Kinetics of Particles Work & Energy

1
3. Kinetics of Particles-
Work & Energy
ME 203
Engineering Mechanics II: Dynamics
Instructor: Masud Rana

2
Work of a Force
r
d

cos
x y
dU F dr
F ds
F dx F dy



 
• Particle displacement:
• Work of force corresponding
to displacement :
F
r
d

Work is a scalar quantity of units [Nm]
If and have same direction
F r
d

dU Fds

If and have opposite directions
F r
d

dU Fds
 
If is perpendicular to
F r
d

0
dU 

3
Work of a Force
Work of a force during a finite displacement:
 
2
1
2 2
1 1
1 2
cos
A
A
s s
t
s s
U F dr
F ds F ds

 
 

 
Work is represented by the area under
the curve of Ft plotted versus s.
 
2
1
1 2
A
x y
A
U F dx F dy
  

Ft = Tangential component
If we use rectangular components:

4
Work of a Force
Work of a constant force in rectilinear motion:
  x
F
U 

 
cos
2
1
Work of the force of gravity:
 
2
1
1 2
2 1
y
y
U W dy
W y y W y
  
     

• Work done by weight = product of weight
W and vertical displacement y
• Work done by weight is +ve when y < 0,
i.e. when body moves down
x y
dU F dx F dy W dy
   

5
Work of a Force
• Magnitude of the force exerted by a spring is
proportional to deflection,
 
lb/in.
or
N/m
constant
spring


k
kx
F
• Work of the force exerted by spring,
2
2
2
1
2
1
2
1
2
1
2
1
kx
kx
dx
kx
U
dx
kx
dx
F
dU
x
x










• Work of the force exerted by spring is positive
when x2 < x1, i.e., when the spring is returning to
its undeformed position.
• Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against x,
  x
F
F
U 



 2
1
2
1
2
1

6
Work of a Force
Forces that do no work:
• Weight of a body when its center of gravity
moves horizontally.
• Reaction at a roller moving along its track
• Reaction at frictionless surface when body in
contact moves along surface
• Reaction at frictionless pin supporting rotating body

7
Particle Kinetic Energy: Principle of Work & Energy
dv
mv
ds
F
ds
dv
mv
dt
ds
ds
dv
m
dt
dv
m
ma
F
t
t
t





• Consider a particle of mass m acted upon by force F

• Integrating from A1 to A2 ,
energy
kinetic
mv
T
T
T
U
mv
mv
dv
v
m
ds
F
v
v
s
s
t










2
2
1
1
2
2
1
2
1
2
1
2
2
2
1
2
1
2
1
work of = change in kinetic energy
F

• Units of work and kinetic energy are the same:
J
m
N
m
s
m
kg
s
m
kg 2
2
2
2
1 
















 mv
T
1 1 2 2
T U T

  Principle of Work & Energy

8
Applications of the Principle of Work and Energy
Determine velocity of pendulum bob at
A2. Consider work & kinetic energy.
Force acts normal to path and
does no work:
P

1 1 2 2
2
2
2
1
0
2
2
T U T
mgl mv
v gl

 
 

• Velocity found without determining acceleration and integrating.
• All quantities are scalars
• Forces which do no work are eliminated

9
Applications of the Principle of Work and Energy
• Principle of work and energy cannot be
applied to directly determine the acceleration
of the pendulum bob.
• Calculating the tension in the cord requires
supplementing the method of work and energy
with an application of Newton’s second law.
• As the bob passes through A2 ,
2
2
2
3
n n
F m a
v
P mg m
l
gl
P mg m mg
l

 
  

gl
v 2
2 
W=mg

10
Power and Efficiency
• rate at which work is done.
v
F
dt
r
d
F
dt
dU
Power










• Dimensions of power are work/time or force*velocity.
Units for power are
W
746
s
lb
ft
550
hp
1
or
s
m
N
1
s
J
1
(watt)
W
1 





•
input
power
output
power
input work
k
output wor
efficiency





11
Sample Problem 13.1
A car weighing 1814.4 kg is driven down a 5o incline at a
speed of 96.54 km/h when the brakes are applied causing a
constant total braking force of 6672 N.
Determine the distance traveled by the car as it comes to a
stop. Use the principle of work and energy.

