Download Kinetics of Particle - Work and Energy Approach

Topics in textbook
 Actual Work
 Definition of Work

Work on Particle
 Potential Energy
 Definition of Energy

Work-Energy Equation
Potential Energy
 Kinetics Energy
 Definition of Energs
1
Three approaches for
solving dynamics
1) Direct Method
1
m
From 2nd Law
(kinetics Eq)



  Fi   dr  m(a  dr )
 i

i
Fi  Fi (a, v , s , t )
Kinematics Eq:
a (v , s , t )  v , s
F2
path
dr

F  a
A
F3
F1
work (and potential energy)
of Force i along the path
2) Work and Energy
From 2nd Law
 F  ma
Newton’s
2nd Law


i   Fi  dr    m(a  dr )
s

 m(a  dr )   m(v dv )
t
t
1
1
mv2 2  mv2 2
2
2
(change of kinetics energy)



F
s  i i   dr  s m(a  dr )
3) Impulse and Momentum
From
2nd
Law


F
s  i i   dt  s m(a dt )

i
 m(a ) dt   m(dv )
  d (mv )  mv  mv
F
dt

m
(
a
)
dt
 i

linear impulse of Force i

2
(linear momentum)
1
2
 F  ma
Newton’s
Work and Energy
From 2nd Law
2nd Law
Usually convenient
when F = F(s), and
you want to find
velocity at final state
(without finding acc.
first).
change of
kinetic energy


i   Fi  dr    m(a  dr )
s

def
Ui 
 F  dr
t
  m(v t dvt )
s
Work of Force i along
the path

Fi
dr
  Fi cos i  ds
 Fi (ds)cos i 
dU1  0
dU 2  0
dU 3  0
1
1
mvB 2  mvA 2
2
2
 TB  TA  TB  A
during small displacement
dU i  Fi
kinetic
energy
at B
 m(a  dr )   m(a ds)
i
Work of a force
i
T
1 2
mv
2
kinetic energy
F2
path
1
kinetic
energy
at A
dr
A
B
F3
F1
Principle of work and Energy
U AB  T  TB  TA
3
Work done on Particle
dU1  0
Work by a force Fi
Fi
dUi  Fi  dr
(dr ) cos i
i
F3
F1
path
 (Fi ) (dr ) cosi
dr
( F ) cos i
dU 2  0
dU 3  0
 (Fi ) cosi (dr )
P
F2
(inactive force)
Work done over particle A
 dU   ( F
i
sum of works
done by all forces
over the particle A
i
 dr )  ( Fi )  dr
Since  F  ma ,
the total work done on object is
5
……
Note on work
F2
path
dr
dUi  Fi  dr
F3
F1
 dU i is positive when
 Fi t  F (cosi ) and s
has the same direction.
 Unit of work is N-m or Joule (J).
 Active force is the force that does the work
 Reactive force = constrain force that does not do the work
6
B
The 10-kg block rest on a smooth incline.
If the spring is originally stretched 0.5 m,
determine the total work done by all forces
acting on the block when a horizontal
force P = 400 N pushes the block up the
plane s = 2 m. The block is not tipping.
A (spring stretched
length is 0.5m)
Pos B
Horizontal Force P: constant
Pos A
U P  ( P cos 30 )(2)  692.8 J
o
Weight W: constant.
UW    mg  (2sin 30 )  98.1J
“Active Force”
Spring Force Fs.: varying Force
2
2
U Fs    ( Fs )(dx)    k ( x  0.5)dx   1 k ( x  0.5) 2   90 J
A
 2

0
0
B
Normal Force NB :
constant
U NB  0
“Inactive Force”
U Overall  692.8  98.1  90
 504.7 J
Work done on Particle
t
Work done over particle A
F4
 dU   ( F
path
i
dr
F3
a
F1
 (ma )  dr
F2
B
B
B
A
A
A
U A B     dU    m  a  dr    m  at ds 
Pos B
TB
Pos A
 dr )
 ( Fi )  dr
sum of works
done by all forces
over the particle A
P
i
B
B
TA
Work done on particle P
during path A->B,
is to increase kinetic
energy of particle
1
  m  vdv    mv 2   1 mvB2  1 mv A2
 2

