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Never stop learning because life never stop Teaching

Thursday 18 September 2014

WORK, POWER & ENERGY - 1st Year Physics

WORK, POWER & ENERGY - 1st Year Physics

    WORK, POWER & ENERGY


    Work

    Work is said to be done when a force causes a displacement to a body on which it acts.
    Work is a scalar quantity. It is the dot product of applied force F and displacement d.



    W = F . d
    W = F d cos θ ………………………… (1)
    Where θ is the angle between F and d.
    Equation (1) can be written as
    W = (F cos θ) d
    i.e., work done is the product of the component of force (F cos θ) in the direction of displacement and the magnitude of

displacement d.
    equation (1) can also be written as
    W = F (d cos θ)
    i.e., work done is the product of magnitude of force F and the component of the displacement (d cos θ) in the direction of

force.

    Unit of Work
    M.K.S system → Joule, BTU, eV
    C.G.S system → Erg
    F.P.S system → Foot Pound
    1 BTU = 1055 joule
    1 eV = 1.60 x 10(-19)

    Important Cases
    Work can be positive or negative depending upon the angle θ between F and d.

    Case I
    When θ = 0º i.e., when F and d have same direction.
    W = F . d
    W = F d cos 0º ………….. {since θ = 0º}
    W = F d …………………….. {since cos 0º = 1}
    Work is positive in this case.

    Case II
    When θ = 180º i.e., when F and d have opposite direction.

    W = F . d
    W = F d cos 180º ………………………. {since θ = 180º}
    W = – F d ………………………………….. {since cos 180º = -1}
    Work is negative in this case

    Case III
    When θ = 90º i.e, when F and d are mutually perpendicular.
    W = F . d
    W = F d cos 90º ………………………. {since θ = 90º}
    W = 0 ……………………………………. {since cos 90º = 0}

    Work Done Against Gravitational Force
    Consider a body of mass ‘m’ placed initially at a height h(i), from the surface of the earth. We displaces it to a height h(f)

from the surface of the earth. Here work is done on the body of mass ‘m’ by displacing it to a height ‘h’ against the

gravitational force.
    W = F . d = F d cos θ
    Here,
    F = W = m g
    d = h(r) – h(i) = h
    θ = 180º
    {since mg and h are in opposite direction}
    Since,
    W = m g h cos 180º
    W = m g h (-1)
    W = – m g h
    Since this work is done against gravitational force, therefore, it is stored in the body as its potential energy (F.E)
    Therefore,
    P . E = m g h

    Power

    Power is defined as the rate of doing work.
    If work ΔW is done in time Δt by a body, then its average power is given by P(av) = ΔW / Δt
    Power of an agency at a certain instant is called instantaneous power.

    Relation Between Power and Velocity
    Suppose a constant force F moves a body through a displacement Δd in time Δt, then
    P = ΔW / Δt
    P = F . Δd / Δt ………………… {since ΔW = F.Δd}
    P = F . Δd / Δt
    P = F . V …………………………… {since Δd / Δt = V}
    Thus power is the dot product of force and velocity.

    Units of Power
    The unit of power in S.I system is watt.
    P = ΔW / Δt = joule / sec = watt
    1 watt is defined as the power of an agency which does work at the rate of 1 joule per second.
    Bigger Units → Mwatt = 10(6) watt
    Gwatt = 10(9) watt
    Kilowatt = 10(3) watt
    In B.E.S system, the unit of power is horse-power (hp).
    1 hp = 550 ft-lb/sec = 746 watt

    Energy

    The ability of a body to perform work is called its energy. The unit of energy in S.I system is joule.