12
Sample Problem 13.1
1 96.54 / 26.817 /
v km h m s
 
 
1 1 2 0
652398 5120.7 0
T U
x

 
 
127.4
x m

   
 
1 2 6672 sin5
5120.7
U x mg x
x
    
 
0
0 2
2 
 T
v
v1=96.54 km/h
v2=0
1 1 2 2
T U T

 
6672 N
N
mg
  
2
2
1 1
1 1
2 2 1814.4 26.817
652398
T mv
Nm
 


13
Sample Problem 13.2
Two blocks are joined by an inextensible cable as shown. If the
system is released from rest, determine the velocity of block A
after it has moved 2 m. Assume that the coefficient of friction
between block A and the plane is mk = 0.25 and that the pulley
is weightless and frictionless. Apply the principle of work and
energy.

14
1 1 2 2
T U T

 
1 0
T 
 
 
1 2
0.25 200 300 9.81 2
4905
A B
A B
U m gx m gx
m m gx
Nm
m
m
   
  
    

x
x
Blocks move by equal amounts
A
m g
m
B
m g
 
2 2
1 1
2 2 2
2 2
1
2 250
A B
A B
T m v m v
m m v v
 
  
2
4905 250
4.43 /
v
v m s


READ SOLUTION IN BOOK

15
Sample Problem 13.3
A spring is used to stop a 60 kg package which is sliding on a
horizontal surface. The spring has a constant k = 20 kN/m and
is held by cables so that it is initially compressed 120 mm. The
package has a velocity of 2.5 m/s in the position shown and the
maximum deflection of the spring is 40 mm.
Determine (a) the coefficient of kinetic friction between the
package and surface and (b) the velocity of the package as it
passes again through the position shown.

16
Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
   0
J
5
.
187
s
m
5
.
2
kg
60 2
2
2
1
2
1
2
1
1 


 T
mv
T
 
      k
k
k
f x
W
U
m
m
m
J
377
m
640
.
0
s
m
81
.
9
kg
60 2
2
1







  
    
   
   J
0
.
112
m
040
.
0
N
3200
N
2400
N
3200
m
160
.
0
m
kN
20
N
2400
m
120
.
0
m
kN
20
2
1
max
min
2
1
2
1
0
max
0
min

















 x
P
P
U
x
x
k
P
kx
P
e
      J
112
J
377
2
1
2
1
2
1 



 

 k
e
f U
U
U m
  0
J
112
J
377
-
J
5
.
187
:
2
2
1
1



 
k
T
U
T
m 20
.
0

k
m

17
Sample Problem 13.3
• Apply the principle of work and energy for the rebound
of the package.
  2
3
2
1
2
3
2
1
3
2 kg
60
0 v
mv
T
T 


     
J
36.5
J
112
J
377
3
2
3
2
3
2






 

 k
e
f U
U
U m
  2
3
2
1
3
3
2
2
kg
60
J
5
.
36
0
:
v
T
U
T



 
s
m
103
.
1
3 
v

18
Sample Problem 13.4
A 2000 lb car starts from rest at point 1 and moves without
friction down the track shown. Determine:
a) the force exerted by the track on the car at point 2, and
b) the minimum safe value of the radius of curvature at point 3.

19
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
 
 
     s
ft
8
.
50
ft
2
.
32
ft
40
2
ft
40
2
2
1
ft
40
0
:
ft
40
2
1
0
2
2
2
2
2
2
2
2
1
1
2
1
2
2
2
2
2
1
2
1














v
s
g
v
v
g
W
W
T
U
T
W
U
v
g
W
mv
T
T
• Apply Newton’s second law to find normal force by
the track at point 2.
:
n
n a
m
F 

 
 