2
2
A
A
def
1
T  mv 2
2
TB
TA
Kinetics Energy
1 2
T  mv
2
path
F2
A
 T is the work done on a particle to
accelerate it from rest to the velocity v
F3
F1
Principle of work and Energy
 Unit of T is N-m or Joule (J)
U AB  TB  TA
Advantage
( TA  U AB  TB )
 Scalar equation. (1 unknown)
 Integral Equation
(not instantaneous eq like 2nd Law)
 No need to find acceleration first
 Get change in velocity directly from active forces.
 it can be applied to system of particles with
frictionless and non-deformable links
12
15
18
How to calculate Work
F2
dr
In general
F3
U i   dU i   Fi  dr
s
s
In xyz-coord
Ui  
path
1
Fi
F   Fi  x iˆ   Fi  y ˆj   Fi  z kˆ
dr  (dx)iˆ  (dy ) ˆj  (dz )kˆ
 Fi x dx   Fi  y dy   Fi z dz
make sure that the direction of  Fi  x
In nt-coord
Ui  
scalar
(be careful
Of +/-)
 Fi  y  Fi  z
is positive according to  x  y  z direction.
 Fi t ds
In r-coord
Ui  
dUi  Fi  dr
  Fi cos i  ds
displacement x
force component
in the direction of
displacement
 Fi (ds)cos i 
displacement in the
direction of force
x force component
make sure that the direction of  Fi t
is positive according to +s direction.
F  Fr eˆr  F eˆ
 Fr  dr   F  rd
dr  (dr )eˆr  (rd )eˆ
19
M3/107) Calculate the work done on 10-kg object with the
constant Force ( F= 8N ) during the curve path AB.
F  8iˆ (const)
x-y
dr   dx  iˆ   dy  ˆj
U F , A B   F  dr 
 
0.75
 8 dx  8(0.75) J
0
mg   mg   ˆj
U mg , A B    mg   dr
y
0

  mg  dy
0.375
 mg (0.375)
x
mg
F=8N (const)
N
U mg , A B  (mg )(0.375)  36.7875 J
U N , A B  0
Does not do the work
U A B  U F , A B  U mg , A B  U N , A B
 42.7875 J
If F is not constant, how to calculate it?
Ans
If F is not constant
U F , A B   F  dr
F  ( x  y )iˆ  ( x 2  y 2 ) ˆj
B
  ( x  y ) dx  ( x 2  y 2 ) dy
A
y
B
  ( x  (0.375  0.667 x 2 )dx
y  0.375  0.667 x
A
2
 0.375  y  2 
  
  y  dy
0.667


A 
B
x
More general case
F  Fx ( x, y)iˆ  Fy ( x, y) ˆj
U F , A B   F  dr
B
  Fx dx  Fy dy
A
B
   Fx ( x, f ( x)) dx  Fy ( x, f ( x)) dy

A
y  f ( x)
Fy ( f 1 ( y), f ( y))dy
or Fy ( x, f ( x))
dy
dx
21
dx
M3/107) Calculate the work done by F during the curve path AB.
n-t
s
F (s)  Ft (s)eˆt  Fn ( s)eˆn
Fn ( s )
Fn doest
not effect
works!
Ft  2s cos 
Fn  2s sin 
engine thrust
F (t )  2 s

