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PH 221‐3A Fall 2009
Review for test 2
Lecture 13
Lecture 13
Chapters 5‐8 (Halliday/Resnick/Walker, Fundamentals of Physics 8th edition)
The potential Energy Curve
If we plot the potential energy U
If
we plot the potential energy U versus x
versus x for a for a
force F that acts along the x‐axis we can get a wealth of information about the motion of a particle on which F is acting. The first parameter p
g
p
that we can determine is the force F(x) using the equation:
dU ( x)
F ( x) = −
dx
An example is given in the figures below. In In fig.a we plot U(x) versus x. In fig.b
g we plot F(x) versus
p
( )
x.
For example at x2 , x3 and x4 the slope of the U(x) vs x curve is zero, thus F = 0. The slope dU/dx between x3 and x4 is negative; Thus F > 0 for the this interval. The slope dU/dx between x2 and x3 is positive; Thus F < 0 for the same interval Turning Points:
The total mechanical energy is Emec = K ( x ) + U ( x )
This energy is constant (equal to 5 J in the figure)
and is thus represented by a horizontal line. We can
solve this equation for K ( x ) and get:
K ( x ) = Emec − U ( x ) At any point x on the x -axis
we can read the value of U ( x ). Then we can solve
th equation
the
ti above
b
andd determine
d t
i K
mv 2
From the definition of K =
the kinetic energy cannot be negative.
2
This property of K allows us to determine which regions of the x-axis
motion is allowed. K ( x) = K ( x) = Emec − U ( x)
If K > 0 → Emech − U ( x) > 0 → U ( x) < Emec Motion
M ti is
i allowed
ll
d
If K < 0 → Emech − U ( x) < 0 → U ( x) > Emec Motion is forbidden
g ppoints
The ppoints at which: Emec
mec = U ( x ) are known as turning
for the motion. For example x1 is the turning point for the
U versus x plot above. At the turning point K = 0
Given the U(x) versus x curve the turning points and the regions for which motion is allowed depends on the value of the mechanical energy Emec
In the picture to the left consider the situation when Emec = 4 J (purple line) The turning points (Emec = U
turning points (E
U ) occur at x
) occur at x1 and x
and x > x
> x5. Motion is allowed for x > x
. Motion is allowed for x > x1 If we If we
reduce Emec to 3 J or 1 J the turning points and regions of allowed motion change accordingly.
Equilibrium Points:
Equilibrium
Points: A position at which the slope dU/dx = 0
A position at which the slope dU/dx 0 and thus F = 0
and thus F 0 is called is called
an equilibrium point. A region for which F = 0 such as the region x > x5 is called a region of neutral equilibrium. If we set Emec = 4 J the kinetic energy K = 0 and any particle moving under the influence of U will be stationary at any point with x > x5 particle moving under the influence of U will be stationary at any point with x > x
Minima in the U versus x curve are positions of stable equilibrium
Maxima in the U
in the U versus x
versus x curve are positions of unstable equilibrium
curve are positions of unstable equilibrium
Note: The blue arrows in the figure indicate the direction of the force F as determined from the equation: dU ( x)
F ( x) = −
dx
Positions of Stable Equilibrium. An example is point x4 where U has a minimum. If we arrange Emec = 1 J then K = 0 at point x4. A particle with Emec = 1 J is stationary at x4. If we displace slightly the particle either to the right or to the left of x4 the force tends to bring it back to the equilibrium position. This equilibrium is stable. Positions of Unstable Equilibrium.
Positions
of Unstable Equilibrium. An example is point x
An example is point x3 where U
where U has a has a
maximum. If we arrange Emec = 3 J then K = 0 at point x3. A particle with Emec = 3 J is stationary at x3. If we displace slightly the particle either to the right or to y
q
the left of x3 the force tends to take it further away from the equilibrium position. This equilibrium is unstable
Work done on a System by an External Force
( fi i i
(no friction involved)
l d)
Up to this point we have considered only isolated systems in which no external forces were present. We will now consider a system in which there are forces external to the id
t
i hi h th
f
t
l t th
system
The system under study is a bowling ball being hurled by a player. The system consists of the Th
d
d i b li b ll b i h l d b
l
Th
i
f h
ball and the earth taken together. The force exerted on the ball by the player is an external force. In this case the mechanical energy Emec of the system is not constant. Instead it changes y
q
done by the external force according to the equation: y
g
q
by an amount equal to the work W
W = ΔEmec = ΔK + ΔU
Work done on a System by an External Force (friction involved)
vo
v
•2nd Newton’s Law for x component F‐fk=ma
d
A
fk
m
Fv
o
B
fk
m
x
•Since a=const, v2 = vo2 +2ad, solving for a and substituting the result in Newton’s Eq.