W
N
g
g
W
v
g
W
a
m
N
W n
5
ft
20
ft
40
2
2
2
2







lb
10000

N

20
Sample Problem 13.4
• Apply principle of work and energy to determine
velocity at point 3.
 
     s
ft
1
.
40
s
ft
2
.
32
ft
25
2
ft
25
2
2
1
ft
25
0
3
2
3
2
3
3
3
1
1






 
v
g
v
v
g
W
W
T
U
T
• Apply Newton’s second law to find minimum radius of
curvature at point 3 such that a positive normal force is
exerted by the track.
:
n
n a
m
F 

 
 
3
3
2
3 ft
25
2


g
g
W
v
g
W
a
m
W n



ft
50
3 


21
Sample Problem 13.5
D has a weight of 600 lb, and C
weighs 800 lb. Determine the
power delivered by the electric
motor M when D:
(a) is moving up at a constant
speed of 8 ft/s
(b) has an instantaneous
velocity of 8 ft/s and an
acceleration of 2.5 ft/s2, both
directed upwards.

22
Sample Problem 13.5
• In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
  
s
lb
ft
1600
s
ft
8
lb
200



 D
Fv
Power
  hp
91
.
2
s
lb
ft
550
hp
1
s
lb
ft
1600 



Power
Free-body C:
:
0


  y
F lb
400
0
lb
800
2 

 T
T
Free-body D:
:
0


  y
F
lb
200
lb
400
lb
600
lb
600
0
lb
600








T
F
T
F

23
Sample Problem 13.5
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.





 2
2
1
2
s
ft
25
.
1
s
ft
5
.
2 D
C
D a
a
a
Free-body C:
:
C
C
y a
m
F 

    lb
5
.
384
25
.
1
2
.
32
800
2
800 

 T
T
Free-body D:
:
D
D
y a
m
F 

   
lb
1
.
262
6
.
46
600
5
.
384
5
.
2
2
.
32
600
600







F
F
T
F
   s
lb
ft
2097
s
ft
8
lb
1
.
262 


 D
Fv
Power
  hp
81
.
3
s
lb
ft
550
hp
1
s
lb
ft
2097 



Power

24
Potential Energy
2
1
2
1 y
W
y
W
U 


• Work of the force of gravity :
W

• Work is independent of path followed; depends
only on the initial and final values of Wy.

 Wy
Vg
potential energy of the body with respect
to force of gravity.
   2
1
2
1 g
g V
V
U 


• Units of work and potential energy are the same:
J
m
N 


Wy
Vg
• Choice of datum from which the elevation y is
measured is arbitrary.

25
Potential Energy
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
2
2
2
1
2
1
2
1
2
1 kx
kx
U 


• The potential energy of the body with respect
to the elastic force,
   2
1
2
1
2
2
1
e
e
e
V
V
U
kx
V




• Note that the preceding expression for Ve is
valid only if the deflection of the spring is
measured from its undeformed position.

26
Conservative Forces
1 2
U 
   
1 2 1 1 2 2
, ,
U V x y V x y
  
A force is conservative if its work is independent of the
path followed by the particle as it moves from A1 to A2.
Gravity forces and elastic spring forces are conservative.
Friction forces are non-conservative.

27
Conservation of Energy
• Work of a conservative force:
2
1
2
1 V
V
U 


• Principle of work and energy:
1
2
2
1 T
T
U 


• It follows that:
constant
2
2
1
1






V
T
E
V
T
V
T
• When a particle moves under the action of
conservative forces, the total mechanical
energy is constant.


W
V
T
W
V
T




1
1
1
1 0
 



W
V
T
V
W
g
g
W
mv
T






2
2
2
2
2
2
1
2 0
2
2
1
• Friction forces are not conservative. Total
mechanical energy of a system involving
friction decreases.
• Mechanical energy is dissipated by friction
into thermal energy. Total energy is constant.

28
Conservation of Energy
• Particle moving along path.
• In the absence of friction, total mechanical energy is constant.
• Potential energy depends only on elevation.
• Particle speed is the same at any given elevation (A, A', A")

29
Sample Problem 13.6
A 20 lb collar slides without friction along a vertical rod. The
spring attached to the collar has an undeflected length of 4 in.
and a constant of 3 lb/in.
If the collar is released from rest at position 1, determine its
velocity after it has moved 6 in. to position 2.

