Work and Energy
Work and Energy
Work and Energy
PARTICLE:
WORK AND ENERGY
KNS1633 Engineering Mechanics
Dr Raudhah Ahmadi
Department of Civil Engineering, UNIMAS
Lecture Outline
• The Work of a Force
• Principle of Work and Energy
• Principle of Work and Energy for a System
of Particles
• Conservative Forces and Potential Energy
• Conservation of Energy
The Work of a Force
• A force F does work on a
particle only when the
particle undergoes a
displacement in the direction
of the force.
• If the particle moves along
the path s from position r to
new position r’,
displacement dr = r’ – r
The Work of a Force
• Magnitude of dr is represented by ds, differential
segment along the path
• If the angle between tails of dr and F is θ, work dU done
by F is a scalar quantity
dU = F ds cos θ
r2 s2
U1 2 F .dr F cos ds
r1 s1
The Work of a Force
Work of a Constant Force Moving Along a Straight Line.
• If the force Fc has a constant magnitude and acts at a
constant angle θ from its straight line path, then the
components of Fc in the direction of displacement is Fc
cos θ
The Work of a Force
The work done by Fc when the particle is displaced from s1
to s2 is determined
s2
U1 2 Fc cos ds
s1
or U1 2 Fc cos ( s2 s1)
U12 Wy
The Work of a Force
Work of a Spring Force.
• The magnitude of force developed in
a spring when the spring is displaced
a distance s from its unstretched
position is Fs = ks.
• Where k = spring stiffness
• If the spring is elongated or
compressed from a position s1 to s2,
the work done on the spring by Fs is
positive, since force and
displacement are in the same
direction.
The Work of a Force
s2 s2
U1 2 Fs ds ks ds
s1 s1
1 2 1 2
ks2 ks1
2 2
U1 2 1 2 1 2
ks2 ks1
2 2
NOTE:
direction of force same with direction of displacement -> +ve work
direction of force opposite to direction of displacement -> -ve work
The Work of a Force UT = (T cos Φ)s (+ve)
No work done!
s1
s2 1 2 1 2
Ft ds mv2 mv1
2 2
Principle of Work and Energy
• Principle of work and energy for the particle,
U12
1 2 1 2
mv2 mv1
2 2
• Term on the LHS is the sum of work done by all the
forces acting on the particle as the particle moves from
point 1 to point 2
• Term on the RHS defines the particle’s final and initial
kinetic energy
• Use homogeneous dimension so that the kinetic energy
has the same units as work (joules (J))
Principle of Work and Energy
T1 U1 2 T2
Ti = ½ mvi2
i = 1, 2
2 i i1 si1 i t si1 i t 2 i i 2
1 si 2 si 2 1
m v 2
( F ) ds ( f ) ds m v 2
1 2 1 2
mv Ps k Ns mv
2 2
• Sliding motion will generate heat, a form of energy which
seems not to be accounted for in the work energy
equation
Example 2
The 17.5-kN automobile is traveling down the 10° inclined
road at a speed of 6 m/s. if the driver jams on the brakes,
causing his wheels to lock, determine how far s his tires
skid on the road. The coefficient of the kinetic friction
between the wheels and the road is μk = 0.5
Solution
Work (Free-Body Diagram).
The normal force NA does no work since it never undergoes
displacement along its line of action. The weight 17.5-kN, is
displaced s sin 10° and does positive work.
+ n
F 0; ; N A 17500 cos 10
N 0
N A 17234.1N
FA 0.5 N A 8617.1N
Principle of Work and Energy.
T1 U1 2 T2
1 17500N
2 9.81m / s
( 6 m / s )
2
17500 N ( s sin 10
) (8617.1N ) s 0
T1 U1 2 T2
s 1
0 (28 3s )(10 )ds (2.50)(10 )(9.81) s (2.50)(103 )v 2
2 3 3
0 2
28(103 ) s (103 ) s3 24.525(103 ) s 1.25(103 )v 2
v (2.78s 0.8s3 )1 / 2
With s = 3 m,
v 5.47m / s
Kinematics.
Since we were able to express the velocity as a function of
displacement using v = ds/dt
ds
(2.78s 0.8s ) 3 1/ 2
dt
3 ds
t
0 ( 2.78s 0.8s 3 )1 / 2
1.79s
Example 4
The platform P is tied down so that the 0.4-m long cords
keep a 1-m long spring compressed 0.6-m when nothing is
on the platform. If a 2-kg brick is placed on the platform and
released from rest after the platform is pushed down 0.1-m,
determine the max height h the block rises in the air,
measure from the ground.
Solution
Work (Free-Body Diagram).
Since block is released from rest and later
reaches its maximum height, the initial and
final velocities are zero. The weight does
negative work and the spring force does
positive work.
T1 U1 2 T2
1 2 1 2 1 2 1 2
mv1 ks2 ks1 Wy mv2
2 2 2 2
Note that here s1 = 0.7 m > s2 = 0.6 m and so the work of
the spring will be positive
1 2
Ve ks
2
Conservative Forces and Potential
Energy
Ve is always positive since, in the
deformed position, the force of
the spring has the capacity for
always doing positive work on
the particle when the spring is
returned to its unstretched
position.
Conservative Forces and Potential
Energy
Potential Function.
If a particle is subjected to both gravitational and elastic
forces, the particle’s potential energy can be expressed as a
potential function
V Vg Ve
Conservation of Energy
When a particle is acted upon by a system of both
conservative and non-conservative forces, the portion of the
work done by the conservative forces can be written in
terms of the difference in their potential energies using
U12 cons. V1 V2
As a result, the principle of work and energy can be written
as
T1 V1 (U1 2 )noncons. T2 V2
Conservation of Energy
• (∑U1-2)noncons represents the work of the non-conservative
forces acting on the particles.
• If only conservative forces are applied to the body, this
term is zero and we have
T1 V1 T2 V2
T V const
Example 5
The gantry structure is used to test the response of an
airplane during a crash. The plane of mass 8-Mg is hoisted
back until θ = 60°, and then pull-back cable AC is released
when the plane is at rest. Determine the speed of the plane
just before clashing into the ground, θ = 15°. Also, what is
the maximum tension developed in the supporting cable
during the motion?
Solution
Potential Energy.
For convenience, the datum has been established at the top
of the gantry.
Conservation of Energy.
TA VA TB VB
0 8000(9.81)( 20 cos 60 )
1
(8000)vB2 8000(9.81)( 20 cos 15 )
2
vB 13.5m / s
Equation of Motion.
The free-body diagram when the plane is at B,
+ Fn man ;
2
(13. 5)
T 8000(9.81) N cos 15 (8000)
20
T 149kN
Example 6
A smooth 2-kg collar C, fits loosely
on the vertical shaft. If the spring is
unstretched when the collar is in
the position A, determine the
speed at which the collar is
moving when y = 1 m if (a) it is
released from rest at A, and (b) it
is released at A with an upward
velocity vA = 2 m/s.
Solution
Part (a)
Potential energy.
For convenience, the datum is established through AB.
When the collar is at C, the gravitational potential energy is
–(mg)y, since the collar is below the datum and the elastic
potential energy is
1 2
ksCB
2
sCB = 0.5 m which represent the stretch in the spring
Conservation of Energy
TA VA TC VC
TA VA TC VC
1 2 1 2 1 2
mv A 0 mvC ksCB mgy
2 2 2
2 1
(2)( 2) 0 (2)vC (3)(0.5) 2 2(9.81)(1)
1 2 1
2 2 2
vC 4.82m / s
THANK
YOU