Fd=1/2mv2‐1/2mvo2+fkd=ΔK+fkd
•Through experiment, we find that the increase in thermal energy by sliding : Th
h
i
fi d h h i
i h
l
b lidi
ΔEth=fkd •Fd is the work W done by external force F on the block floor system
y
y
•Hence, W=ΔEmec+ΔEth energy statement for the work done on a system by an external force when friction is involved.
Problem 42. A worker pushed a 27 kg block 9.2 m along a level floor at constant speed with a force directed 32 degree below the horizontal. If the coefficient of kinetic friction between block and floor was 0.20, what were (a) the work done by the worker’s force and (b) the ,
( )
y
( )
increase in thermal energy of the block‐floor system?
r
. Since the velocity is constant, a = 0 and the horizontal component of the worker's
push F cos θ (where θ = 32°) must equal the friction force magnitude fk = μk FN. Also, the
vertical forces must cancel,
cancel implying
FN.=Fsinθ +mg, on the other hand Fcos θ=μk FN
which is solved to find F = 71 N.
(a) The work done on the block by the worker is, using Eq. 7-7,
b gb
g
W = Fd cosθ = 71 N 9.2 m cos32° = 5.6 × 102 J .
)(9. m)) = 5.6
. × 102 J.
(b) Since
S ce fk = μk (mg + F ssin θ ), we find
d ΔEth = f k d = ((60 N)(9.2
Conservation of Energy
Energy cannot magically appear or disappear. It can be transferred to or from objects and systems by means of work done on a system. The total energy E of a system can change only by amounts of energy that are transferred to or from the system.
W=ΔE=ΔEmec+ΔEthh+ΔEint
i
The law of conservation of energy is not something we have derived from basic physics principles. Rather it is a law based on countless experiments. We have never found an exception to it
exception to it.
Isolated System
If a system is isolated from its environment, there can be no energy transfers to or from it The total energy E of an isolated system cannot change
it. The total energy E of an isolated system cannot change.
Problem 61. A stone with a weight of 5.29 N is launched vertically from ground level with an initial speed of 20.0 m/s, and the air drag on it is 0.265 N throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the
are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?
(a) The maximum height reached is h. The thermal energy generated by air resistance as
the stone rises to this height is ΔEthh = fh by Eq.
Eq 8-31.
8 31 We use energy conservation in the
form of Eq. 8-33 (with W = 0):
K f + U f + ΔE th = Ki + U i
and we take the potential energy to be zero at the throwing point (ground level). The
1
initial kinetic energy is Ki = mv02 , the initial potential energy is Ui = 0, the final kinetic
2
energy is
i Kf = 0,
0 andd the
th fi
finall potential
t ti l energy is
i Uf = wh,
h where
h w = mg is
i the
th weight
i ht off
1
the stone. Thus, wh + fh = mv02 , and we solve for the height:
2
mv02
v02
.
h=
=
2( w + f ) 2 g (1 + f / w)
Numerically, we have, with m = (5.29 N)/(9.80 m/s2)=0.54 kg,
(20.0 m/s)2
h=
= 19.4 m .
2(9.80 m/s2 )(1 + 0.265/5.29)
(b) We notice that the force of the air is downward on the trip up and upward on the trip
down, since it is opposite to the direction of motion. Over the entire trip the increase in
1
thermal energy is ΔEth = 2fh. The final kinetic energy is K f = mv 2 , where v is the speed
2
of the stone just before it hits the ground. The final potential energy is Uf = 0. Thus, using
Eq. 8-31 (with W = 0), we find
1 2
1
mv + 2 fh = mv02 .
2
2
W substitute
We
b tit t the
th expression
i found
f d for
f h to
t obtain
bt i
2 fv02
1
1
= mv 2 − mv02
2 g (1 + f / w) 2
2
which leads to
⎛
2 fv02
2 fv02
2 f ⎞ 2 w− f
2
= v0 −
= v02 ⎜1 −
v =v −
⎟ = v0
w((1 + f / w)
w
+
f
w+ f
mgg ((1 + f / w)
⎝
⎠
2
2
0
where w was substituted for mg and some algebraic manipulations were carried out.