30
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1:   
0
lb
ft
2
0
lb
in.
24
lb
in.
24
in.
4
in.
8
in.
lb
3
1
1
2
2
1
2
1
2
1













T
V
V
V
kx
V
g
e
e
Position 2:   
  
2
2
2
2
2
2
2
1
2
2
2
2
1
2
2
2
1
311
.
0
2
.
32
20
2
1
lb
ft
5
.
5
lb
in.
66
120
54
lb
in.
120
in.
6
lb
20
lb
in.
54
in.
4
in.
0
1
in.
lb
3
v
v
mv
T
V
V
V
Wy
V
kx
V
g
e
g
e
























Conservation of Energy:
lb
ft
5
.
5
0.311
lb
ft
2
0 2
2
2
2
1
1








v
V
T
V
T

 s
ft
91
.
4
2
v

31
Sample Problem 13.7
The 0.5 lb pellet is pushed against the
spring and released from rest at A.
Neglecting friction, determine the
smallest deflection of the spring for which
the pellet will travel around the loop and
remain in contact with the loop at all
times.
• For pellet to remain in contact, force
exerted on pellet must be greater
than or equal to zero.
• Set force exerted by loop on pellet to
zero & solve for the minimum
velocity at D.
• Apply principle of conservation of
energy between A & D.

32
Sample Problem 13.7
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
:
n
n ma
F 

 
   2
2
2
2
s
ft
4
.
64
s
ft
32.2
ft
2 




rg
v
r
v
m
mg
ma
W
D
D
n
• Apply the principle of conservation of energy between
points A and D.
 
0
18
ft
lb
36
0
1
2
2
2
1
2
2
1
1







T
x
x
kx
V
V
V g
e
  
  lb
ft
5
.
0
s
ft
4
.
64
s
ft
2
.
32
lb
5
.
0
2
1
lb
ft
2
ft
4
lb
5
.
0
0
2
2
2
2
2
1
2
2











D
g
e
mv
T
Wy
V
V
V
2
5
.
0
18
0 2
2
2
1
1






x
V
T
V
T
in.
47
.
4
ft
3727
.
0 

x

33
Principle of Impulse and Momentum
• Newton’s second law:
  
 v
m
v
m
dt
d
F



linear momentum
2
2
1
1
2
1 force
the
of
impulse
2
1
v
m
v
m
F
dt
F
t
t











Imp
Imp
• Final momentum = initial momentum +
impulse of force during time interval
• Units of impulse is [Ns]=[kg m/s]
 
1
2
2
1
v
m
v
m
dt
F
v
m
d
dt
F
t
t










34
2
1
1 2
t
t
mv Fdt mv
 

Principle of impulse & momentum:
Can be resolved
into components:
       
2 2
1 1
1 2 1 2
,
t t
x x x y y y
t t
mv F dt mv mv F dt mv
   
 
2
1
1 2
t
t
mv Fdt mv
 

When several forces act on a particle:
When more than 1 particle is involved, we can either consider
each particle separately, or we can add the momentum and
impulse for the entire system of particles: 2
1
1 2
t
t
mv Fdt mv
 
  

Note: forces of action & reaction exerted by the particles on
each other have impulses that cancel out. Only the impulses of
external forces need to be considered.

35
If sum of external forces = 0 1 2
mv mv

 
Total momentum is conserved
1 2
0 A A B B
mv mv
m v m v

 
 
Boats move in opposite directions
A B
B A
v m
v m
 
2
1
1 2
t
t
mv Fdt mv
 
  

For entire system of particles:

36
Impulsive Motion
2
1 v
m
t
F
v
m





 
Impulsive force: a force that acts on a particle during a very short
time interval, but is large enough to cause a significant change in
momentum
When impulsive forces act on a tennis ball:
Non-impulsive forces: forces for which is small, and
therefore can be neglected (such as weight)
F t


37
Sample Problem 13.10
A car weighing 1814.4 kg is driven down a 5° incline at a
speed of 96.54 km/h when the brakes are applied causing a
constant total breaking force of 6672 N.
Determine the time required for the car to come to a stop.
Use the principle of impulse and momentum.

