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Dynamics and Kinematics

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
Y

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

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.

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