Therefore,
w− f
5 29 N − 00.265
5.29
265 N
v = v0
= (20.0 m/s)
= 19.0 m/s .
w+ f
5.29 N + 0.265 N
GENERAL PHYSICS PH 221-3A (Dr. S. Mirov)
Test 2 (10/10/07)
Sample Test 2
STUDENT NAME: _Key____________________ STUDENT id #: ___________________________
-------------------------------------------------------------------------------------------------------------------------------------------
ALL QUESTIONS ARE WORTH 20 POINTS. WORK OUT FIVE PROBLEMS.
NOTE: Clearly write out solutions and answers (circle the answers) by section for each part (a., b., c., etc.)
Important Formulas:
1. Motion along a straight line with a constant acceleration
vaver. speed = [dist. taken]/[time trav.]=S/t;
vaver.vel. = Δx/Δt;
vins =dx/Δt;
aaver.= Δvaver. vel./Δt;
a = dv/Δt;
v = vo + at; x= 1/2(vo+v)t; x = vot + 1/2 at2; v2 = vo2 + 2ax (if xo=0 at to=0)
2.
Free fall motion (with positive direction ↑)
g = 9.80 m/s2;
y = vaver. t
vaver.= (v+vo)/2;
v = vo - gt; y = vo t - 1/2 g t2; v2 = vo2 – 2gy (if yo=0 at to=0)
3
3.
M ti
Motion
in
i a plane
l
4.
Projectile motion (with positive direction ↑)
5.
Uniform circular Motion
vx = vo cosθ;
vy = vo sinθ;
x = vox t+ 1/2 ax t2; y = voy t + 1/2 ay t2; vx = vox + at; vy = voy + at;
vx = vox = vo cosθ;
x = vox t;
xmax = (2 vo2 sinθ cosθ)/g = (vo2 sin2θ)/g for yin = yfin;
vy = voy - gt = vo sinθ - gt;
y = voy t - 1/2 gt2;
a=v2/r,
T=2πr/v
6. Relative motion
r
r
r
v PA = v PB + v BA
r
r
a PA = a PB
7.
Component method of vector addition
A =
A x2 + A y2 ; θ = tan-1 ⏐Ay /Ax⏐;
r
r
The scalar product A a ⋅ b = a b c o s φ
r
r
a ⋅ b = ( a x iˆ + a y ˆj + a z kˆ ) ⋅ ( b x iˆ + b y ˆj + b z kˆ )
r r
a ⋅ b = a xbx + a yb y + a zbz
r
r
The vector product a × b = ( a x iˆ + a y ˆj + a z kˆ ) × ( b x iˆ + b y ˆj + b z kˆ )
A = A1 + A2 ; Ax= Ax1 + Ax2 and Ay = Ay1 + Ay2;
iˆ
r
r
r
r
a × b = −b × a = ax
bx
kˆ
ˆj
a
y
by
a
z
= iˆ
bz
a
a
y
by
z
bz
a
− ˆj
x
bx
a
z
bz
+ kˆ
a
x
bx
a
y
by
=
= ( a y b z − b y a z ) iˆ + ( a z b x − b z a x ) ˆj + ( a x b y − b x a y ) kˆ
1.
Second Newton’s Law
ma=Fnet ;
2.
Kinetic friction fk =μkN;
3
3.
St ti friction
Static
f i ti
fs =μsN;
N
4.
Universal Law of Gravitation: F=GMm/r2; G=6.67x10-11 Nm2/kg2;
5.
Drag coefficient
6.
Terminal speed
7.
Centripetal force: Fc=mv2/r
8.
Speed of the satellite in a circular orbit: v2=GME/r
9.
The work done by a constant force acting on an object:
10. Kinetic energy:
D =
1
C ρ Av
2
vt =
2m g
C ρ A
K =
1
m v
2
2
W
r r
= F d cosφ = F ⋅ d
2
11. Total mechanical energy: E=K+U
12. The work-energy theorem: W=Kf-Ko; Wnc=ΔK+ΔU=Ef-Eo
13. The principle of conservation of mechanical energy: when Wnc=0, Ef=Eo
14. Work done by the gravitational force:
W
g
= m gd cosφ
1.
Work done in Lifting and Lowering the object:
ΔK = K
f
− K
i
= W
a
+ W g ; if K
= K i; W
f
= −W
g
F x = − k x ( H o o k 's l a w )
2.
Spring Force:
3.
Work done by a spring force:
W
s
1
1
k x i2 −
k x o2 ; i f x i = 0 a n d x
2
2
=
x
4.
a
Work done by a variable force:
W =
f
= x; W
s
= −
f
∫
F ( x )dx
xi
r r
W
dW
; P =
; P = F v cosφ = F ⋅ v
Δt
dt
Pa v g =
5.