38
Principle of impulse
and momentum:
Taking components parallel to the incline:
 
    
1 sin5 0
1814.4 26.817 1814.4 9.81sin5 6672 0
mv mg t Ft
t t
   
    
9.5 s
t 
2
1
1 2
t
t
mv Fdt mv
 


39
A 113.4 g baseball is pitched with a velocity of 24.4 m/s. After
the ball is hit by the bat, it has a velocity of 36.6 m/s in the
direction shown. If the bat and ball are in contact for 0.015 s,
determine the average impulsive force exerted on the ball
during the impact. Use the principle of impulse and momentum
24.4 m/s
36.6 m/s

40
Principle of impulse and momentum:
     
1 2 cos40
0.1134 24.4 0.015 0.1134 36.6 cos40
396.4 N
x
x
x
mv F t mv
F
F
    
   

   
2
0 sin40
0 0.015 0.1134 36.6 sin40
177.9 N
y
y
y
F t mv
F
F
   
  

   
396.4 177.9
434.5 N
F i j
F
 

1 2
mv F t mv
  
x-direction:
24.4 m/s
36.6 m/s
y-direction:
396.4 N
177.9 N
434.5 N
24.2°

41
A 10 kg package drops from a chute into a 24 kg cart with a
velocity of 3 m/s. Knowing that the cart is initially at rest and
can roll freely, determine (a) the final velocity of the cart, (b)
the impulse exerted by the cart on the package, and (c) the
fraction of the initial energy lost in the impact. Use the principle
of impulse and momentum.
Principle of impulse and
momentum:
2
1
1 2
t
t
mv Fdt mv
 


42
Apply the principle of impulse and momentum to the package-cart system:
  2
2
1
1 v
m
m
v
m c
p
p




  
Imp
m/s
742
.
0
2 
v
1 2
mv F t mv
  
 
x-direction:  
    
1 2
2
cos30 0
10 3 cos30 10 25
p p c
m v m m v
v
   
  

43
Apply the same principle to the package alone to determine the impulse
exerted on it from the change in its momentum.
1 2
p p
m v F t m v
  

x direction:
    
1 2
2
cos30
10 3 cos30 10
p x p
x
m v F t m v
F t v
   
    s
N
56
.
18 


t
Fx
y components:
  
1 sin30 0
10 3 sin30 0
p y
y
m v F t
F t
    
     s
N
15 

t
Fy
   
1 2 18.56 15
23.9 N s
F t i j
F t
     
  
Imp

44
To determine the fraction of energy lost, calculate initial and
final energies:
  
    
2
2
1 1
1 1
2 2
2
2
1 1
2 2
2 2
10 3 45 J
10 25 0.742 9.63 J
p
p c
T m v
T m m v
  
    
1 2
1
45 9.63
0.786
45
T T
T
 
 
Initial energy:
Final energy:

45
• Impact: Collision between two bodies which occurs
during a small time interval and during which the
bodies exert large forces on each other.
• Line of Impact: Common normal to the surfaces
in contact during impact.
Impact
• Central Impact: Impact for which the mass
centers of the two bodies lie on the line of
impact; otherwise, it is an eccentric impact.
• Direct Impact: Impact for which the velocities of
the two bodies are directed along the line of
impact.
• Oblique Impact: Impact for which one or both of
the bodies move along a line other than the line
of impact.

46
Direct Central Impact
• Bodies moving in the same straight line:
• Upon impact the bodies undergo a
period of deformation, at the end of which,
they are in contact and moving at a
common velocity.
• A period of restitution follows during
which the bodies either regain their
original shape or remain permanently
deformed.
• Wish to determine the final velocities of the
two bodies. The total momentum of the
two body system is preserved,
A A B B A A B B
m v m v m v m v
 
  
• A second relation between the final
velocities is required.
A B
v v


47
Direct Central Impact
1
0 









e
u
v
v
u
Pdt
Rdt
n
restitutio
of
t
coefficien
e
A
A
• A similar analysis of particle B yields
B
B
v
u
u
v
e




• Combining the relations leads to the desired
second relation between the final velocities.
 
