Transcript Kinetic Energy and Work

Chapter 7
Kinetic Energy and Work
Work and Energy
Energy: scalar quantity associated with a state (or condition)
of one or more objects.
Work and energy are scalars, measured in N·m or Joules, J,
1J=kg∙m2/s2
Energy can exist in many forms - mechanical, electrical,
nuclear, thermal, chemical….
Work and Energy
Energy is a conserved quantity - the total
amount of energy in the universe is constant.
Energy can be converted from one type to
another but never destroyed.
Work and energy concepts can simplify solutions of
mechanical problems - they can be used in an
alternative analysis
Work done by a constant force
Work done on an object by a constant force is
defined to be the product of the magnitude of the
displacement and the component of the force
parallel to the displacement
W = FII · d
Where FII is the component of the force F parallel
to the displacement d
Work
In other words W = F d cosq
Where q is the angle between F and d
F
q
d
If q is > 90o, work is negative. A decelerating car has
negative work done on it by its engine.
The unit of work is called Joule (J), 1 J = 1 N·m
1J=kg∙m2/s2
Work - on and by
A person pushes block 30 m along the ground
by exerting force of 25 N on the trolley. How
much work does the person do on the trolley?
W = F d = 25N x 30m = 750 Nm
Trolley does -750 Nm work on the person
Ftp
Fpt
d
Mechanical Energy
Mechanical energy (energy associated with
masses) can be thought of as having two
components: kinetic and potential
– Kinetic energy is energy of motion
– Potential energy is energy of position
Kinetic Energy
In many situations, energy can be considered as
“the ability to do work”
Energy can exist in different forms
Kinetic energy is the energy associated with the
motion of an object
A moving object can do work on another object
– E.g hammer does work on nail.
Kinetic Energy
Consider an object with mass m moving in a straight
line with initial velocity vi. To accelerate it uniformly to
a speed vf a constant net force F is exerted on it parallel
to motion over a distance d.
v 2f  v i2  2a(x f  x i )  v i2  2ad
a
v 2f  v i2
2d
Work done on object
W=Fd=mad
So
W m
v 2f  v i2
2d
(NII)
d
Kinetic Energy
If we rearrange this we obtain
1 2 1 2
W  mv f  mv i
2
2
We define the quantity ½mv2 to be the translational
kinetic energy (KE) of the object
W = DKE
This is the ‘Work-Energy Theorem’:
“The net work done on an object is equal
to its change in kinetic energy”
The Work-Energy Theorem
W = DKE
• Note
– The net work done on an object is the work
done by the net force.
– Units of energy and work must be the same (J)
Energy: scalar quantity associated with a state (or
condition) of one or more objects.
I. Kinetic energy.
II. Work.
III. Work - Kinetic energy theorem.
IV. Work done by a constant force
- Gravitational force
V. Work done by a variable force.
- Spring force.
1D-Analysis
3D-Analysis
VI.
Power
I. Kinetic energy
Energy associated with the state of motion of an object.
1 2
K  mv
2
Units:
1 Joule = 1J = 1 kgm2/s2
II. Work
Energy transferred “to” or “from” an
object by means of a force acting on the
object.
To +W
From -W
- Constant force:
v 2f
 v02
Fx  max
 2a x d  a x 
v 2f  v02
2d
1
2
2 1
 Fx  max  m(v f  v0 )
2
d
1
 m(v 2f  v02 )  K f  Ki  Fx d  W  Fx d
2
Work done by the force = Energy transfer due to the force.
-To calculate the work done on an object by a force during
a displacement, we use only the force component along
the object’s displacement. The force component perpendicular
to the displacement does zero work.
 
W  Fxd  Fd cos   F  d
-Assumptions: 1) F = constant force
2) Object is particle-like (rigid object, all parts
of the object must move together).
A force does +W when it has a vector component in the
same direction as the displacement, and –W when it has
a vector component in the opposite direction. W=0 when it
has no such vector component.
  90  W
180    90  W
  90  0
Net work done by several forces = Sum of works done by
individual forces.
Calculation: 1) Wnet= W1+W2+W3+…
2) Fnet  Wnet=Fnet d
II. Work-Kinetic Energy Theorem
DK  K f  K i  W
Change in the kinetic energy of the
particle = Net work done on the particle
III. Work done by a constant force
- Gravitational force:
W  mgd cos 
Rising object: W= mgd cos180º = -mgd 
Fg transfers mgd energy from the object’s kinetic energy.
Falling down object: W= mgd cos 0º = +mgd 
Fg transfers mgd energy to the object’s kinetic energy.
External applied force + Gravitational force:
DK  K f  K i  Wa  Wg
Object stationary before and
after the lift: Wa+Wg=0
The applied force transfers the same amount of
energy to the object as the gravitational force
transfers from the object.
IV. Work done by a variable force
- Spring force:


