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CHAPTER 14
KINEMATICS OF A PARTICLE
(WORK AND ENERGY)
Chapter Objectives
To provide an introduction o the basic quantities and idealisatios of mechanics
To give the statement of Newton’s Law of Motion
To examine the standard procedures for performing numerical calculations
To present a general guide for solving problems
14.1 The work of a force

A force F does work when the particle undergoes a displacement in the direction
of the force.
Eg : particles moves along the paths, from
position r to r’, dr = r’ – r, where dr = ds. If
the angle , the work dU is done by F is a
scalar quantity
dU = F.ds cos 
Definition of the dot product dU = F.dr
dU = F.ds cos 
= F.dr
dU = o when F perpendiculars.
dr = ds cos 
Work of a variable force.
If the particle undergoes finite displacement
along its path, s 1 to s 2 , work is determined
by integration.
u 1 2 =

s2
r1
F .dr 

s2
s1
f cos .
If ( F cos  vs s ) was plotted, working
components as the area under the curve,
from position s 1 to s 2 .
Work of constant force moving along a straight line
If the F c has a constant magnitude and acts at a constant angle  from its straight line
path, so work done F c is
u 1 2 = F c cos 

s2
s1
d s
= F c cos  ( s 2 - s 1 ).
Work F c represents the area of the rectangle.
Work of weight
The particle moves along the path s, from s 1 to s 2 .At the immediate point, the
displacement dr = dx i + dy j + dz k .
Since w = -w j
u 1 2 =
 f .dr
r2
=

=

r1
( wj ).( dxi  dyj  dzk )
1
2
1
1
 wdy  w( y 2  y1 )
u 1 2 = -w (y 2 - y 1 )
This, work done is equal to the magnitude of
the particle’s weight times its vertical displacement.
Work of spring
The magnitude of force developed in a linear elastic spring when the spring is displaced a
distance s from its unstreched position is Fs = ks , where k is the spring stiffness. If the
spring is elongated or compressed from a position s1 to a further position s2 , the work
done on the spring by Fs is positive, since in each case, the force and displacement are in
the same direction.
U1-2 =

s2
s1
Fsds
=

s2
s1
ksds
= 1 ks22  1 ks12 .
2
2
When the particle is attach to a spring, then the force Fs exerted on the particle is
opposite to that exerted on the spring.
The force will do negative work on the particle when the particle is moving so as a
further elongate ( or compress ) the spring.
U1-2 = - ( 1 ks22  1 ks12 ).
2
2
 If both same direction – positive work .opposite direction – negative work.
14.2 Principle work and energy
If the particle has a means, m and is subjected to a system of external forces, represented
by the resultant FR = Ft = ma t.
Applying the kinematics equation:
at  v. dv

s2
s1
Ft ds =

v2
v1
ds
mvdv.
2
2
= 1 mv2  1 mv1
2
2
U 1 2
= the sum of the work done by all the forces acting as the particle
moves from point 1 -2.
T = 1 mv 2
T = particle final kinetic energy
2
1 mv 2 = particle initial kinetic energy
2

T1 + U1-2 = T2
Note : i ) Ft = m.at , to obtain at , integrate at = v. dv
ds
.
ii ) F n = m.an cannot be used, since these force do no work on the particle on
the forces directed normal to the path.
Procedure for analysis
The principle of work and energy is used to solve the kinetic problems that involve
velocity , force and displacement.
FBD : Draw FBD in order to account for all the forces.
Principle of w  E :

T1 + U1-2 = T2

Kinetic energy at the initial / final point always positive T = 1 mv 2
2

A forces does work when it moves through its displacement in the direction of the
force.( +ve same direction )

Force that are functions of displacement must be integrate to obtain the work.

The work of a weight – weight magnitude and the vertical displacement.
Uw =  wy ( +  ).

The work of spring , us = 1 hs2 .
2
14.3 Principle of Work and Energy for a system of particles
i th  particles
m i  mass
F i  resultant external force
f i  resultant internal force
Which applies in the tangential direction, the principle of work and energy:
s
s
1 m v 2  i 2 ( Fi) ds  i 2 ( fi) ds  1 m v 2
si1 t si1 t
2 i i
2 i i2
For all of the i th particles.
 12 m v
2
i i1

