Results 1 to 4 of 4

Math Help - [SOLVED] Work, Power and Energy - Mechanics

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    148

    Cool [SOLVED] Work, Power and Energy - Mechanics

    Q) A lorry of mass 15 000 kg moves with constant speed 14ms−1 from the top to the bottom of a straight hill of length 900 m. The top of the hill is 18m above the level of the bottom of the hill. The total work done by the resistive forces acting on the lorry, including the braking force, is 4.8 106 J.

    Find

    (i) the loss in gravitational potential energy of the lorry
    (ii) the work done by the driving force.


    My Attempt

    The answer to part (i) is mgh = 2700000J

    But I am stuck at (ii)

    I used the formula

    WDF - Work Done By Resistance = 2700000

    WDF = 2700000+4.8x10^6
    WDF = 7500000J

    But the correct answer is 2.1 x 10^6J

    How? Am I using a wrong method?
    Last edited by unstopabl3; May 17th 2010 at 12:02 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    question is kind badly put, i think

    they mean driving force besides the weight

    total work is zero, because velocity is constant

    so WDF+Wweight=Wresistent

    WDF+2700000=4800000

    if i answered this i would consider the weight also a driving force
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    148
    Huh? This is a past paper question so I doubt the examiner would make a mistake.

    I am still not sure what you did or what to do with this sort of question!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jul 2009
    Posts
    68
    W_{total}=\Delta E_c
    as the velocity is constant the \Delta E_c=0

    so the total work is zero

    there are 3 forces
    one is the weight which contributes for the truck to move forward and its work its equal to -\Delta E_p (as it contributes for the truck moving forward its work is positive) (Wweight= -\Delta E_p=2700000J)
    the second is the 'driving' force, it also contributes for the truck moving forward so its work is also positive (this is waht you want to know)
    the third is the resistive forces which contribute to diminish the truck speed. so its work is negative (Wresistent=-4800000J)

    as the total work is zero
    W_{total}=WDF+Wweight+Wresistent=0
    Last edited by Haytham; May 18th 2010 at 03:34 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mechanics - Energy, work and power
    Posted in the Math Topics Forum
    Replies: 10
    Last Post: February 23rd 2013, 07:13 AM
  2. [SOLVED] Generalised energy and energy (lagrangian mechanics)?
    Posted in the Advanced Applied Math Forum
    Replies: 2
    Last Post: January 16th 2012, 03:03 AM
  3. Work, energy, power?
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: January 27th 2010, 11:35 AM
  4. Mechanics: Potential energy
    Posted in the Advanced Applied Math Forum
    Replies: 3
    Last Post: March 7th 2009, 11:37 PM
  5. work , energy and power
    Posted in the Math Topics Forum
    Replies: 5
    Last Post: October 2nd 2007, 12:10 PM

Search Tags


/mathhelpforum @mathhelpforum