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Percent change in KE

  1. Oct 17, 2005 #1
    % change in KE

    A sled is being pulled across a horizontal patch of snow. Friction is negligible. The pulling force points in the same direction as the sled's displacement, which is along the +x axis. As a result, the kinetic energy of the sled increases by 36.7%. By what percentage would the sled's kinetic energy have increased if this force had pointed 62.2° above the +x axis?

    i'm not sure how to even begin this problem... i'm guessing i should use the change in KE = Work but i'm not sure how to go about it. help? thanks!
     
  2. jcsd
  3. Oct 17, 2005 #2

    Päällikkö

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    You are correct: W = Fx.
    We can assume that only the horizontal component of force increases the kinetic energy.

    What is the horizontal component of F in case one (where F is horizontal)? How about in case two (where F is tilted)?

    Can you figure out how to get an equation describing the situation (increase in kinetic energy) in case one? How about in case two?
     
  4. Oct 17, 2005 #3
    if X is the horizontal component then x*cos(62.2) is the hypotenuse for the second case, right? i'm not sure how to make an equation for that. i know that PE at the bottom of the hill is 0 and therefore it must have a higher KE at the bottom and has a higher PE at the top. i'm really not sure how to make an equation out of that though
     
  5. Oct 17, 2005 #4

    Päällikkö

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    Hill? What hill? The hypotenuse is F. What is F's horizontal component?
     
  6. Oct 17, 2005 #5
    F*cos(62.2) is F's horizontal component... how can i use that in an equation?
     
  7. Oct 17, 2005 #6

    Päällikkö

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    Only the horizontal F increases kinetic energy, unless the sled takes off (as in: airplanes do), but we assumed it doesn't.

    Anyways, this is the situation:
    K1 + W = K2
    In the beginning, the sled has some kinetic energy (K1). Some (positive) work is done to the sled (W) and its kinetic energy increases (K2).

    Can you use the information given in the problem to get two equations (case one and case two)?
     
  8. Oct 17, 2005 #7
    does this make sense... for situation one --> (1/2)*m*(v^2) + (f*d) = .367 * (1/2)*m*(v^2) and for situation two --> (1/2)*m*(v^2) + (f*d*cos(62.2)) = (1/2)*m*(v^2) ??? how can i use those
     
  9. Oct 17, 2005 #8

    Päällikkö

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    Close, but needs some adjusting:
    (1/2)*m*(v^2) + (f*d) = .367 * (1/2)*m*(v^2)
    Multiplying by 0,367 would mean the work was negative (as the kinetic energy decreases). What should you multiply the final kinetic energy with?

    (1/2)*m*(v^2) + (f*d*cos(62.2)) = (1/2)*m*(v^2)
    You must distinguish the energy in the beginning and in the end. Your equation gives fdcos(62,2) = 0, which is not the case. As you will be wanting the ratio between energy in the beginning and the end, I suggest multiplying the final energy by some A (which would be related to the asked percentage).

    As you are not asked for velocity, using (1/2)mv^2 is unnecessary. Use K.
     
  10. Oct 17, 2005 #9
    instead of multplying by .367 should i divide by .367?
     
  11. Oct 17, 2005 #10

    Päällikkö

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    As there's an 36,7 % increase: Einitial + 0,367Einitial = Efinal

    So, you should multiply it by 1,367.
     
    Last edited: Oct 17, 2005
  12. Oct 17, 2005 #11
    so now i have the two equations...
    K1 + f*d = 1.367*K2 and
    K1 + f*d*(cos(62.2)) = A*K2
    but i have a lot of variables so what do i solve for?
     
  13. Oct 17, 2005 #12

    Päällikkö

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    K2 is actually K1.
    A is the asked quantity. Solve for it.

    Well, now it's just mathematics. Give it a shot, ask for more help if you can't get it solved :smile:.
     
  14. Oct 17, 2005 #13
    what about fd though

    aren't there still 3 variables... fd, A, and K1
     
  15. Oct 17, 2005 #14

    Päällikkö

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    fd will cancel out.
     
  16. Oct 17, 2005 #15
    from equation one i got K1 = fd/.367 and i plugged that into equation two and eventually got A=1.46... I then multiplied that by 100 to get 146%, but thats wrong... i'm not sure what i did wrong
     
  17. Oct 17, 2005 #16

    Päällikkö

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    Ok, K = fd / .367.
    K + fdcos(62.2) = AK
    fd / .367 + fdcos(62.2) = A(fd / .367)
    fd(1/.367 + cos(62.2)) = fdA / .367 _____ | fd cancels out
    A = 1 + .367cos(62.2) = 1.1712

    => 17.1 %
    I hope that's correct :smile:.
     
  18. Oct 17, 2005 #17
    oh so you have to take out that one before putting into a percentage... thanks!
     
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