U F , A B   F  dr   Ft ( s)ds
  2 s cos  ds
M3/107) Calculate the work done by F during the curve path AB.
r-
 FC eˆr
F  F
FCC eˆr
U F , A B   F  dr
rA
rB
F (const)
r- reference point
d 0
(dr )eˆr  rd eˆ
central force
    Feˆr  0eˆ  
 (dr )eˆr  (rd )eˆ 
B
    FC  dr   FC
A
r  reˆr
  FC  rB  rA 
 
dr   dr  eˆr  r  d  eˆ     dr  eˆr  r eˆ
B
 dr
A
 FC (rA  rB )
23
M3/121) The 0.2-kg slider moves freely along the fixed curved rod from A to B
in the vertical plane under the action of the constant 5-N tension in the cord. If
the slider is released from rest at A, calculate its velocity v as it reaches B.
mg
F
Work-Energy Eq.
0
general
position
U AB  12 mv
TB  T A
N
2
B
Does not
do the work
eˆr
F
r- coordinate
reference point
U mg   mg (0.25)  0.4905 J
B
U F    Fdr
A
 F (rA  rB )
 F
rB
 dr   F (r
B
 rA )
rA
 5( 0.62  0.252  0.15)  2.5
vB 

2U A B
m
2
 2.0095  4.482745
0.2
What does it mean
if U A B  0?
U A B  U F  U mg  U N
0
 2.5   0.4905  2.0095 J
25
Work on frictionless connected particles
Only the external
forces are needed to
calculate the total
work on a system of
particles.
(If frictions exist, the
sum of action and
reaction of the
friction may not be
zero.)
 internal force R and –R will have the same displacement.
 So, the sum of these works are zero.
33
The system starts from rest at
Configuration 1. Find the velocity of A
at configuration where d = 0.5 m . F is
20 N (constant)
initial state
mA g
mA g
mB g
mB g
U A 12  TA
U B 12  TB
N causes
no work!
Final state
0.5
Fd  mA gd 

Tdy  TA
0
0.5
mB gd 

Tdy  TB
0
System selection is not so good
(you have to calculate Tension T for its work)
U sys
mA g
mB g
12
 TA
1
1
Fd  mA gd  mB gd  TA  mAv 2  mB v 2
2
2
34
M3/131) The ball is released from position A with a velocity of 3m/s and
swings in a vertical plane. At the bottom position, the cord strikes the fixed bar
at B, and the ball continues to swing in the dashed arc. Calcuate the velocity
v of the ball as it passes position C.
Work-Energy Eq.
U AC  12TmvC  12TmvA
2
C
T
U AC  U mg 
AC
2
A
 UT  AC
does
no work
system
U mg  mg (hA  hB )  mg (1.2cos 60o  0.8)
 mg (0.2)  mg (0.2)
mg
vC 
2
1 2
2
U

mv

12.924

3.595
m/s
A
 mg
m
2

35
F2
Power
path
 Power is defined as time rate of work
d F r 
dU
P

 F v
dt
dt
dr
A
F3
F1
(scalar quantity)

For a machine, power tells how
much work it can do in a period of
time.
(small machine can deliver lots
of energy given enough time)
 Unit of power: Watt (W) = J/s = N-m/s
37
Mechanical Efficiency
 Mechanical Efficiency
useful power output

power input
energy output

energy input
1
If energy applied to the
machine occurs during the
same time interval at which
it is removed.
Poutput
 F v
 Since machine consists of moving
parts which may have frictions, so extra
energy or power is needed to overcome
the frictions.
38
A car has a mass of 2 Mg and an engine efficiency of  = 0.65. The car
uniformly accelerates at 5 m/s2, starting from rest. During that
constant acceleration, the wind outside creates a drag resistance on the car
of FD = 1.2v2 N, where v is the velocity in m/s. Find the engine output
input when t=4 s.
a

Poutput
Pinput
Poutput  F  v
Poutput
t 4
  F t 4  v t 4 
x
 10480  20  209.6 kW
Pinput
t 4

