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Energy questions?

  1. Jan 14, 2006 #1
    Energy questions??

    ok i have two questions which i am completely stumped on and i need some help...

    At riverside park in Agawam, Massachusetts, a ride called the cyclone is a giant roller coaster that ascends a 34.1 m hill and then drops 21.9 m before ascending the next hill. THe train of cars has a mass of 4727 kg. a)how much work is required to get an empty train of cars from the ground to the top of the hill. b)what power must be generated to bring the train to the top of the hill in 30 s. c)how much potential energy is converted into kinetic energy from the top of the hill to the bottom of the 21.9 m drop?

    a) (4727 kg)(10m/s^2)(34.1m)=1611907 N.m (ithink thats right)
    b) 1611907N.m/30s=53730.23 watts (i also think this is right)
    c) Help please??



    other question

    Using her snowmobile, Midge pulls a 60.0 kh skier up a ski slope inclined at an angle of 12.0 degrees to the horizontal. The snowmobile exerts a force of 200.0 N parallel to the hill. If the coeficient of friction between the skis and hte snow is 0.120, how fast is the skier moving afster he has been pulled for 100.0 m starting from rest? (ignore the effects of the static friction that must be overcome to initially start the skier in motion.) Use the law of conservation of energy.

    i know the formula for conservation of energy is KEo+PEo=KEf+PEf and i think for this problem i would have to find the net work by using the equation of KEf-KEo= net work... i also know the KE=1/2mv^2 and PE=mgh my problem is how do i use this information to solve this problem????
     
  2. jcsd
  3. Jan 15, 2006 #2
    Hi yosup
    could u tell me how do i post my query?
     
  4. Jan 15, 2006 #3

    Pengwuino

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    Well you basically know the potential energy at the top and the potential energy at the bottom. This change in potential energy has been converted into kinetic energy.
     
  5. Jan 15, 2006 #4
    okay so i guess i would do this:

    (4727 kg)(10m/s^2)(21.9 m)=1035213J
    1611907J-1035213J= 576694J (is that right)

    can anyone help with the second question
     
  6. Jan 15, 2006 #5
    im kind of new at this... why does it say my post has been moved??

    CAN SOMEONE PLEASE HELP ME OUT WITH THIS QUESTION?!


    EDIT: n/m i see i should have posted this in the homework question section
     
    Last edited: Jan 15, 2006
  7. Jan 15, 2006 #6
    come on please im sure someone here knows how to do these problems :confused:
     
  8. Jan 16, 2006 #7
    C'mon its been two days; isnt there someone who could help me with these questions? Your help would be greatly appreciated.
     
  9. Jan 16, 2006 #8

    Hootenanny

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    Question b is correct, assuming question a is correct. Question c is also appears correct. It would be a simple question if done using newtonian laws of motion; however, I'mnot sure how to proceed using conservation of energy.
     
  10. Jan 16, 2006 #9
    ok... but how about the other question thats the one that im having a lot of trouble with
     
  11. Jan 17, 2006 #10

    Hootenanny

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    I'm talking about the second question. Perhaps someone else could give a hint?
     
  12. Jan 17, 2006 #11

    HallsofIvy

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    penguino explained how to do c) six minutes after you posted the question!

    What second question? You posted three questions. You gave answers to the first two and were told that they were correct. (You are taking g= 10 m/s2? 9.8 or even 9.81 is more accurate.)
     
  13. Jan 17, 2006 #12
    other question

    Using her snowmobile, Midge pulls a 60.0 kh skier up a ski slope inclined at an angle of 12.0 degrees to the horizontal. The snowmobile exerts a force of 200.0 N parallel to the hill. If the coeficient of friction between the skis and hte snow is 0.120, how fast is the skier moving afster he has been pulled for 100.0 m starting from rest? (ignore the effects of the static friction that must be overcome to initially start the skier in motion.) Use the law of conservation of energy.

    i know the formula for conservation of energy is KEo+PEo=KEf+PEf and i think for this problem i would have to find the net work by using the equation of KEf-KEo= net work... i also know the KE=1/2mv^2 and PE=mgh my problem is how do i use this information to solve this problem????


    this question is kind of urgent... could someone please help!!!!!!?
     
  14. Jan 17, 2006 #13

    Hootenanny

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    using newtonian motion equations; its a simple matter of resolving forces and using F=ma, v=u+at, v^2 = u^2 + 2as, s = ut + 1/2 at^2 etc.
     
  15. Jan 17, 2006 #14
    well i did:

    60kg x sin 12= 12.47kg
    200N x .12 x cos 12=23.47N
    200N-23.47N= 176.53N
    176.53N/12.47kg= 14.52 m/s^2
    Vf^2=0+2(14.2m/s^2)(100m)
    Vf=53.3 m/s

    Now I know this is wrong could you please explain to me where i went wrong
     
  16. Jan 17, 2006 #15
    Please this is urgent someone answer
     
  17. Jan 17, 2006 #16
    AFAIK, what you are doing with mass in the first operation is meaningless in this context. Also, you aren't finding [what I'm guessing to be] the force of friction properly in the second line. The question asked you to use the conservation of energy in the first place!

    The snowmobile exerts 200N of force parallel to the slope for 100m, thus [itex]W_s = F_{||}d = (200)(100) = 200 000 J[/itex]

    They give you the coefficient of friction as 0.120 and the mass (60 kg) so you can find the force of friction parallel to the displacement with some trig, and therefore [itex]W_f[/tex] should be straightforward. Friction does negative work ([itex]\cos 180 = -1[/itex]) so subtract it from [itex]W_s[/itex] to get net work.

    Remember [itex]\Sigma W = \Delta KE[/itex]
     
    Last edited: Jan 17, 2006
  18. Jan 17, 2006 #17
    c'mon i have 60 more minutes to get this done i think i figured out some stuff but im not sure WILL SOMEONE PLEASE HELP

    i did

    60kg x 10m/s^2 x .12 x cos 12= 70.4
    200N-70.4N=129.6 N
    129.6N x 100m=12960 J

    then im not sure what else to do PLEASE PLEASE HELP ME
     
  19. Jan 17, 2006 #18

    G01

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    (i am somewhat rushing and assuming your answer above is correct....)

    Here you found the total work done.
    Now someone told you that the total work is equal to the change in kinetic energy. So if:

    [tex] 12960J = \Delta K =.5mv_f^2 - 0 [/tex]

    (0 is the initial kinetic energy.)You can solve for vf can't you?
     
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