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    Dynamics and Kinematics

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    For detailed description with figs. please refer the attachment.

    Prob. 1 : A box slides down a ramp with two straight segments and on leaving the ramp it slides on a rough horizontal surface and then impacting a spring. To determine kinetic energy, velocity at different points and the compression of the spring.

    Prob. 2 : A projectile is launched which lands on a hill. To determine max. height, horizontal range, time of flight etc.

    Prob. 3 : To determine velocity, acceleration at different points of a two pulley system (free end of the string being pulled by hand)

    Prob. 4 : To determine velocity, acceleration at different points of a two pulley system (a load hanging from the free end of the string)

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    Solution Preview

    Please refer attachment.

    Problem 1

    A 3 kg

    h1= 3m

    h B
    μk = 0.2 k = 3000 N/m
    h2 =3m

    C 10m D

    a) At the starting point A the box has only potential energy equal to mgh = 3x9.8x6 = 176.4 Joule

    As the box slides down the smooth ramp, its potential energy is progressively converted into kinetic energy. At point B it is 3m lower than A. Hence, PE equivalent to 3m is converted into KE. Hence,
    KE at B = mgh1 = 3x9.8x3 = 88.2 J

    b) At point C whole of PE is converted into KE. Hence, ½ m(vC)2 = mgh

    Or vC = √2gh = √2x9.8x6 = 10.8 m/s

    c) Segment AB is straight. Hence its radius of curvature is infinity.

    d) Between points C and D, the box is subjected to a decelerating frictional force equal to μkmg = 0.2 x 3 x 9.8 = 5.88 N

    Deceleration = Decelerating force/Mass = 5.88/3 = 1.96 m/s2

    Applying v2-u2 = 2as between points C and D we get : (vD)2 - (vC)2 = 2as

    Or (vD)2 - (10.8)2 = 2 x (-1.96) x 10 or vD = 8.8 m/s

    e) Kinetic energy of the box at D = ½ m(vD)2 = ½ x 3 x 8.82 = 116 J

    As the box strikes the spring, its KE is converted into the PE of the spring. If the spring is compressed by x, then the PE stored in the spring = ½ kx2
    At maximum compression, whole of KE of the box is converted into the PE of the spring. Hence,

    ½ kx2 = 116 or x2 = 2 x 116/3000 = 0.077 m or 7.7 cm

    Problem 2

    u = 80 m/s h
    u sinθ
    θ = 45O 50m
    u cosθ X

    The problem is solved from first principles.

    Initial velocity can be resolved into components as follows :

    uy = u sinθ = 80 sin45O = 56.6 m/s

    ux = u cosθ = 80 cos45O = 56.6 m/s

    a) The projectile has no force acting on it in the horizontal direction (ignoring air resistance). Hence, its ...

    Solution Summary

    Box sliding down a ramp with two straight segments in horizontal surfaces are determined. The projectile launching time of flight is determined.