F  kd
Hooke’s law
k = spring constant  measures spring’s stiffness. Units: N/m
Hooke’s law:
1D  Fx  kx
Fx
xi
Δx
xf
x
Fj
Work done by a spring force:
•
•
Assumptions: •
•
Spring is massless  mspring << mblock
Ideal spring  obeys Hooke’s law exactly.
Contact between the block and floor is frictionless.
Block is particle-like.
- Calculation:
1) The block displacement must be divided
into many segments of infinitesimal width, Δx.
2) F(x) ≈ cte within each short Δx segment.
Fx
xi
Δx
xf
x
Fj
Ws   F j Dx  Dx  0 

xf
xi
xf
xf
xi
xi
F dx   (kx) dx  k 
xf
xi
 1  2 2
   k ( x f  xi )
 2 
Ws>0  Block ends up closer to the
relaxed position (x=0) than it was initially.
Ws<0  Block ends up further away from
x=0.
if xi  0
Ws=0  Block ends up at x=0.
1
1
2
Ws  k xi  k x 2f
2
2
1
Ws   k x 2f
2
 
 1  2
x dx    k  x
 2 
Work done by an applied force + spring force:
DK  K f  K i  Wa  Ws
Block stationary before and after the displacement:
Wa= -Ws  The work done by the applied force displacing
the block is the negative of the work done by the spring force.
Work done by a general variable force:
1D-Analysis
DW j  F j , avg Dx
W 
 DW   F
j
j , avg Dx
better approximation  Dx  0
xf
W 
lim F
Dx  0
j , avg Dx
 W 
 F ( x)dx
xi
Geometrically: Work is the area between the curve F(x)
and the x-axis.
3D-Analysis

F  Fxiˆ  Fy ˆj  Fz kˆ ;

dr  dx iˆ  dy ˆj  dz kˆ
Fx  F ( x), Fy  F ( y ), Fz  F ( z )
rf
xf
yf
zf
 
dW  F  dr  Fx dx  Fy dy  Fz dz  W   dW   Fx dx   Fy dy   Fz dz
ri
xi
Work-Kinetic Energy Theorem - Variable force
xf
xf
xi
xi
W   F ( x)dx   ma dx
ma dx  m
dv
dv
dx  m v dx  mvdv
dt
dx
 dv dv dx dv 
 v
 
 dt dx dt dx 
vf
vf
vi
vi
W   mv dv  m  v dv 
1 2 1 2
mv f  mvi  K f  K i  DK
2
2
yi
zi
V. Power
Time rate at which the applied force does work.
- Average power: amount of work done
in an amount of time Δt by a force.
-Instantaneous power:
-instantaneous time rate of doing work.
Pavg
W

Dt
dW
P
dt
Units: 1 watt= 1 W = 1J/s
1 kilowatt-hour = 1000W·h = 1000J/s x 3600s = 3.6 x 106 J
= 3.6 MJ
 
dW F cos  dx
 dx 
P

 F cos     Fv cos   F  v
dt
dt
 dt 
F
φ
x
In the figure (a) below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m across
a frictionless floor. Find an expression for the speed vf at the end of
that distance if the block’s initial velocity is:
(a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that
the block is initially moving at 1m/s to the right, but now the 2N force
is directed downward to the left. Find an expression for the speed of
the block at the end of the 1m distance.
(a)
N
Fy
Fx
Fx
Fy
mg
mg
(a ) v0  0  DK  0.5mv 2f
(2 N )(1m) cos q  0.5(4kg)v
 v f  cos q m / s
 
W  F  d  ( F cos q )d
N
W  DK  0.5m(v 2f  v02 )
(b) v0  1m / s  DK  0.5mv2f  2 J
2
f
(2 N ) cos q  0.5(4kg)v 2f  2 J
 v f  1  cos q m / s
In the figure (a) below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m across
a frictionless floor. Find an expression for the speed vf at the end of
that distance if the block’s initial velocity is:
(a) 0 and (b) 1m/s to the right. (c) The situation in Fig.(b) is similar in that
the block is initially moving at 1m/s to the right, but now the 2N force
is directed downward to the left. Find an expression for the speed of
the block at the end of the 1m distance.
(a)
N
Fy
(b)
Fx
Fx
Fy
mg
mg
(a ) v0  0  DK  0.5mv 2f
(2 N )(1m) cos q  0.5(4kg)v
 v f  cos q m / s
 
W  F  d  ( F cos q )d
N
W  DK  0.5m(v 2f  v02 )
(b) v0  1m / s  DK  0.5mv2f  2 J
2
f
(2 N ) cos q  0.5(4kg)v 2f  2 J
 v f  1  cos q m / s
In the figure (a) below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m across
a frictionless floor. Find an expression for the speed vf at the end of
that distance if the block’s initial velocity is:
(a) 0 and (b) 1m/s to the right. (c) The situation in (b) is similar in that
the block is initially moving at 1m/s to the right, but now the 2N force
is directed downward to the left. Find an expression for the speed of
the block at the end of the 1m distance.
N
Fy
Fx
N
Fx
Fy
mg
mg
 
