Results 1 to 6 of 6

Math Help - Gravity-Work Question

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    6

    Gravity-Work Question

    How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by peon123 View Post
    How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
    I believe it is W = Fscos\theta

    F = mg
    s = 50
    cos\theta = cos(42) (this works for the horizontal angle)

    W = 24 \times 9.81 \times cos(42) \times 50 = 8748,3J  = 8.7kJ

    I think that's right, if you have an answer please post it because there could be another way to do it
    Last edited by e^(i*pi); May 29th 2009 at 03:22 PM. Reason: see post 5
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by peon123 View Post
    How much work is done by gravity in causing a 24 kg rock to tumble 50m down a slope at an angle 48 degrees to the verticle, to one decimal place?
    Work done is equal to Fs, where F is the force acting on the particle and s is the displacement.

    If I remember my mechanics correctly, then as the question is asking you the work done by gravity, it is the horizontal displacement that you are after, which is equal to 50 \sin{48}.

    Then as we are interested in the direction of gravity, would it be 24g.50 \sin{48}?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    This works out to be 8739.4J or 8.7kJ.

    Just to note, I have taken g as 9.8, as opposed to 9.81.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Just realised that we used the same equations

    Quote Originally Posted by e^(i*pi) View Post
    W = 24 \times 9.81 \times cos(42) \times 50 = 87483J ~ 8.7kJ
    There's a typo in your answer, it should be  8748.3J, not 87483J
    Follow Math Help Forum on Facebook and Google+

  6. #6
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by craig View Post
    Just realised that we used the same equations



    There's a typo in your answer, it should be  8748.3J, not 87483J
    Yeah, I checked that on my calculator, makes sense since sin(90-x) = cos(x)

    Thanks for that, I worked it out in kJ first but had to expand for the 1dp
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: May 3rd 2011, 06:42 AM
  2. gravity question
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: March 13th 2011, 12:34 PM
  3. Calculate the work against gravity..?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 30th 2010, 07:38 AM
  4. Replies: 0
    Last Post: November 27th 2009, 04:40 PM
  5. Quick question about position due to gravity.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 1st 2009, 09:17 PM

Search Tags


/mathhelpforum @mathhelpforum