Help with my coursework?

Discussion in 'General Off-Topic Chat' started by kokiri_link, Feb 27, 2008.

  1. kokiri_link

    kokiri_link GBAtemp Regular

    Nov 16, 2006
    So i need some help with some coursework....

    A lift cage of mass 550kg accelerates upwards from rest to a velocity of 6m/s whilst travelling a distance of 14m. The frictional resistance to motion is 225N. Making use of the the principle of conservation of energy, determine :

    i) the work done
    ii)the tension in the lifting cable
    iii)the maximum power developed

    any help?
  2. martin88

    martin88 GBAtemp Advanced Fan

    Dec 18, 2005
    I can't calculate the acceleration if don't know how long it took to go from 0m/s to 6m/s.
  3. eldaddio

    eldaddio Advanced Member

    Jan 29, 2008
    You'll need something like F= (mu)R combined with F=ma and Work Done = Force x Distance.
  4. Masta_mind257

    Masta_mind257 GBAtemp Regular

    Jul 3, 2006
    In ur mother
    off topic but there are easy ways 2 remember formulas, my teach goes
    f= ma - fat mans arse
    work = fXd - work is for donkeys i cant rembr the others lol
  5. FAST6191

    FAST6191 Techromancer

    pip Reporter
    Nov 21, 2005
    My usual policy of not giving you exact answers but reasoning applies here.
    The questions also seem to be about as well written as a usual question of this type and requires several rather large assumptions (which you should state and maybe even explain: if nothing else it goes to method marks).
    I have assumed this question is of a level that would normally be taught to 16 year olds through to first year of an engineering degree but tell me if it is any different here.

    I am assuming it accelerates to 6 m/s over the course of the 14m. Acceleration must be linear as no function is given for anything meaning basic equations of linear motion. My symbols might be different so
    s= displacement
    v=final velocity
    u=initial velocity
    a= acceleration.

    v2 =u2 +2as here as you have no time value
    s= ut +1/2at^2 or its cousin s= vt - 1/2at^2, it does not really matter as you have v and u. You will end up with a quadratic to solve here.

    Here I would read on and determine what I needed for the later questions and write it down but do it however you like. You might like to draw a diagram too as some of this has the potential to cause issues because you are going to be rearranging equations where sign counts.

    1) I assume this means calculate the energy expended during the course of movement.
    break it down and add it up (1st law of thermodynamics referenced there tells you this). Work done = energy. The components here being gravity and friction.

    work against gravity
    Yawn, gravitational potential energy that I assume you have done several times before.
    weight.height = m.g.h = work against gravity

    work against friction
    I assume friction is constant, what sort of resistance it is at this point is entirely inconsequential.
    Extra force would have to be provided to overcome friction.
    force.distance moved= work done. Force given friction does not seem to be causing acceleration = friction (albeit in the opposite direction so I should really say equal in magnitude).

    2) an odd one. Normally such a question is posed with multiple angled threads to determine weight and it is assumed the acceleration is near 0 (not to mention the maths gets fairly complex). The cable I assume is light and inextensible (again it does not make the maths nice).
    For this reason I assume it is simply the mass having been suspended and the tension now measured.
    Resistance to motion is mentioned but as I have assumed this is a stationary object I am willing to neglect it (it is a simple extra step when resolving forces anyhow), if it says it is friction (from a shaft of something) rather than air then this will have an effect.

    Tension is simply a force so you add it like you did the first one.
    weight=mass.gravitational field strength.

    I have a feeling my assumptions could be called into question so again reference the first question and find the components and add them together if necessary.

    3) Maximum power. Energy used per time unit, when it is at the highest?
    Almost a trick question. Constant force (constant acceleration+constant friction) = constant power.

    To get time I say use the equation
    get a using
    v^2= u^2 +2as

    use the newly found a to get t via the quadratic formula (remember to make it equal 0) (u aka b in the standard quadratic = 0 which is nice). If you are getting odd results remember s is negative.

    power = energy expended/time taken
  6. kokiri_link

    kokiri_link GBAtemp Regular

    Nov 16, 2006
    genius buddy. that will help alot cheers [​IMG]