    Kinetic Energy
    The energy possessed by a body by virtue of its motion is called it kinetic energy.
    K.E = 1/2 mv2
    m = mass,
    v = velocity

    Prove K.E = 1/2 mv2

    Proof
    Kinetic energy of a moving body is measured by the amount of work that a moving body can do against an unbalanced force

before coming to rest.
    Consider a body of mas ‘m’ thrown upward in the gravitational field with velocity v. It comes to rest after attaining height

‘h’. We are interested in finding ‘h’.
    Therefore, we use
    2 a S = vf2 – vi2 ………………………. (1)
    Here a = -g
    S = h = ?
    vi = v (magnitude of v)
    vf = 0
    Therefore,
    (1) => 2(-g) = (0)2 – (v)2
    -2 g h = -v2
    2 g h = v2
    h = v2/2g
    Therefore, Work done by the body due to its motion = F . d
    = F d cos θ
    Here
    F = m g
    d = h = v2 / 2g
    θ = 0º
    Therefore, Work done by the body due to its motion = (mg) (v2/2g) cos0º
    = mg x v2 / 2g
    = 1/2 m v2
    And we know that this work done by the body due to its motion.
    Therefore,
    K.E = 1/2 m v2

    Potential Energy
    When a body is moved against a field of force, the energy stored in it is called its potential energy.
    If a body of mass ‘m’ is lifted to a height ‘h’ by applying a force equal to its weight then its potential energy is given by
    P.E = m g h
    Potential energy is possessed by
    1. A spring when it is compressed
    2. A charge when it is moved against electrostatic force.

    Prove P.E = m g h OR Ug = m g h

    Proof
    Consider a ball of mass ‘m’ taken very slowly to the height ‘h’. Therefore, work done by external force is
    Wex = Fex . S = Fex S cos θ ……………………………. (1)
    Since ball is lifted very slowly, therefore, external force in this case must be equal to the weight of the body i.e., mg.
    Therefore,
    Fex = m g
    S = h
    θ = 0º ……………………. {since Fex and h have same direction}
    Therefore,
    (1) => Wex = m g h cos 0º
    Wex = m g h …………………………………………………………. (2)
    Work done by the gravitational force is
    Wg = Fg . S = Fg S cosθ …………………………………………. (3)
    Since,
    Fg = m g
    S = h
    θ = 180º …………………. {since Fg and h have opposite direction}
    Therefore,
    (3) => Wg = m g h cos 180º
    Wg = m g h (-1)
    Wg = – m g h …………………………………………………………. (4)
    Comparing (2) and (4), we get
    Wg = -Wex
    Or
    Wex = – Wg
    The work done on a body by an external force against the gravitational force is stored in it as its gravitational potential

energy (Ug).
    Therefore,
    Ug = Wex
    Ug = – Wg ………………………… {since Wex = -Wg}
    Ug = -(-m g h) ………………… {since Wg = – m g h}
    Ug = m g h ……………………………………….. Proved

    Absolute Potential Energy
    In gravitational field, absolute potential energy of a body at a point is defined as the amount of work done in moving it

from that point to a point where the gravitational field is zero.

    Determination of Absolute Potential Energy
    Consider a body of mass ‘m’ which is lifted from point 1 to point N in the gravitational field. The distance between 1 and N

is so large that the value of g is not constant between the two points. Hence to calculate the work done against the force of

gravity, the simple formula W = F .d cannot be applied.
    Therefore, in order to calculate work done from 1 to N, we divide the entire displacement into a large number of small

displacement intervals of equal length Δr. The interval Δr is taken so small that the value of g remains constant during this

interval.
    Diagram Coming Soon
    Now we calculate the work done in moving the body from point 1 to point 2. For this work the value of constant force F

may be taken as the average of the forces acting at the ends of interval Δr. At point 1 force is F1 and at point 2, force is F2.

    Law of Conservation of Energy

    Statement
    Energy can neither be created nor be destroyed, however, it can be transformed from one form to another.

    Explanation
    According to this law energy may change its form within the system but we cannot get one form of energy without spending

some other form of energy. A loss in one form of energy is accompanied by an increase in other forms of energy. The total

energy remains constant.

    Proof
    For the verification of this law in case of mechanical energy (Kinetic and potential energy). Let us consider a body of mass

‘m’ placed at a point P which is at a height ‘h’ from the surface of the earth. We find total energy at point P, point O and

point Q. Point Q is at a distance of (h-x) from the surface of earth.

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