W  F  d  ( F cos q )d
W  DK  0.5m(v 2f  v02 )
(b) v0  1m / s  DK  0.5mv  2 J
(c) v0  1m / s  DK  0.5mv2f  2 J
(2 N )(1m) cos q  0.5(4kg)v  2 J
 (2 N ) cos q  0.5(4kg)v 2f  2 J
 v f  1  cos q m / s
 v f  1  cos q m / s
2
f
2
f
A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning
at t=0, a steady wind pushes on the lunchbox in the negative
direction of x, Fig. below. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force
from the wind do on the lunch box from t=1s to t=5s?
Motion  concave downward
parabola
xt
1 2
t
10
dx
2
v
 1 t
dt
10
dv
2
a
   0.2m / s 2
dt
10
F  cte  ma  (2kg)( 0.2m / s 2 )  0.4 N
(a) t  1s  v f  0.8m / s
W  F  x  (0.4 N )(t  0.1t 2 )
K f  0.5(2kg)(0.8m / s ) 2  0.64 J
A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning
at t=0, a steady wind pushes on the lunchbox in the negative
direction of x, Fig. below. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force
from the wind do on the lunch box from t=1s to t=5s?
Motion  concave downward
parabola
xt
1 2
t
10
dx
2
 1 t
dt
10
dv
2
a
   0.2m / s 2
dt
10
v
F  cte  ma  (2kg)( 0.2m / s 2 )  0.4 N
W  F  x  (0.4 N )(t  0.1t 2 )
(c) W  DK  K f (5s )  K f (1s )  0  0.64  0.64 J
(b) t  5s  v f  0
K f  0J
In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m
y
up a frictionless ramp. (a) During the
x
displacement, what is the net force done
N
on the book by Fa, the gravitational force
F
on the book and the normal force on the
F
mg
book? (b) If the book has zero kinetic
energy at the start of the displacement,
 
what is the speed at the end of the
N  d W  0
displacement?
Only Fgx , Fax do work
gx
gy
(a ) W  WFa x  WFg x
or
Wnet
 
 Fnet  d
Fnet  Fax  Fgx  20 cos 30  mg sin 30
Wnet  (17.32 N  14.7 N )0.5m  1.31J
N12. In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m
y
up a frictionless ramp. (a) During the
x
displacement, what is the net force done
N
on the book by Fa, the gravitational force
F
on the book and the normal force on the
F
mg
book? (b) If the book has zero kinetic
energy at the start of the displacement,
 
what is the speed at the end of the
N  d W  0
displacement?
Only Fgx , Fax do work
gx
gy
(b) K 0  0  W  DK  K f
W  1.31J  0.5mv 2f  v f  0.93m / s
(a) Estimate the work done represented by the graph below
in displacing the particle from x=1 to x=3m. (b) The curve is given
by F=a/x2, with a=9Nm2. Calculate the work using integration
(a) W  Area under curve 
(11.5squares)(0.5m)(1N )
 5.75 J
3
9
1
1
(b) W   2 dx  9   9(  1)  6 J
3
 x 1
1 x
3
An elevator has a mass of 4500kg and can carry a maximum
load of 1800kg. If the cab is moving upward at full load at 3.8m/s,
what power is required of the force moving the cab to maintain
that speed?
Fa
mtotal  4500kg  1800kg  6300kg

 
 
Fa  mg  0  ma  Fnet
mg
 Fa  mg  (6300kg)(9.8m / s )  61.74kN
2
 
P  F  v  (61.74kN )(3.8m / s )  234.61kW
A single force acts on a body that moves along an x-axis.
The figure below shows the velocity component versus time for
the body. For each of the intervals
v
AB, BC, CD, and DE, give the sign
B
C
(plus or minus) of the work done by
D
A
t the force, or state that the work is zero.
E

1
W  DK  K f  K 0  m v 2f  v02
2
AB  vB  v A  W  0
BC  vC  vB  W  0

CD  vD  vC  W  0
DE  vE  0, vD  0  W  0
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T
T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
(a) Pulley 1 : v  cte  Fnet  0  2T  mg  0  T  98 N
mg
Hand  cord : T  F  0  F 
 98 N
2
In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
(b) To rise “m” 0.02m, two segments of the cord must be shorten by that
amount. Thus, the amount of the string pulled down at the left end is: 0.04m
(c) WF  F  d  (98 N )(0.04m)  3.92 J
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
(d ) WFg  mgd  (0.02m)(20kg)(9.8m / s 2 )  3.92 J
WF+WFg=0
There is no change in kinetic energy.