Kinetics of particles work energy method

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P
INTRODUCTION
 In the first part of this chapter we shall use the Work Energy principle
method for analyzing kinetics of a moving particle or a system of particles.
This is an alternate approach to Newton's Second Law. This method
eliminates the determination of acceleration and thereby at times results in
quicker solution.
 In the second part of this chapter we will learn the Conservation of Energy
method for solving certain special problems involving conservative forces.
WORK OF A FORCE
 Work is a scalar quantity. It is defined as the product of the force and the
displacement in the direction of the force. It is denoted by letter U. Units of
work are N.m or Joule (J).
 Consider a block acted upon by a constant force F acting at an angle  as
shown. Let this force cause the block to displays by s.
 then, Work by force
cosU F s  ……. [11.1 (a)]
 A special case, when 0  i.e. force acts along the placement, then work by
force
U F s  …… [11.2 (b)]
 Also when 90  i.e. force is  to the displacement, then work by force = 0
Kinetics of Particles-Work Energy Method

2
WORK OF A SPRING FORCE
 Consider an undeformed spring of stiffness k as shown in Fig. (a). Let the
spring be deformed by some external agency, not shown in the figure, by an
amount X1, as shown in Fig. (b).
 Let this be position (1) of the spring. Now let the spring get further
deformed such that its new deformation measure from the neutral position
be x2. This is position of the spring. (Fig.).
 We know that the force in spring is variable as it is proportional to its
deformation x and is directed towards the neutral position i.e. spring force
F kx 
 The work done by the spring between position (1) and position (2) is
 
2
1
2 2
2 1
x 1
U= -kxdx=- k -
2
x
x x
 2 2
1 2
1
or work byspring U= k -
2 x x …… [11. 2]
 Here, k is the spring constant to be taken in N/m, x1 and x2 are the
deformations in the spring in position (1) and position (2) respectively and
should be taken in metre to get the work in the units of N.m (joule).

3
WORK OF A WEIGHT FORCE
 Consider a particle of mass m i.e. weight = mg move along a curved path in
a vertical plane from position (1) to position (2).
 Let h be the vertical displacement between the two positions. Since the
weight force which acts in the vertical direction has undergone a vertical
displacement. h, the work done by weight force,
U=±mgh …… [11.3]
 In simple words if the final position of the particle is below the initial
position work by weight is positive. If the final position is above the initial
position, work by weight is negative.
WORK OF A FRICTION FORCE
 Consider a block of mass m slide down
distance s on an inclined rough plane. If s
and k are the coefficient of static and kinetic
friction, the block's motion would be resisted
by the frictional force = k N .
 The work of a friction force is
kU N s   ……. [11.4]
 here, s though implies displacement, would be equal to the actual distance
moved by the particle since frictional force always orients itself so as to
oppose motion. Due to this, work by friction is always negative.

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KINETIC ENERGY
 It is defined as the energy possessed by the particle by virtue of its motion.
If the particle is static i.e. not in motion it will not possess any kinetic
energy. It is denoted by letter T.
 If a particle of mass m has a velocity v at a given instant, its Kinetic Energy
is
21
2
T mv ……. [11.5]
 Kinetic Energy is a scalar quantity and its S.L unit is Newton-metre (N.m)
or Joule(J).
WORK ENERGY PRINCIPLE
 Consider a collar of mass m travel from position (1) with a velocity v1 and
reach position (2) with a new velocity v2. The collar slides over a smooth
curved guide kept in a horizontal plane, under the action of force F at angle
 with the tangent to the path.
 For an instant during its motion, applying equation of Newton's Second
Law,
weget Fcosα= ma
dv dv
Fcosα=m since at=
ds dt
dv ds
Fcosα=m × ........heredsis thesmallarclength
ds dt
traveled by thecollarin dt time.
dv ds
Fcosα=mv since v=
ds dt
or Fcosαds=mvdv
x
t
F ma





5
Integrating between position (1) where 1 1s s and v v  and position
(2) where 2 2s s and v v 
2 2
1 1
cos
s v
s v
F ds m v dv  
Since the L.H.S represents the work done by a force we get,
2 2
2 11 2
2 2
2 1 2 2
2 2
1 1 2 2
1 1
2 2
1 1
2 2
1 1
, weget
2 2
mv mv
mv U mv
if mv T and mv T
U 

