Get better study experience with the jambite for andriod app! jambite app download
  • Physics Tutoring Session #12 Work, Energy and Power on Mon. 7th March, 2016 @ 5:00pm - 6:00pm,

    Physics

    Chima Sunday (Tutor)
    07-03-2016 15:34:00 +0000

    Work W is said to be done, when a force F move a body to a distance s, in the direction of the applied force F. (When F and s are parallel we have a maximum workdone)
    Therefor W = F × s.
    S.I unit is in joule.
    Definition of a joule: a joule is define as the workdone when a force of 1Newton moves a body through a distance of 1metre in the direction of the applied force.
    But when the applied force F is at an angle say θ We resolve the force to its horizontal component
    Therefor W = Fscosθ
    Example(1)

    A force of 150N is attached to a mass of 200kg at an angle of 30 to drag it though a horizontal distance of 10m. How much work is done by the force.
    A. 1200J
    B. 1300J
    C. 2000J
    D. 1299J
    E. 1290J
    (Post utme, futo 2013/2014)
    solution
    From the formular,
    W = Fscosθ
    F = 150N
    S = 10m
    θ = 30o
    W = 150 × 10 × cos30 = 1299J.
    Option D.

    If we pull a body of mass m up in an incline plane of angle θ and of height h metre, Then the work done would be that of against the weight plus friction. i.e work done is that against friction plus the one against gravity. But we need to resolve the weight to its horizontal component to obtain Wsinθ. recall that frictional for F = μR.
    where coefficient of static friction, and R the normal Recation.
    there total workdone W in moving the body a distance of s is given as
    W = (Wsinθ + μR)s.

    Energy
    this is define as the ability to do work. Its a scalar quantity and has same unit with work. We group Energy into Renewable(inexhaustible) and Non-Renewable(exaustible)
    Energy appears in may forms e.g are sound energy, wind energy, electrical energy, mechanical energy, nuclear enegy, heat energy,solar energy, chemical energy etc.
    The principle of conservation of energy state thus that energy can neither be created nor destroyed but can be transformed from one form to another.

    Mechanical Energy is made up of
    i)Kinetic Energy: which is the energy possesed by a body due to its motion. It si given by
    K.E = (1/2)mv2

    ii)Potential Energy
    This is the energy posses by a body by the virtue of its position h. It is given as
    P.E = mg × h

    Conservation law on Mechanical Energy
    This is illustrated when a body is dropped from a height say h, its P.E is maximum at height h and K.E = 0, when it falls at a distance say x metres, it losses P.E, But the loss in P.E is converted to K.E. given the total energy T = P.E + K.E
    i.e T = mgh1 + (1/2)mv2
    T = mgx + (1/2)mv2

    Example(2)
    A 5kg mass is dropped from a height of 30m above the ground. Determine the velocity of the mass when it is 18m above the ground. (g = 10ms2)
    Solution
    total energy T = F(mg) × dist = 5 *10 * 30 = 1500J
    recall that T = mgh1 + (1/2)mv2
    m = 5kg, g = 10.
    therefor T = 5*10*18 + 5*v2
    by subject formular v2 = (T - mgh1 )1/m = 2(1500 - 5*10*18)1/5 = 240
    v = 15.5m/s.

    Alternatively we can still get the same resultknowing that the lost in P.E = gain in K.E
    So loss in P.E = mg(h - h1), h = 30m and h1 = 18
    So loss in P.E = gain in K.E
    mg(h - h1) = (1/2)mv2
    by subject formular
    v2 = 2g(h - h1) = 2 * 10 * (30 - 18) = 240
    therefor v = 15.5m/s

    POWER
    This is define as the rate of doing work
    Power p = work/time
    it is measured in watts (W)
    It can also be express as P = (F * s)/t = F * s/t = F*v, were v is velocity.
    Example(3)
    If a pump ejects 30kg of water in 60s through a vertical height of 3m. Calculate the minimum power of the pump. (g = 10m/s2)
    Solution
    Work done = P.E = mgh = 30*10*2 = 600J
    Power P = WORKDONE/TIME = 600/60 = 10W.

    Please do go through the script above and document your questions accordingly, and also prepare for a similar questions culled from JAMB and from other exams. See You in Class at 5:00PM

    Don't forget to fill our the survey after the session:
    Survey Feedback Link - http://goo.gl/forms/Tasr40Ck1Z



    1 40 0

  • Stephanie Samuel

    ok

    1 07-03-2016 15:41:00 +0000

  • Chima Sunday (Tutor)

    Hello @all welcome to today's tutorial. @Stephanie Samuel thanks for being on time in today's class. Please do feel free to ask any question you may have and also go through the script above. then after that that do try the questions i will send subsequently. once again welcome @ all!

    1 07-03-2016 16:06:00 +0000

  • Chima Sunday (Tutor)

    @all. please let see if you can answer this question.
    A 70kg man ascends a flight of stairs of height 4m in 7s. The power expended by the man is?
    A. 40W
    B. 100W
    C. 280W
    D. 400W
    (OAU post utme).

    1 07-03-2016 16:07:00 +0000

  • Favour Francis

    D

    1 07-03-2016 16:14:00 +0000

  • Chima Sunday (Tutor)

    Hope i gat some student in here. @Stephanie Samuel hope your still there. are?

    1 07-03-2016 16:15:00 +0000

  • Chima Sunday (Tutor)

    Good work @Favour Francis. thats a correct answer keep it up okay

    1 07-03-2016 16:17:00 +0000

  • Adeogun Raodat

    F=ma ,70*4=280/7=40

    1 07-03-2016 16:18:00 +0000

  • Chima Sunday (Tutor)

    Can you try this one also.?
    A nail is pulled from a wall with a string tied to the nail. If the string is inclined at an angle 60o to the wall and the tension T in the string is 200N, Calculate the workdone in pulling the nail off from 1cm depth inside the wall
    A. √3J
    B. 2J
    C. 1J
    D. 100J
    (Tutor Question)

    1 07-03-2016 16:18:00 +0000

  • Favour Francis

    1

    1 07-03-2016 16:21:00 +0000

  • Favour Francis

    c

    1 07-03-2016 16:22:00 +0000