Poutput


209.6
 322.46 kW
0.65
[  :  Fx  max ] FC  1.2v  ma
2
Constant
acceleration:
FC  ma  1.2v 2
 10480 N
a  5 (constant)
v  at
v t 4  (5)(4)  20 m/s
39
A 50-N load (B) is lifted up by the motor from rest until the distance is
10 m. The motor M has an efficiency of 0.76 and exerts a constant force
of 30 N. Find the power supplied to the motor at that instant. Neglect
the mass of the pulleys and cable.
P

v=?
sP
F = 30 N
(const)
output
Pinput 
Pinput
LM  2sP  C
50N
s = 10 m
(start
from rest)

dsP
d
vM  LM  2
 2vP  2vB
dt
dt
Energy
  F  ma 
Approach
B
 TBma
TA
2UFAB 50
1
(2 F2)(Fs)50mgs  mv 2
a
 9.81
2
v
50 N
375.6

 494.2 W
0.76
 (30)(2  6.26)  375.6 W
Poutput  F  v
2F= 2(30)
Poutput
5029.81
 2 Fs  mgs 
v  u  2as
2
2
m
 6.26
v  2(9.81)10
Work and Energy
kinetic
energy
at B
1
 1

2

mv

mvA2 
B

2
 2

U AB  TB  TA
F1
kinetic
energy
at A
path
dr
B
A
F2
F3
summation of all forces
U F1 , A B
Work from all
other forces
(not spring &
gravitation)
U F2 , A B
Work from
Gravity Force
Gravitational
Potential
Energy
U F3 , A B
Work from
spring
Elastic
Potential
Energy
We found that ….
It is much easier to solve dynamic
problem, if we think the work done
by spring and gravity force in the
form of Potential Energy
42
Work of Gravity Force
dUW  W (dh)
>0
UW ,12  mg (h1  h2 )
Vg ,1
Only depends on
position at
final state (2)
Only depends on
position at
initial state (2)
any path
1
energy level (higher)
W=mg
Vg ,2
2
h1
h
 mg (h2  h1 )  (mgh2  mgh1 )
Work done by W , only depends on the initial
state position and final state position only,
i.e. , it does not depends on actual path
energy level (lower)
h2
Think in Term of “Potential Energy”
(for convenience)
Fixed reference line
UW , anyPath 12   (Vg ,2  Vg ,1 )
Work done by
Gravity Force:
from position 1
to position 2
U  V
Work = “Energy in Transfer”
def
Vg  mgh
point function
Potential Energy
- Energy from gravity field
Vg ,2  Vg ,1 :
Change in Internal Energy: + (increasing)
 V
Energy “Emission”:
from position 1 to position 2
h1  h2  Vg ,1  Vg ,2
when change in g is significant
Define V g as negative of work done from the position to
r 

 GM earth 
F 
m
2

 r

g
GM earth
 Rearth 
2
 9.81
mgR 2
mgR 2
Vg (r )    2 dr  
r
r
r
 the potential energy at r is
mg  Rearth 
Vg  
r
2
 Vg from r1 to r2
Vg  
GM earth
0
Rearth
 mgR 2   mgR 2 
Vg   


r2  
r1 

44
Work of Spring Force
E1
 Lo  L1    Lo  L2 
L1
1
 Ve,1  Ve,2
L
dU Fs  Fs dL
Only depends on
position at
initial state (1)
2
U Fs ,12   k ( Lo  L)dL
1
L2
L2
1

1
1

2
   k ( L  Lo ) 2    
k
(
L

L
)

k ( L1  Lo ) 2 

2
o
2
 L1
2
2

Lo natural length
(unstretched length)
E2
Fs
Only depends on
position at
final state (2)
Work done by Spring , depends only on the
initial state and final state only, i.e. ,
it does not depends on actual path
2
any path
def
Think in Term of “Energy”
(for convenience)
U Fs , anyPath 12
1
Ve  kx 2
2
x : distance , stretched or
compressed from natural
length
point function
  (Ve,2  Ve,1 )  (Ve ,1  Ve ,2 )
Work done by Spring Force:
from position 1 to position 2
U  V
Energy Emission:
from position 1 to position 2
Work-Energy Equation
FBD **
(Use Energy Concept)
FBD
Not
Recommended
Method in this
course
N
N
Work-Energy Equation
Work-Energy Equation
(1st Form)
U AB  T

