lecture note chapter 7 work and energy 1 kinetic energy
TRANSCRIPT
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Lecture Note Chapter 7 Work and Energy 1 Kinetic energy and work I
We consider the simplest case where the acceleration a is constant. Newton’s second law
cosFma or cosm
Fa
Here we use
cos1
2222 Fdm
advv if
where vf is the final velocity and vi is the initial velocity. Then we get
dF cos)(2
22 Fdvvm
if
(1) The kinetic energy is defined by
2
2
1mvK
((Note)) Units In SI units, the units of work is J (Joule)
1 J (Joule) = kg m2/s2 = N m 1 N = kg m/s2
In cgs units, the unit of work is erg.
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1 erg = g cm2/s2 = dyne cm 1 dyne = g cm/s2
(2) The work is defined by
dF W (work done by a constant force) Then Eq.(1) can be rewritten as
netif WKKK
This is called as work-kinetic energy theorem, simply, work-energy theorem: 2 Work: general case 2.1 Definition
We now consider the work in the general case,
The work along a path is given by
rF dW
where the integral is made along a path (path integral).
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2.2 Work done by the gravitational force (an example of Wc) Conservative force
For the rising case, the work done by the gravitational force is –mgh (<0). For the falling case, the work done by the gravitational force is mgh (>0). When the particles goes up and then falls down, the work done by the gravitational force is
-mgh + mgh = 0.
2.3 Work done by the frictional force (an example of Wnc)
Nonconservative force
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rf ddW knc , work done by the frictional force. This work is always negative. That is
not dependent on the direction of the movement. The frictional force is one of the non-conservative forces.
0 rF dWnc , which does not depend on the direction of motion.
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3. Work-energy theorem
The work done in the displacement by the force is defined as
B
A
dBAW rF)( (1)
where the limits A and B stand for the positions rA and rB. The substitution of the force F defined by
dt
dm
vF
into Eq.(1) leads to
B
A
ddt
dmBAW r
v)( .
Now
dtdtdt
dd v
rr
so that
B
A
dtdt
dmBAW v
v)(
where the limits A and B now stand for the times tA and tB when the particle is at the positions rA and rB. Here we can rearrange the integrand
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vvv
vvv
vv dt
d
dt
d
dt
d
dt
dv
dt
d2)(2
So that
2222
2
1
2
1)(
2)(
2)( AB
B
A
B
A
mvmvvdm
dtvdt
dmBAW
where the limits A and B now stand for the velocities vA and vB when the particle is at the positions rA and rB. This expression is rewritten as
AB KKBAW )( where K is the kinetic energy of the particle. In summary, the work-energy theorem is given by
WK where
B
A
dW
vvmK AB
r
r
rF
)(2
1 22
4 Net work
The force F is the sum of Fc and Fnc,
ncc FFF
where Fc is the conserved force (such as a gravitational force) and Fnc is the non-conserved force (such as friction).
The net work Wnet is the sum of Wc and Wnc
nccnet WWW
where
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B
A
B
A
dW
dW
ncnc
cc
r
r
r
r
rF
rF
Then we have the work-energy theorem:
cncnet WWWK
5 Potential energy U
We define a potential energy. This is a scalar function associated with a conservative force.
)( ifc UUUW
where Wc is the work done by the conservative force, Ui is the potential energy at the initial state and Uf is the potential energy at the final state. (a) For the one dimensional system, we have
)()(')'( i
x
x
cc xUxUdxxFWi
Here we take a derivative of both sides with respect to x, we get
dx
xdUxFc
)()(
((Example)) For the gravitational force (conservative force), the potential energy is given by
U(z) = mgz, since
mgdz
zdUzFc
)()(
(b) For the three dimensional system, we have
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f
i
cc dW rF
The conservative force Fc is expressed by the potential energy as
),,(y
U
y
U
x
UgradUUc
F
((Note)) In Chapter 8 we will give a proof of this expression using Stokes theorem. 6 Spring Force 6.1 Hooke’s law
A particle is subject to a linear restoring force in the x direction. A linear restoring force (spring force) is one that is directly proportional to the displacement (x) measured from the equilibrium point. The spring force is defined by
F=-kx, or iF kx where k is a positive constant. The unit of k is N/m. This is called a Hooke’s law. For sufficiently small displacements, the spring force may be produced by a stretched or compressed spring. The sign of the force is such that the particle is always attracted toward the origin x = 0.
