermal


    -E thermal = Eo - Ef
    - E thermal = PEo - KEf
    E thermal = Kef-PEo
    E thermal = -530.025 J

    Alright i hope this helps and anyone correct me if im wrong... thanks
    Last edited: Oct 19, 2005
  4. Oct 19, 2005 #3
    Thanks. Hmm , my class hasn't learned how to do that yet. Is there any way to do it with the PE + KE = KE + PE + Fr (d)???
  5. Oct 19, 2005 #4
    thats exactly what im doing except im not using the distance...

    (PEo + KEo) - (PEf+KEf) = - Work Friction (which is equal to Fr (d) ) Oo

    if you need to do further work with the friction you can use the result i got and do

    Work friction = Force friction (d)

  6. Oct 19, 2005 #5
    Doh! You are right, seiya, the answer is 530. The friction work is Friction X d. For some reason I thought I was just looking for the force of friction. Stupid me.

    Thank you for the help! :)
  7. Oct 19, 2005 #6
    you are welcome , just helping while thinking about my problem and hoping someone will give me a hint :p
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