Results 1 to 5 of 5

Math Help - Energy

  1. #1
    Super Member
    Joined
    Sep 2007
    Posts
    528
    Awards
    1

    Energy

    A box of mass 10kg is resting on a sloping platform which is inclined at 20^\circ to the horizontal. The coefficient of friction between the box and the platform is 0.2. The box is released from rest and slides down the platform. Calculate:

    a) the speed of the box after it has been moving for 3 seconds.
    b) the potential energy lost by the box during this time.
    __________________

    I've done part (a) but I cannot get my head around energy. Can I have help on part (b) please. Thanks.
    Last edited by Simplicity; April 30th 2008 at 09:38 AM. Reason: Added [MATH] Codes
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Kai
    Kai is offline
    Junior Member
    Joined
    Apr 2008
    Posts
    59
    Hmm, lost in PE = gain in KE + work done against force of friction ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Danshader's Avatar
    Joined
    Mar 2008
    From
    http://en.wikipedia.org/wiki/Malaysia now stop asking me where is malaysia...
    Posts
    158
    hi,

    Part 1)
    Weight of the mass, W = mg = 10\times 9.81 N = 98.1 N
    normal force acting on the mass by the platform, N = m\times g\times cos(20^o)=98.1\times cos(20^o) = 92.1838 N
    force from the weight that is pulling the mass parallel down the slope F = m\times g\times sin(20^o) = 98.1\times sin(20^o) = 33.55217N
    frictional force, F_{r} = \mu \times N = 0.2\times 92.1838 = 18.43676 N
    resultant force pulling the mass down parallel to the slope,  R =F - R_r = 15.11541 N

    from Newton's second law, F = ma
    R = ma
     F - R_r = 10\times a
     15.11541 = 10\times a
    acceleration, a = \frac{15.11541}{10} = 1.511541 ms^{-2}

    Part 2)
    Using principle of conservation of energy,
    Gain in kinetic energy + other energy lost(due to friction) = Lost in gravitational potential energy
    \frac{1}{2} mv^2 + \mu N S= mgh

    where,
    v = final velocity of mass after 3 sec under acceleration
    S = distance moved by the mass during the 3 second acceleration
    h = vertical distance moved by the mass

    Finding the final velocity of the mass:
    using v = u + at:
    u = 0
    a = 1.511451
    t = 3

    v = 0\times 3 + 1.511451\times 3
    v =4.534353 ms^{-1}

    Kinetic energy obtained:
     \frac {1}{2}m\times v^2 = \frac{1}{2} \times 10 \times 4.534353^2
     \frac {1}{2}m\times v^2 =102.8017856 Joule

    Finding the distance moved parallel to the plane:
    using S =ut+\frac{1}{2}a\times t^2
    u = 0
    t = 3
    a = 1.511451

    S =ut+\frac{1}{2}a\times t^2
    S =0\times 3+\frac{1}{2}\times 1.511451\times 3^2
    S =0\times 3+\frac{1}{2}\times 1.511451\times 3^2
    S = 6.8015295 m

    Other energy lost(due to friction):
    LostInFriction = \mu \times N \times S
    LostInFriction = 0.2\times 92.1838 \times 6.8015295
    LostInFriction = 125.398167Joule


    Substituting into the principle of conservation of energy:
    Gain in kinetic energy + other energy lost(due to friction) = Lost in gravitational potential energy
    102.8017856 + 125.398167 = Lost in potential energy
    Lost in potential energy = 228.2121958 Joule

    There is another method to obtain the lost in gravitational potential energy. Much faster but if you are to answer in an exam this method of using the principle of conservation of energy is the best method to use. Good luck
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,856
    Thanks
    321
    Awards
    1
    Quote Originally Posted by Danshader View Post
    There is another method to obtain the lost in gravitational potential energy.
    I must admit that I'm at a loss. What method are you thinking of?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Danshader's Avatar
    Joined
    Mar 2008
    From
    http://en.wikipedia.org/wiki/Malaysia now stop asking me where is malaysia...
    Posts
    158
    directly finding the gravitational potential energy:

    Lost in gravitational potential energy = mgh

    we already found that the mass will travel 6.801529 m down the plane. this corresponds to a drop in 6.801529\times sin(20^o) = 2.362599m vertically


    hence from:
    Lost in gravitational potential energy = mgh
    = 10 \times 9.81 \times 2.362599
    =228.2060985 Joules
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Generalised energy and energy (lagrangian mechanics)?
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: January 16th 2012, 03:03 AM
  2. kinetic energy during energy levels
    Posted in the Math Topics Forum
    Replies: 0
    Last Post: December 27th 2010, 01:12 AM
  3. Energy
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 19th 2009, 11:13 AM
  4. Energy
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 15th 2009, 03:30 PM
  5. Replies: 7
    Last Post: April 17th 2006, 01:04 PM

Search Tags


/mathhelpforum @mathhelpforum