 
  
 
1 1 2 2T U T  ……. [a]
 Equation (a) relates the change in kinetic energy of the particle to the work
done by the force on it. This relation can be extended to a system of forces
acting on a particle and hence Work Energy Principle states "For a particle
moving under the action of forces, the total work done by these forces
is equal to the change in its kinetic energy”. Equation (a) is therefore
expressed as
1 1 2 2T U T  
……. [b]

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APPLICATION OF WORK ENERGY PRINCIPLE
 Work Energy principle is a simpler approach to the kinetics of a moving
particle or a system of particles. It involves the use of the scalar equation
(b) viz. 1 1 2 2T U T  where T1 and T2 represent the particle's kinetic
energy in position (1) and (2) respectively and U1-2 represents the work done
by various forces acting on it. This concept does not involve the calculation
of the particle s acceleration which is the case in Newton's Second Law
method of solving kinetics. This principle is useful whenever the problem
involves known or unknown parameters like forces, mass, velocity and
displacement. Knowing the forces and the displacement of the particle, the
particle's velocity in the new. Position can be found out, or in some cases
knowing the particle's initial and final velocities and the active forces, the
particle's displacement can be worked out.
 If the system involves more than one particle, the principle can be applied
to the system of particles also. In such cases, the total kinetic energy would
be sum of the individual kinetic energies of different particles and the work
done would be the summation of the work done by various forces on 1he
individual particles.

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POWER AND EFFICIENCY
 Consider a case of two persons in a race, set to climb the stairs and reach
the top of a 10 storied building. Here both the persons would be doing an
equal amount of work in reaching the top, but if one person reaches earlier
than the other, he would be said to have exerted a greater power than the
other one as he has completed the work in lesser time. Thus the rate at
which the work done is equally important.
 Power is defined as the rate of doing work
i.e.
dU
power =
dt
…….. [11.7 (a)]
 If the rate of doing work is constant, then
 
F×s
power = .......sincework =F×s
t
s
or power = F×v .........sincev= ,..... 11.7(c)
t
 Power is a scalar quantity and its S.I unit is N.m/s or J/sec or Watt
1N.m/s = 1J/s = l W
 For any machine doing work, its efficiency is defined as
 
Output work Power output
η= or η=
Input work Power input
Power output
%η= ×100 ........ 11.8
Power input
 All machines have efficiency less than 1 or 100 % because of various energy
losses. Major energy losses are the frictional losses. Man has always strived
to increase the efficiency of machines it uses by reducing the various
energy losses it undergoes during its working.

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ENERGY
 The capacity of particle to do work is defined as the energy possessed by
the particle. Energy, exists in nature in various forms viz.,
1) Mechanical
2) Electrical
3) Heat
4) Sound
5) Light
6) Chemical etc.
 Mechanical energy, which is the sum of the potential energy (v) and
kinetic energy (T), is of interest in study of kinetics.
 Potential Energy
 It is defined as the energy possessed by a particle by virtue of its position
with respect to a datum, or by virtue of elastic forces acting on it. It is
denoted by letter V. Potential energy is a scalar quantity and its S.I. units
are Newton-metre (N.m) or Joule (J).
 Consider a particle of mass m moving on a vertical curved surface as
shown. If the ground is chosen as the datum, the potential energy due to
its position is given by
V mgh ……. [11.9 (a)]
1 1
2 2
3 3
here, V mgh
V mgh
V mgh




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 Consider the same block now acted upon by spring force. If x is the
deformation of the spring from the neutral position, the potential energy
due to elastic force is given by
21
2
V kx  ……. [11.9 (b)]
2
1 1
2
2 2
1
here,
2
1
and
2
V
V
kx
kx
 
 
Kinetic Energy
 This form of mechanical energy has been explained in the earlier part of the
chapter.

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CONSERVATION OF ENERGY
 In mechanics we define conservative forces to be those forces who do work
independent of the path followed by the particle on which they act. Work of
a spring force, weight force come under the category of conservative forces.
Non conservative forces are those forces whose work depends on the path
followed by the particle on which they act. Frictional force is a non-
conservative force.
 When a particle is acted upon by only conservative forces, the mechanical
energy of the particle remains constants
i.e T+V=constant …….. [11.10 (a)]
1 1 2 2or T +V =T +V …….. [11.10 (b)]
 Equation 11.10 (b) is referred to as Conservation of Energy Equation and
can be alternatively used to solve problems in kinetics involving
conservative forces only.

