Results 1 to 6 of 6

Math Help - kinetic energy

  1. #1
    Junior Member
    Joined
    Feb 2008
    Posts
    71

    kinetic energy

    Okay so i'll go right to it (been sitting about 2 hours with this question);

    An icecube is sliding 3 meters friction free off the ledge of a house roof which has the angle of 27 degrees. What speed will it attain at the edge of the roof?

    Just imagine a triangle with the hyp of 3 meters and the angle at the edge is 27degrees. The one and only force that can meddle with the ice cube must be gravity (9.82m/s(square))

    So please solve this for me so I can understand what im doing wrong.


    The answer is v=square(2*9.82+3*sin(27))

    I've got no clue where the 2 comes from...
    Last edited by mr fantastic; January 6th 2009 at 01:10 PM. Reason: Removed potentially offensive language
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Last_Singularity's Avatar
    Joined
    Dec 2008
    Posts
    157
    Quote Originally Posted by Hanga View Post
    Okay so i'll go right to it (been sitting about 2 hours with this question);

    An icecube is sliding 3 meters friction free off the ledge of a house roof which has the angle of 27 degrees. What speed will it attain at the edge of the roof?

    Just imagine a triangle with the hyp of 3 meters and the angle at the edge is 27degrees. The one and only force that can meddle with the ice cube must be gravity (9.82m/s(square))

    So PLEASE for the LOVE OF GOD solve this for me so I can understand wtf im doing wrong.


    The answer is v=square(2*9.82+3*sin(27))

    I've got no clue where the 2 comes from...
    Are you sure that v=square(2*9.82+3*sin(27))?

    Using conservation of energy, the result seems to be:
    KE = PE \rightarrow \frac{1}{2}mv^2 = mgh

    \frac{1}{2}v^2=gh

    v = \sqrt{2gh} = \sqrt{2(9.81)(3)(sin27)}

    Alternatively, you can just use kinematics and the answer appears as clear as day:
    v_f^2 = v_0^2 + 2ax where v_f, v_0, a, x are the final velocity, initial velocity, acceleration, and distance traveled, respectively.
    v_f^2 = 0 + 2(9.81sin(27))(30) \rightarrow v_f = \sqrt{2(9.81)(3)(sin27)}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    It looks like you are using the formula v^2 = u^2 + 2as, although I am not sure where you numbers have come from?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Last_Singularity's Avatar
    Joined
    Dec 2008
    Posts
    157
    Quote Originally Posted by craig View Post
    It looks like you are using the formula v^2 = u^2 + 2as, although I am not sure where you numbers have come from?
    Oops, I meant 3 instead of 30 in
    <br />
v_f^2 = 0 + 2(9.81sin(27))(30) \rightarrow v_f = \sqrt{2(9.81)(3)(sin27)}<br />

    Sorry about that
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Last_Singularity View Post
    Oops, I meant 3 instead of 30 in
    <br />
v_f^2 = 0 + 2(9.81sin(27))(30) \rightarrow v_f = \sqrt{2(9.81)(3)(sin27)}<br />

    Sorry about that
    Sorry that was aimed at the OP :S

    But I'll accept thanks for spotting the error lol
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2008
    Posts
    71
    wow dude! I really understand exactly what I was doing wrong here.

    This is a very hard question from my book and I acually thought that it would involve simple trigonometry, silly me.

    Thanks alot mate! You just made my day!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. kinetic energy during energy levels
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: December 27th 2010, 01:12 AM
  2. kinetic energy PDE
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: January 28th 2010, 07:55 AM
  3. Baseball - kinetic energy
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: December 9th 2009, 05:55 PM
  4. kinetic energy
    Posted in the Math Topics Forum
    Replies: 7
    Last Post: January 23rd 2008, 10:29 AM
  5. kinetic energy
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: August 16th 2006, 11:29 AM

Search Tags


/mathhelpforum @mathhelpforum