   ( Fi) t ds    ( fi) t ds   1 mi vi 2
2
si 1
si 1
si 2
ii 2
T 1 + U 1 2 = T 2
system’s ferial
kinetic energy
work done by all the
external or internal forces.
system’s initial kinetic energy.
2
Work of Friction Caused by Sliding.
v
v
P
P
The cases where a body is sliding over the
surface on another body in the presence of
friction.
s
Applied force P just balance the resultant
friction force  k .N .
W
P
1 mv 2  Ps   k N s  1 mv 2
2
2
F = kN
is satisfied when P =  k N
N
The sliding motion will generate heat, a form of energy which seems not to be accounted
for in the W&E equation.
See Example:
-
14.2
-
14.3
-
14.4
-
14.5
-
14.6
14.4 Power and Efficiency
Defined as the amount of work performance per unit of time.
Power
du
dt
F.dr
=
dt
dr
= F.
dt
P=
where du = F.dr
or P = F.V.
Power is a scalar, where in the formulation, V represents the velocity of the point which
is acted upon by the force, F.
I WaH = I J
s
= I N. m
s
=Iw
Note:
The term ‘power’ provides the useful basis for determining the type of motor or
machine which is required to do a certain amount of work in a given time. For example,
two pumps may each be able to empty a reservoir if given enough time, however the
pump having the larger power will be complete the job sooner.
The mechanical efficiency of a machine is defined as the ratio
of the output of useful power produce by the machine to the
input of power supplied to the machine.
Efficiency
E
If the energy applied,
E
power  output
power  input
energy  output
energy  input
Since machines consist of a series of moving parts, frictional forces will always be
developed within the machine, and as result, extra energy or power is needed to
overcome these forces.
“The efficiency of a machine is always less than 1”
Procedure for Analysis
o
o
o
o
Determine the external force F
If accelerating , F  m.a
One F and V here been found, P = F.V = Fr cos
In some problem F per unit time, P = du
dt
See Example:
-
14.7
-
14.8
14.5 Conservative forces and potential energy
Conservative force

When the work done by a force in moving a particle from one point to another is
independent of the path followed by the particle, the force is called a conservative
force.

e.g. : - weight of the particle
: - the force of an elastic spring

weight – depends on the particle’s vertical displacement
spring – depends only on the extension/compression.

in contrast – force of friction exerted on a moving objects – depends on the
path/neoconservative.
Potential Energy

Energy - capacity for doing works.
- from the motion of particle – kinetic energy.
- from the position of particle ( fixed datum / reference ) - potential
energy
(potential energy due to gravity ( weight ) and elastic spring is important.)

Gravitational potential energy: Vg = Wy

2
Elastic potential energy: v e = 1 k s
2
[ v e always +ve, the spring has the
capacity for always doing positive
work ] when the spring back to
outstretch position.

Potential Function
If a particle is subjected to both gravitational and elastic forces, - potential function.
v = v g + ve
U 1 2 = V 1 - V 2
work done by a conservative force.
E.g. : potential function for a particle of
weight W, suspended from a spring can be
expressed in term of its position, s ,
measured from a datum.
V = Vg + Ve
= - ws + 1 ks2
2
If the particle moves from s1 to lower s2,
2
2
u 12  v1  v2  (ws1  1 ks1 )  (ws2  1 ks2 )
2
2
2
2
= w( s 2  s1 )  ( 1 ks2  1 ks1 )
2
2
14.6 Conservation of Energy
When the particle acted upon by a system of both conservative and neoconservative,
work done by conservative position.
T1  V1  (U12 )noncons  T2  V2
If ‘noncons’ is zero => T1  V1  T2  V2
Note : conservative force ( not follow the path ) – weight / spring neoconservative
force ( follow exactly the path ) – friction.
T1  V1  T2  V2  conservation of mechanical energy, conservation energy.
Note : during the motion, sun of potential and kinetic energies remains constant. Far
this occur, kinetic energy must be transformed into potential energy and
vice versa.
E.g. : The ball of weight ‘w’ is dropped from a height, ‘h’ above the ground (datum).
 at initial position, mechanical energy
E = T1 + V1
2
= 1 mv1  (ve  v g )
2
2
= 1 mv1  wh
2
= 0 + wh
= wh.
When the ball has fallen a distance h , its speed can be determined by using:
2
v 2  v0  2ac ( y  yo )
= 2a c ( y  y o )
= 2 g (h )
2
 v = 2 g (h )
2
2
= 2 gh
Therefore:
E  V2  T2
= w( h )  1 m( gh ) 2
2
2
w
= w( h )  1
( gh ) 2
2
2g
= wh.
When the ball strikes the ground, P.E. (0) and
v 2  vo  2ac ( y  yo )
= 2.g.(h)
2
= 2 gh
Total energy:
E  V3  T3
= 0
= wh.
1w
( 2 gh ) 2
2g