11
EXERCISE 1
1. A bullet of mass 30 gm moving with a velocity of 226 m/s strikes a wooden
1og and penetrates through a distance of 185 mm. Calculate the average
retarding force offered by the log in stopping the bullet.
2. A 1200 kg automobile is driven down a 100 incline at speed of 54kmph.
When the brakes are applied, a constant braking force of 5000 N acts and
brings the automobile to a halt Within a certain distance. Find this
distance.
3. A block of mass 60 kg at A is being pushed up a inc1ined plane ( 0.2  ) by
applying a constant force P = 500 N. knowing that the speed of the block at
A is 3 m/s, determine i) the speed of the block at B (ii) if the force P is now
reduced and P = 300 N acts during its motion from B,C. find the distance x
travelled by the block as it comes to a halt at C.
4. Two masses B kg and 3 kg are initially, held at rest in the position shown.
Determine the speed of the 8 kg block as it hits the ground. Neglect friction
at the pulley.

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5. From the top of a building 20 m high, a ball is projected at 15 m/s at an
angle of 300 upwards to the horizontal. Find the magnitude of the velocity
of the ball as it hits the ground. Use Work Energy Principle.
6. At a certain instant a body of mass 15 kg is failing vertically down at a
speed of 25 m/s. what upward vertical force will stop the body in 2
seconds?
7. A stone weighing 15 N dropped from a height of 25 m buries itself 3OO mm
deep in the sand. Find the average resistance to penetration and the time of
penetration.
8. Find the velocity of block A and B when A has
traveled 1.2 m up the inclined plane starting
from rest. Mass of A is 10 kg and that of B is
50 kg. Coefficient of friction between block A
and inclined plane is 0.25. Pulleys are
massless and frictionless. Use Work energy principle.
9. The 3 kg smooth collar is attached to a spring of spring constant, k = 6
N/m that has an upstretched length 2.8 m. Determine its speed at A, when
it is drawn to point B and released from rest.
10. A 5 kg steel collar is attached to a spring of k =
800 N/m and a free length of O.7 m. If the
collar is released from rest at A, determine the
speed of the collar as it passes through B and
C. If the collar finally comes to halt at D, find
the distance CD.

13
11. A spring of stiffness k is placed horizontally and a ball of mass m strikes
the spring with a velocity v. find the maximum compression of the spring.
Take m = 5 kg, k = 500 N/m, v = 3 m/s.
12. A 5 kg mass drops 2m upon a spring whose modulus is 10 N/mm. What
will be the speed of the block when the spring is deformed 100.?
13. A small block of mass 5 kg is released at A from rest on a frictionless
circular surface AB. The block then travels on the rough horizontal surface
BC whose 0.3 and 0.2s k   . A spring having stiffness k = 900 N/m is
placed at C to bring the block to a halt. Find the maximum velocity attained
by the block and also the maximum compression undergone by the spring
14. A block of mass m = 8O kg is compressed against a spring as shown in
figure. How far from point B (distance
x) will the block strike on the plane at
point A. Take free length of spring as
0.9 m, k = 0.2 and spring stiffness
as K = 40 x l02 N/m.
15. A hammer of mass 4 kg slips off from the hands of a carpenter working on
the roof top at A. If
0.2k 
determine
the velocity of the hammer at the edge
B of the roof and also find the
location x of point C where the
hammer strikes the ground.

14
16. Marbles having a mass of 5gm fall from rest at A through the glass tube
and accumulate in the can at
'C'. Determine the placement
'R' of the can from the end of
the tube and the speed at
which the marbles fall into the
can. Neglect the size of the can.
17. A spring is compressed by 0.3 m and held by a latch mechanism. When the
latch is released it propels a package of 500N weight from position A to
position B on the conveyor. If k . = 0.2 between the package and the incline
and the desired speed of the package at B is 4 m/s, determine the stiffness
value k of the spring which should be provided.
18. A collar of mass 10 kg moves in a vertical guide as
shown. Neglecting friction between guide and the
collar, find its velocity when it passes through
position (2), after starting from rest in position (1).
The spring constant is 2OO N/m and the free length
of the spring is 200mm.
19. Two springs each having a stiffness of O.6 N/cm are connected to a ball B
having a mass of 3kg in a horizontal position producing an initial tension of
1.8 N in each spring. If the ball is allowed to fall from rest, what will be its
velocity after it falls through a height of 15cm? The length of each spring in
the horizontal position is 2Ocm.