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A plot of the spring force is shown as a function of x.
The spring force is a conservative force.
kxdxdxkxd iirF )(
)(2
1)()( 22
if
x
x
x
x
xxkkxdxdfiWf
i
f
i
rF
Note that
0)(2
1)(
2
1)()( 2222 fiif xxkxxkifWfiW ,
which means the spring force is a conservative force. Then the potential energy U can be defined as
kxdx
xdUxFc
)()(
or
2
0 2
1kxkxdxU
x
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The energy conservation:
From the work-energy theorem
UWK we have
0)( UK or 0E where E is the total energy and is defined by
22
2
1
2
1kxmvUKE (Energy conservation law)
and
iiifff EkxmvkxmvE 2222
2
1
2
1
2
1
2
1
Where
K (= 2
2
1mv ) is the kinetic energy
U (= 2
2
1kx ) is a potential energy.
((Note)) Derivation of the energy conservation law for the simple harmonics.
Approach from the Newton's second law
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We start from the Newton's second law for the simple harmonics
kxFvm (1) Multiplying both sides of Eq.(1) by xv , we get
xkxFvmv or
0)2
1
2( 22 kxvm
dt
d
Then we have the energy conservation law for the simple harmonics
Ekxmv 22
2
1
2
1 = const (2)
6.2 Energy conservation in the spring system
Here we consider the energy conservation law for the spring,
22
2
1
2
1kxmvUKE
At x = 0, 2max2
1mvE
At x = xmax (amplitude) 2max2
1kxE
or
2max
2max 2
1
2
1kxmvE
or
maxmax xv
From the energy conservation law, we have
22
2
1
2
11 kx
Emv
E
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or
12
max
2
max
x
x
v
v
This is a circle in the x/xmax vs v/vmax plane. The center is at the origin. In general, the v-x plane is called a phase space.
Fig. Phase space (x, v) for the simple harmonics.
6.3 Determination of the spring constant k
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Free-body diagram
In equilibrium, we have
mgkx 0
or
k
mgx 0
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In SI units, the spring constant is in the units of N/m. In dynamics, we set up an equation of motion,
)()( 02
2
xxkk
mgxkkxmg
dt
xdm
(simple harmonics, see Chapter 15 for detail)
)( 02
2
2
xxdt
xd
where is the angular frequency
m
k
The solution of the second order differential equation is
)cos(0 tAxx
10 20 30 40 50wt
-1.0
-0.5
0.5
1.0
xA
The period T is given by
k
mT
22
((Note)) The dimension of m/k
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2
2
1][ s
ms
mkg
kg
m
Nkg
k
m
6.4 Example Problem 7-67 (SP-7) (10-th edition)
A spring with a pointer attached is hanging next to a scale marked in millimeters. Three different packages are hung from the spring, in turn, as shown in Fig. (a) Which mark on the scale will the pointer indicate when no package is hung from the spring? (b) What is the weight W of the third package?
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m1g = 110 N m2g = 240 N x1 = 40 mm x2 = 60 mm x3 = 30 mm Determination of the spring constant
)(
)(
022
011
xxkgm
xxkgm
or )()( 1212 xxkgmm
The we have
mNm
N
xx
gmmk /105.6
10)4060(
)110240()( 33
12
12
(a) 011 xxk
gm and mm
k
gm9.161
Then we have
x0 = x1-16.9 = 23.1 mm
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since x1 = 40.0 mm. (b) NxxkW 9.44)( 03
where x3 = 30 mm and x0 = 23.1 mm. 7 Power P 7.1 Average power
The power is the time rate of transfer energy. The average power is defined as
t
WPavg
(average power)
where
rB
rA
dW rF
7.2 Instantaneous power
If the force F causes a particle to undergo a displacement dr, the work done is
rF ddW . Since
dtd vr the instantaneous power (or simply power) provided by the force id
vFr
F dt
d
dt
dWP (instantaneous power)
From this expression, W can be derived as
2
1
)()( 21
t
t
dttPttW
((Units)) In SI units, the units of power is W (Watt)
1W = 1 J/s 1 horsepower=1 hcp = 746 W kWh =(103 J/s) 3600 s = 3.6 x 106 J
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8. Example 8.1 Problem 7-42*** (SP-7) (10-th edition)
Figure shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.20 m, so that the cart slides from x1 = 3.00 m to x2 = 1.00 m. During the move, the tension in the cord is a constant 25.0 N. What is the change in the kinetic energy of the cart during the move?
