Work, Energy and Power

Work, Energy and Power

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Mechanics

Physical World, Units and Measurements
Motion in a Straight Line
Motion in a Plane
Laws of Motion
Work, Energy and Power
Gravitation
Mechanical Properties of Solids
Mechanical Properties of Fluids
System of Particles and Rotational Motion
  1. Two bodies of masses m and 4 m are moving with equal K.E. The ratio of their linear momenta is​​








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    m1 = m, m2 = 4 m ​
    K . E1 = K . E2

    1
    		m1v12 =
    1
    		m2v22 ;
    1
    		mv12 =
    1
    		mv22
    2222

    v1
    		 = 2 ⇒ v1 = 2v2
    v2

    Linear momentum of first body
    Linear momentum of second body

    =
    m1v1
    		 =
    m .2v2
    		 =
    1
    m2v24mv22

    Correct Option: C

    m1 = m, m2 = 4 m ​
    K . E1 = K . E2

    1
    		m1v12 =
    1
    		m2v22 ;
    1
    		mv12 =
    1
    		mv22
    2222

    v1
    		 = 2 ⇒ v1 = 2v2
    v2

    Linear momentum of first body
    Linear momentum of second body

    =
    m1v1
    		 =
    m .2v2
    		 =
    1
    m2v24mv22

  1. The kinetic energy acquired by a mass (m) in travelling distance (s) starting from rest under the action of a constant force is directly proportional to








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    K.E. =
    1
    		mv2
    2

    Further, ​ v2 = u2 + 2as = 0 + 2ad = 2ad
    = 2(F / m)d
    Hence , K.E. =
    1
    		m × 2(F / m)d = Fd
    2

    or, K.E. acquired = Work done ​ 
    = F × d = constant. ​i.e., it is independent of mass m.

    Correct Option: D

    K.E. =
    1
    		mv2
    2

    Further, ​ v2 = u2 + 2as = 0 + 2ad = 2ad
    = 2(F / m)d
    Hence , K.E. =
    1
    		m × 2(F / m)d = Fd
    2

    or, K.E. acquired = Work done ​ 
    = F × d = constant. ​i.e., it is independent of mass m.

  1. If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is








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    Initial momentum (p1) = p; Final momentum (p2) = 1.5 p and initial kinetic energy (K1) = K. ​

    Kinetic energy (K) =
    p2
    		 ∝ p2
    2m


    or,
    K1
    		 =
    p1
    		2 =
    p
    		2 =
    1
    K2p21.5 p2.25

    or, K2 = 2.25 K. ​
    Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.

    Correct Option: C

    Initial momentum (p1) = p; Final momentum (p2) = 1.5 p and initial kinetic energy (K1) = K. ​

    Kinetic energy (K) =
    p2
    		 ∝ p2
    2m


    or,
    K1
    		 =
    p1
    		2 =
    p
    		2 =
    1
    K2p21.5 p2.25

    or, K2 = 2.25 K. ​
    Therefore, increase in kinetic energy is 2.25 K – K = 1.25 K or 125%.

  1. Consider a car moving along a straight horizantal road with a speed of 72 km/h. If the coefficient of static friction between road and tyres is 0.5, the shortest distance in which the car can be stopped is​​








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    Force due to friction = kinetic energy

    μ mg s =
    1
    		mv2
    2

    [ Here, v = 72 km/h
    =
    72000
    		 = 20 m/s ]
    60 × 60

    or , s =
    v2
    		 =
    20 × 20
    		 = 40 m
    2μg2 × 0.5 × 10

    Correct Option: B

    Force due to friction = kinetic energy

    μ mg s =
    1
    		mv2
    2

    [ Here, v = 72 km/h
    =
    72000
    		 = 20 m/s ]
    60 × 60

    or , s =
    v2
    		 =
    20 × 20
    		 = 40 m
    2μg2 × 0.5 × 10

  1. Two masses of 1g and 9g are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is








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    1
    		(1)v12 =
    1
    		(9)v22
    22

    ⇒
    v12
    		 = 9 or
    v1
    		 = 3
    v22v2

    Ratio of their linear momenta =
    m1v1
    		 =
    1
    		 × (3) =
    1
    m2v293

    Correct Option: C

    1
    		(1)v12 =
    1
    		(9)v22
    22

    ⇒
    v12
    		 = 9 or
    v1
    		 = 3
    v22v2

    Ratio of their linear momenta =
    m1v1
    		 =
    1
    		 × (3) =
    1
    m2v293