U AB  U A B  T
*
Virtual work by nonconservative forces.
 U A B  V   VB  VA 

Energy Concept
(2nd Form)
U *AB  V  T
Work-Energy Equation
(1st Form)
U AB  T
FBD
Work-Energy Equation
(2nd Form)
U AB  V  T
*
N
= E
M3/173) The 0.6-kg slider is released from rest at A and slides down
under the influence of its own weight and of the spring of k = 120 N/m.
Determine the speed of the slider and the normal force at point B. The
U
U
0
unstrecthed length of the spring is 200 mm.
*
AB
N , AB
U A*  B  V  T
F
N
y  2x
VA  TA  VB  TB
mg  5.866
2
1
1
1
1
mghA  kxA2  mvA2  mghB  kxB 2  mvB2
2
2
2
2
gravitational
potential datum
hA  0.5
x A  ( 0.52  0.252  0.2)
xB  (0.25  0.2)
vB  5.9234 m/s
At position B
FB  k (0.25  0.2)
man

mg
mat
NB
N B  mg  Fs  m
vB2
an
N B  84.09 N
3
2 2

  df  

1  
 
  dx  


2
d y
dx 2
1  (4 x) 

2
4
3
2

x 0

1
m
4
49
Advantage
 Integral Equation
(not instantaneous equation like 2nd Law)
 Scalar equation. (easy to handle with1 unknown)
 Get change in velocity directly.
(No need to find acceleration first)
 Handle with only active forces.
 it can be applied to system of particles with
frictionless and non-deformable links
We will see this later, when applying at system of particles
57
Work on frictionless connected particles
A
B
C
O
58
initial state
The system starts from rest at
Configuration 1. Find the velocity of A
at configuration where d = 0.5 m . F is
20 N (constant)
Final state
T
T
mA g
T
mB g
T
F
mA g
mB g
F
Object A
0.5
U A* ,12  VA  TA
Fd 

Tdy  VA  TA
0
Object B
Ny
U B* ,12  VA  TA
0.5
  Tdy  VA  TA
0
Nx
*
U sys
,12  Vsys  Tsys
F
Fd  VA  TA
T is internal force
(excluding from Work Calculation)
We have no interest in T, thus object separation
(separating object A and B) is not good in this problem.
60
64
M3/158) If the system is released from rest,
determine the speeds of both masses after B have
move 1 m. Neglect friction and the masses of pulleys.
SA
SB
System: block A + block B + cord+ 2 Pulleys
Position A: at rest
(assume)
up
Position B: block B moves down as 1 meter
U
*
sys , AB
 V  T
2
2
 sB   
3
3
2
hA   sin 20o
3
3vA  2vB  0
3aA  2aB  0
datum
1
 1

mA g  hA   mB g  hB    mAv A2  0    mB vB2  0   0
2
 2

s A  
L  3S A  2S B  C1
sB  1
hB  1
2
v A   vB
3
2
1  2  1
10.9924  mA   vB   mB vB2  0
2  3  2
unsolvable
M3/158) If the system is released from rest,
determine the speeds of both masses after B have
move 1 m. Neglect friction and the masses of pulleys.
SA
SB
System: block A + block B + cord+ 2 Pulleys
Position A: at rest
(assume)
up
Position B: block B moves down as 1 meter
U
*
sys , AB
L  3S A  2S B  C1
3vA  2vB  0
3aA  2aB  0
 V  T
datum
1
 1