((Solution))
h1 = 1.20 m. x1 = 3.0 m x2 = 1.0 m. T = 25.0 N
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J
hxhxT
dxhx
xTdxTdW
hx
x
x
x
7262.41
)(
.cos
cos
222
221
2
122
22
rT
8.2 Problem 7-52*** (SP-7) (10-th edition)
A funny car accelerates from rest through a measured track distance in time T with the engine operating at a constant power P. If the track crew can increase the engine power by a differential amount dP, what is the change in the time required for the run.
((Solution))
dt
dWP
When P is constant,
222
0
)]([2
1)]0([
2
1)]([
2
1tvmtvmtvmKPtPdtW
t
using the work-energy theorem. Then we have
m
Pt
dt
dxtv
2)(
or
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LTm
Pdtt
m
Pdxx
T
2/3
0 3
222
or
23
8
9mLPT = const
This can be rewritten as
constmLTPPT )8
9ln(ln3ln)ln( 23
or
03
T
T
P
P
or
T
T
P
P
3
9. Advanced problem Serway Problem 6-52 and 7-14
A particle is attached between the identical springs on a horizontal frictionless table. Both springs have spring constant k and are initially unstressed. (a) If the particle is pulled a distance x along a direction perpendicular to the initial configuration of the springs, as shown in Fig., show that the force exerted by the springs on the particle is
iLx
Lkx ˆ)1(2
22 F
(b) Determine the amount of work done by this force in moving the particle from x = A to 0.
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(a)
22 Lxs , 22
cosLx
x
s
x
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)1
1(2)(2cos)(22222
22
Lxkx
Lx
xxLxkLskFx
(b)
A
AA
x dxLx
Lkxdx
Lx
LkxdxFW
022
0
22
0
)1(2)1(2
or
)](2[ 222 LALLAkW (c) The potential U
)22( 2222
0
LxLxLkdxFUx
x
((Mathematica))
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F 2 k x 1 L
x2 L2
2 k x 1 L
L2 x2
J1 SimplifyA
0Fx, A 0, L 0
k A2 2 L L A2 L2
rule1 L 1, k 1L 1, k 1
F1 F . rule1
2 x 1 1
1 x2
PlotF1, x, 2, 2, PlotStyle Red, Thick,
Background LightGray, AxesLabel "x", "F"
-2 -1 1 2x
-2
-1
1
2
F
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Potential energy
U1 Simplify 0
xFx, k 0, L 0, x 0
k 2 L2 x2 2 L L2 x2
U2 U1 . rule1
2 x2 2 1 x2
PlotU2, x, 2, 2, PlotStyle Red, Thick,
Background LightGray, AxesLabel "x", "U"
-2 -1 1 2x
0.2
0.4
0.6
0.8
1.0
1.2
1.4
U
10. Problems of HW-7, SP-7, and so on 10.1 Problem 7-20 (SP-07) (10-th edition)
A block is set up a frictionless ramp along which an x axis extends upward. Figure gives the kinetic energy of the block as a function of position x; the scale of the figure’s vertical axis is set by Ks = 40.0 J. If the block’s initial speed is 4.00 m/s, what is the normal force on the block?
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((Solution)) vi = 4.00 m/s Ks = 40.0 J.
The work-energy theorem
xmgKK
KKmvmvK
xmgUWK
s
si
c
sin2
1
2
1
sin
22
((Note)) One can apply directly the energy conservation law to this problem. The slope of K vs x is –mg sin.
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The normal force N is given by
cosmgN ____________________________________________________________________ 10.2 Problem 7-38 (SP-07) (10-th edition) A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an x axis is applied to the block. The force is given by
NxxF i)5.2()( 2 , where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.0 m? (b) What is the maximum kinetic energy of the block between x = 0 and x = 2.0 m? ((Solution)) m = 1.50 kg. vi = 0. The work energy theorem
f
i
f
i
x
x
x
x
fif dxxdxxFWKKKK )5.2()( 2
Since
Ki = 0, and 2
2
1ff mvK
xdx
Wd
xdx
dW
2
05.2
2
2
2
10.3 Problem 7-57 (SP-07) (10-th edition)
A 230 kg crate hangs from the end of a rope of length L = 12.0 m. You push horizontally on the crate with a varying force F to move its distance d = 4.00 m to the side (see Fig). (a) What is the magnitude of F when the crate is in this final position?