B
A
A
B v
v
e
v
v 




• Perfectly plastic impact, e = 0: v
v
v A
B 



  v
m
m
v
m
v
m B
A
B
B
A
A 



• Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.
B
A
A
B v
v
v
v 




• Period of deformation: u
m
Pdt
v
m A
A
A 
 
• Period of restitution: A
A
A v
m
Rdt
u
m 

 
e depends on materials, impact velocity, shape & size of colliding surfaces.

48
Coefficient of Restitution
• Plastic impact (e = 0): In a plastic impact, the
relative separation velocity is zero. The particles
stick together and move with a common velocity
after the impact.
• Elastic impact (e = 1): In a perfectly elastic collision, no
energy is lost and the relative separation velocity equals the
relative approach velocity of the particles. In practical
situations, this condition cannot be achieved.
In general, e has a value between 0 and 1. The two limiting
conditions are:
Some typical values of e are:
Steel on steel: 0.5 – 0.8 Wood on wood: 0.4 – 0.6
Lead on lead: 0.12 – 0.18 Glass on glass: 0.93 – 0.95
 
B
A
A
B v
v
e
v
v 





49
Coefficient of Restitution
The quality of a tennis ball is
measured by the height of its
bounce. This can be quantified
by the coefficient of restitution.
If the height from which the ball is dropped and the height of
its resulting bounce are known, we can determine the
coefficient of restitution. How?
 
B
A
A
B v
v
e
v
v 




During a collision, some of the initial kinetic energy will be lost
in the form of heat, sound, or due to localized deformation.

50
Problems Involving Energy and Momentum
• Three methods for the analysis of kinetics problems:
- Direct application of Newton’s second law
- Method of work and energy
- Method of impulse and momentum
• Select the method best suited for the problem or part of a
problem under consideration.

51
A 20-Mg railroad car moving at a speed of 0.5 m/s to the
right collides with a 35-Mg car which is at rest. If after
collision the 35-Mg car is observed to move to the right at a
speed of 0.3 m/s, determine the coefficient of restitution.

52
A A B B A A B B
m v m v m v m v
 
  
0.3 ( 0.025)
0.65
0.5
B A
A B
v v
e
v v
 
  
  

       
20000 0.5 20000 35000 0.3
A
v
 
0.025 /
A
v m s
  

53
A 30 kg block is dropped from a height of 2 m onto the the
10 kg pan of a spring scale. Assuming the impact to be
perfectly plastic, determine the maximum deflection of the
pan. The constant of the spring is k = 20 kN/m.

55
• Apply principle of conservation of energy to
determine velocity of the block at instant of impact.
   
    
     s
m
26
.
6
0
30
J
588
0
0
30
J
588
2
81
.
9
30
0
2
2
2
2
1
2
2
1
1
2
2
2
2
1
2
2
2
1
2
1
1














A
A
A
A
A
A
v
v
V
T
V
T
V
v
v
m
T
y
W
V
T
• Determine velocity after impact from conservation of
momentum.
     
     s
m
70
.
4
10
30
0
26
.
6
30 3
3
3
2
2







v
v
v
m
m
v
m
v
m B
A
B
B
A
A

56
Initial spring deflection due to
pan weight:
   m
10
91
.
4
10
20
81
.
9
10 3
3
3






k
W
x B
• Apply principle of conservation of energy to
determine maximum deflection of spring.
    
  
  
   
    2
4
3
2
1
3
4
2
4
3
2
1
3
4
2
4
2
1
4
4
2
3
3
2
1
2
3
2
1
3
2
2
1
2
3
2
1
3
10
20
10
91
.
4
392
10
20
392
0
J
241
.
0
10
91
.
4
10
20
0
J
442
7
.
4
10
30
x
x
x
x
x
kx
h
W
W
V
V
V
T
kx
V
V
V
v
m
m
T
B
A
e
g
e
g
B
A

































   
m
230
.
0
10
20
10
91
.
4
392
0
241
.
0
442
4
2
4
3
2
1
3
4
4
4
3
3












x
x
x
V
T
V
T
m
10
91
.
4
m
230
.
0 3
3
4





 x
x
h m
225
.
0

h

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