15
20. A wagon of mass 50 tonne, starts from rest and travels 3 km down a 1%
grade and strikes a post with bumper string as shown in figure. If the
rolling resistance of the track is 50 N/t, find the velocity with which the
wagon strikes the post. Also find amount by which the spring will be
compressed, if the bumper spring compresses 1mm per 20 kN force.
21. A block weighing 500 N is released from rest and slides down the inclined
plane. It soon comes in contact with a spring of k = 2000 N/m which gets
compressed as the clock hits it. Find the maximum compression of the
spring.
22. A block of mass 5 kg is released from rest on an inclined plane as shown in
figure. Find maximum compression
of the spring, If the spring constant is
1 N/mm and coefficient of the friction
between the block and the inclined
plane is 0.2.
23. Two 500 kg stones are being dragged on a horizontal surface by means of
the truck T. If the towing cable
passes over a smal1 pulley at
A, determine the speed of the
stones when 60  . The stones
are at rest at 30  and the
ruck exerts the constant force
F = 5 kN on the cable at B as shown. Take 0.4  between stone D and
ground and neglect friction between stone C and ground

16
24. A ball of weight 25 N is suspended from 12OO mm long
elastic cord. The ball is now pulled vertically down by 200 mm
and then released. Determine the speed of the ball as it
strikes the ceiling. Neglect initial deformation of the elastic
cord due to self-wt. of the ball.
25. In a bungee jumping event a man of mass 65 kg has a 20 m elastic cord
tied to his round ankles. The man jumps from a bridge and falls freely for
20 m before the elastic cord begins to stretch. The man reaches 42 m below
the bridge before he starts rising, upwards. a) Determine the stiffness of the
elastic cord. b) How much below the bridge does the man acquire maximum
velocity and the value of the maximum velocity?
26. A tram car weighs 120kN, the tractive resistance being 5N/kN. What power
will be required to propel the car at a uniform speed of 20kmph.
1) On leve1 surface
2) Up an incline of 1 in 300
3) Down an inclination of 1 in 300 take efficiency of motor as 80 %.
27. A lift of total mass 500 kg at rest snaps off accidentally and falls freely for 6
m till it comes in contact with a set of four nested springs and is soon
brought to a halt. Each nested spring consists of an outer spring of
stiffness 10kN/m and an inner spring of stiffness 15kN/m. The inner
spring is lower by 0.2 m than the outer spring as shown. Find the
maximum deformation of the outer spring as the lift is brought safely to a
halt.

17
EXERCISE 2
Theory Questions
Q.1 Explain
1) work of a force
2) work of a spring force
3) work of a weight force
4) work of a friction force
Q.2 State Work Energy Principle.
Q.3 Derive the relation of Work Energy Principle
Q.4 Explain Power and Efficiency.
Q.5 Explain in detail Mechanical Energy.
Q.6 Define Work, Power and Energy. Explain Principle of Conservation of
Energy
Q.7 Differentiate between conservative and non-conservative forces giving
examples.

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UNIVERSITY QUESTIONS
1. State and prove work Energy principle. (5 Marks)
2. A spring is used to stop 100 kg package which is moving down a 30°
incline. The 10 spring has constant k = 30 kN/m is held by cables so that it
is initially compressed by 90 mm. If the velocities of package is 5 m/s,
when it is 9 m from spring. Determine the maximum additional
deformation of spring in bringing the package to rest. Assume coefficient of
friction between block and incline as 0-2. (10 Marks)
3. Explain work energy principle. (6 Marks)
4. Explain instantaneous centre of rotation. (6 Marks)
5. A motorist travelling at a speed of 90kmph suddenly the brakes and comes
to rest after skidding 100m. Determine the time required for the car to stop
and coefficient of kinetic friction between the tires and the road. (4 Marks)
6. The mass of A is 23kg and mass of B is 36kg. The coefficient are 0.4
between A and B, and 0.2 between ground and block B. Assume smooth
drum. Determine the maximum mass of M at impending motion. (8 Marks)

Kinetics of particles work energy method

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Kinetics of particles work energy method