mA g  hA   mB g  hB    mAv A2  0    mB vB2  0   0
2
 2

2
2
 sB   
3
3
2
hA   sin 20o
3
s A  
sB  1
hB  1
2
1
 2  1
10.9924  mA   vB   mB vB2  0
2  3  2
2
v A   vB
3
vB2  0.85285
vB  0.85285
vB  0.85285
vA  0.61566
H14/16) Block A rest on a surface which has
friction. Determine the distance d cylinder B
must move down so that A has a speed of
vA  2 m/s starting from rest.
sA
20 N
k  0.3
sB
System: block A + block B + cord+ 2 Pulleys
50 N
Position A: at rest
Position B: block B moves down as d meter
U
*
Sys , AB
 E
*
U Sys
, AB  ( k N )(2d )
1
1
2
E  mA v A  mB vB 2  (mB g )d
2
2
  k N  2d  
1
1
mA (2)2  mB (1) 2  mB g (d )
2
2
1 (4mA  mB )
d
 0.1744 m
2 (mB g  2 k mA g )
L  s A  2 sB
0  sA  2sB
sB : d  sA  2d
vA  2vB  0
vA : 2  vB  1
67
3/168) The system is released from rest with =180, where the uncompressed
spring of stiffness k= 900 N/m is just touch the underside of 4-kg collar.
Determine the angle  corresponding to the maximum spring compression.
O2-1
*
U sys
, AB  V  T
O2-2
O1
r
L
datum
r
VA  TA  VB  TB
 /2
 /2
1
1
m1 g (h1, A )  2m2 g (h2, A )  kx A2  m1 g (h1, B )  2m2 g (h2, B )  kxB2
2
2
 /2
System: O1+O2+O3+4 rods
Position A: at rest with =180
h1, A
 180 
 2(0.2) sin 
  0.4
 2 
h2, A
 180 
 (0.2  0.3) sin 
  0.5
 2 
Position B: maximum compression
 * (v*  0 for all object)
h1  2r sin
h2, B
* 
 (0.2  0.3)sin  
 2


* 
xB   2(0.2) sin    2(0.2) 
 2


 


* 
*  1
 * 
4(9.81)(0.4) 1  sin   (2)(3)(9.81)(0.5) 1  sin    900  0.4 1  sin  
2
2 2
2 


 

2
h2  (r  L) sin
* 
h1, B  2(0.2)sin  
 2

2


* 
 *  45.13
1  sin   0 or 1  sin  
2
2
72


 *  0 or 43.8o
2
68
Power
F2
path
 Power is defined as time rate of work
d F r 
dU
P

 F v
dt
dt
dr
A
F3
F1
(scalar quantity)

For a machine, power tells how much
work it can do in a period of time.
(small machine can deliver lots
of energy given enough time)
 Unit of power: Watt (W) = J/s = N-m/s
70
Mechanical Efficiency
 Mechanical Efficiency
useful power output

power input
energy output

energy input
1
If energy applied to the
machine occurs during the
same time interval at which
it is removed.
Poutput
 F v
 Since machine consists of moving
parts which may have frictions, so extra
energy or power is needed to overcome
the frictions.
71
A 50-N load (B) is lifted up by the motor from rest until the distance is
10 m. The motor M has an efficiency of 0.76 and exerts a constant force
of 30 N. Find the power supplied to the motor at that instant. Neglect
the mass of the pulleys and cable.
v=?

sP
F = 30 N
(const)
Poutput
Pinput 
Pinput
Poutput  F  v

375.6

 494.2 W
0.76
(  0.76 )
Poutput ? (30)(6.26) No!
 (30)(2  6.26)  375.6 W
LM  2sP  C
50N
2F= 2(30)
Poutput
Ny
Nx
s = 10 m
(start from rest)
vM 
Energy
  F  ma 
Approach
F = 30 N
(const)
ds
d
LM  2 P  2vP  2vB
dt
dt
U2A*F
(VBma
VA )  (TB  TA )
 B50
1 2
2
F

50
a( F
 )(2  20) mgs
9.81 2 mv
50
9.81

v
50 N
50 N
v  u  2as
2
2

2 2 Fs  mgs

  6.26
v m 2(9.81)10
Summary

Make sure you write FBD (no FBD, no score)
*
A B
 V  T
U A B  T

U

Scalar Equation (Only 1 unknown)

or
Equation itself is not hard to solve, but calculating
work may be more difficult than you thought.
74
Recommended Problem
M3/155
M3/144
M3/166
M3/160
M3/168
H14/93 , H 14/96
75
77