Power:
6.
Potential energy:
7.
Gravitational Potential Energy:
ΔU = mg(y
f
Δ U = −W ; Δ U = −
∫
x
f
xi
F ( x )d x
− y i ) = m g Δ y ; if y i = 0 a n d U
1
kx
2
8.
Elastic potential Energy:
U (x) =
9.
Potential energy curves:
F (x) = −
i
= 0; U ( y ) = m gy
2
dU (x)
; K (x) = E
dx
− U (x)
m ec
10. Work done on a system by an external force:
F r ic tio n is n o t in v o lv e d W = Δ E
m ec
= ΔK + ΔU
W h e n k in e tic f r ic tio n f o r c e a c ts w ith in th e s y s te m W = Δ E
ΔE
th
m ec
+ ΔE
th
= fkd
11 Conservation of energy:
11.
12. Power:
Pa v g =
W = ΔE = ΔE
m ec
+ ΔE
th
+ Δ E in t
f o r is o la te d s y s te m ( W = 0 ) Δ E
ΔE
dE
; P =
Δt
dt
m ec
+ ΔE
th
+ Δ E in t = 0
1
kx
2
2
1.
4T
4T
61.2 kg
16
2
‐2
=4x10
4x103 m
6.
7.
(a)
(b)
(c)
(d)
A 700 g block is released at height ho above a vertical spring of constant k=400
N/m and negligible mass. The block sticks to the spring and momentarily stops
after compressing the spring 19.0 cm. How much work is done
by the block on the spring?
by the spring on the block?
What is the value of ho?
If the block were released from height 2.00 ho above the spring, what would be the
maximum
i
compression
i
of
f the
h spring?
i
?
We place the reference position for evaluating gravitational potential energy at the
relaxed position of the spring. We use x for the spring's compression, measured positively
downwards (so x > 0 means it is compressed).
(a) With x = 0.190
0 190 m
m,
Ws = −
1
k x 2 = − 7 .2 2 J ≈ − 7 . 2 J
2
for the work done by the spring force.
force Using Newton
Newton's
s third law
law, we see that the work
done on the spring is 7.2 J.
(b) As noted above, Ws = –7.2 J.
(c) Energy conservation leads to
Ki + U
i
= K
f
+U
m g h0 = − m g x +
f
1
kx 2
2
which (with m = 0.70 kg) yields h0 = 0.86 m.
(d) With a new value for the height h 0′ = 2 h 0 = 1 . 7 2 m , we solve for a new value of x
using the quadratic formula (taking its positive root so that x > 0).
1
m g h 0′ = − m g x +
kx 2 ⇒
2
which yields x = 0.26 m.
x =
mg +
bm g g
k
2
+ 2 m g k h 0′
8. A particle moving along the x axis is acted upon by a single force F = F0e-kx,
where F0 and k are constants. The particle is released from rest at x = 0. What is
the maximum kinetic energy it will attain?
According
g to the work energy
gy theorem W = K f − Ki = K f since we started from rest (Ki = 0))
f
− kx
⎛ e− kx f 1 ⎞
⎡
⎤
e
− kx
On the other hand W = ∫ F ( x)dx → K =∫ F ( x)dx = Fo ∫ e dx = Fo ⎢−
⎥ =Fo ⎜ − k + k ⎟
0
0
0
k
⎣
⎦0
⎝
⎠
k
Max K f occurs at a point when K ' ( x) = 0, K ' ( x) = Fo (e− kx
) = 0 only for x=∞
x
xf
→ Kmax =
Fo
Joules
k
xf
xf
9. The potential energy of a 0.20-kg particle moving along the x axis is given by
U(x) = (8.0J/m2)x2 + (2.0J/m4)x4 . When the particle is at x = 1.0m it is traveling in the positive x
direction with a speed
p
of 5.0m/s. At what x coordinate it will stop
p momentarily
y to turn around.
( a ) Calculate total mechanical energy of the particle at coordinate x=1.0 m
mv 2 0.2 × 5.02
U (1.0m ) = 8.0 J + 2.0 J = 10.0 J ; K (1.0m) =
=
= 2.5J
2
2
→ Emec = 12.5 J
(b) The particle will stop momentarily to turn around when K ( x ) = Emec − U ( x ) = 0
→ 8.0 x 2 + 2.0 x 4 = 12.5
−8.0 ± 8.02 + 4(2.0)(12.5)
x =
4.0
x 2 = ( −2.0
2 0 ± 3.2)
3 2)m 2
2
x 2 = 1.2m 2
x = 1.1m