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During the crate’s displacement, what are (b) the total work done on it, (c) the work done by the gravitational force on the crate, and (d) the work done by the pull on the crate from the rope? (e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate. (f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?
m = 230 kg, L = 12.0 m, d = 4.0 m, 0= arcsin(d/L) = 19.5° F varies with the angle .
(a)
FT
mgT
sin
0cos
tanmgF .
(b) The total work
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0 WK (work-energy theorem)
(c) )cos1( 0 mgLUW gg work due to gravity.
(d) 0 TFg WWWK . WT = 0
)cos1( 0 mgLWW gF
______________________________________________________________________ ((Note-1)
We calculate the work done by the force F, tension T, and gravitational force (mg) separately.
d
T
F
Fg
dr
(i) Work done by the force F
F varies with the angle ,
tanmgF . Since Lddr , we have
dmgLdmgLdFLFdrddWF sincostancoscos rF . Then we get
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)cos1(sin 0
0
0
mgLdmgLWF .
(ii) Work done by the force T
Since the tension T is always perpendicular to the displacement dr,
WT = 0 (iii) Work done by the gravitational force (Fg = mg)
dmgLmgdrdmdWg sin)2
cos( rg
or
)cos1(sin 0
0
0
mgLdmgLWg
The total work done is
0)cos1()cos1( 00 mgLmgLWWWW gTFtotal
((Note-2))
The result 0totalW is evident since
0)( gtotal dW FTFr
and
0 gFTF
along any point on the path. 10.4 Problem 7-33*** (from SP-7) (10-th edition)
The block in Fig. lies on a horizontal frictionless surface, and the spring constant is 50 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 3.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is reached, what are
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(a) the position of the block, (b) the work that has been done in the block by the applied force, and (c) the work that has been done on the block by the spring force?
During the block’s displacement, what are (d) the block’s position when its kinetic energy is maximum and (e) the value that maximum kinetic energy?
((Solution)) k = 50 N/m F = 3.0 N x = 0 and v = 0 initially. The work-energy theorem
nc
ncncc
WUKE
WUWWK
In the present case, we have
220
2
2
1
2
1
2
1kxkxkxU
FxWnc
with x0 = 0. or
2
2
1kxFxUWK nc
(a) When the stopping point is reached, K = 0
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mk
Fxx
kxFxK
12.02
2
10
max
2maxmax
(b)
JFxWnc 36.0312.0max
(c)
JFxkxUWc 36.02
1max
2max
(d)
mk
Fx
kxFdx
Kd
kxFxK
06.0
0)(
2
1
1
2
(e)
Jk
FkxFxK 09.0
22
1)(
22
11max
10.5 Problem 7-79 (HW-7 Hint) (10-th edition)
A 2.0 kg lunchbox is sent sliding over a frictionless surface, in the positive direction of an x axis along the surface. Beginning at time t = 0, a steady wind pushes on the lunchbox in the negative direction of the x axis. Figure shows the position x of the lunchbox as a function of time t as the wind pushes on the lunchbox. From the graph, estimate the kinetic energy of the lunchbox at (a) t = 1.0 s and (b) t = 5.0 s. (b) How much work does the force from the wind do on the lunchbox from t = 1.0 s to t = 5.0 s.
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((Solution)) m = 2.0 kg
FxvvmWK )(2
1 20
2
or
FxmvmvK 20
2
2
1
2
1
((Note)) Equation of motion
axvv
tm
Ftvattvx
vatvm
Fa
Fma
2
22
1
20
2
20
20
0
11. Summary Work-energy thorem ((from Wikipedia))
In physics, mechanical work is the amount of energy transferred by a force. Like energy, it is a scalar quantity, with SI units of joules. The term work was first coined in the 1830s by the French mathematician Gaspard-Gustave Coriolis.
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According to the work-energy theorem if an external force acts upon an object, causing its kinetic energy to change from Ek1 to Ek2, then the mechanical work (W) is given by
)2
1( 2
12 mvEEEW kkk
where m is the mass of the object and v is the object's velocity.