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    Work, Energy and Power
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    A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to,    

    • (i) t1/2 

    • (ii) t  

    • (iii) t 3/2

    • (iv) t2

    B.

    (ii) t  

    Tips: -

    From,

    v = u + at

    v = 0 + at = at 

    As power, P = F × v

    ∴            P = (ma) × at = ma2t

    As m and a are constants, therefore, P ∝ t
    258 Views
    Answer
    A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 m s–1?
     
    Given, 

    Radius of the rain drop, r = 2 mm = 2 × 10–3 m

    Volume of the rain drop, V = πr3

                                       =  × 3.14 × (2 × 10-3)3 m-3
    Density of water, ρ = 103 kg m–3
     
    Mass of the rain drop, m = ρV 

                                     =  × 3.14 × (2 × 10-3)3 × 103 kg
    Gravitational force, F = mg

                                =  × 3.14 × (2 × 10-3)3 × 103 × 9.8  N
    The work done by the gravitational force on the drop in the first half of its journey, 

                            WI = F . s 

                                 = × 3.14 × (2 × 10-3)3 × 103 × 9.8 × 250
                                 =  0.082 J 

    This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey.


    i.e.,     WII = 0.082 J 

    According to the law of conservation of energy,

    If no resistive force is present, then the total energy of the rain drop will remain the same. 

    ∴ Total energy at the top, 

         ET = mgh + 0 

             = (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × 500 × 10-5 

              = 0.164 J 

    Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.

    ∴Total energy at the ground is, 

          EG = (1/2) mv2 + 0 

               = (1/2) × (4/3) × 3.14 × (2 × 10-3)3 × 103 × 9.8 × (10)2 

               = 1.675 × 10-3 J 

    ∴ Resistive force = EG – ET = –0.162 J
    164 Views
    Answer
    Answer carefully, with reasons:
    What are the answers to (a) and (b) for an inelastic collision?
    In an inelastic collision, the linear momentum of the system remains conserved but not the kinetic energy.
    174 Views
    Answer
    Answer carefully, with reasons:
    If the potential energy of two billiard balls depends only on the separation distance between their centers, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
    Elastic.

    In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.
    202 Views
    Answer
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    A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to: 
    • (i) t1/2
    •  (ii) t 
    • (iii) t3/2
    • (iv) t2

    C.

    (iii) t3/2
    As

    Power, P = force × velocity 

             P = [MLT-2] [LT-1] = [ML2T-3] 

    As,     P = [ML2T-3] 

               = constant  

    ∴  L2T-3 = constant 

     L2/T3 = constant

    ∴        L2 ∝ T3 

          L ∝ T3/2

    Tips: -

    As 

    Power, P = force × velocity 

             P = [MLT-2] [LT-1] = [ML2T-3] 

    As,     P = [ML2T-3] 

               = constant  

    ∴  L2T-3 = constant 

    rightwards double arrow L2/T3 = constant

    ∴        L2 ∝ T3 

    rightwards double arrow      L ∝ T3/2
    226 Views
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