Download 2 - Purdue Physics - Purdue University

PHYSICS 149: Lecture 17
• Chapter 6: Conservation of Energy
–6
6.7
7 Elastic Potential Energy
– 6.8 Power
• Chapter 7: Linear Momentum
– 7.1 A Vector Conservation Law
– 7.2 Momentum
Lecture 17
Purdue University, Physics 149
1
ILQ 1
A force of 5 N is applied to the end of a spring, and
it stretches 10 cm
cm. How much "farther"
farther will it stretch
if an "additional" 2.5 N of force are applied?
A)
B)
C)
D)
2.5 cm
5 cm
10 cm
15 cm
Lecture 17
Purdue University, Physics 149
2
ILQ 2
A mass is attached to the bottom of a vertical
spring.
p g This causes the spring
p g to stretch and the
mass to move downward. Does the potential energy
of the spring increase or decrease? Does the
gravitational potential energy of the mass increase
of decrease?
A)
B)
C)
D)
Lecture 17
PE of spring decreases; PE of mass increases
both decrease
b
both
h iincrease
PE of spring increases; PE of mass decreases
Purdue University, Physics 149
3
ILQ 3
A simple catapult, consisting of a leather pouch attached to
rubber bands tied to two forks of a wooden Y, has a spring
constant k and is used to shoot a pebble horizontally. When
the catapult is stretched by a distance d, it gives the pebble
a speed v.
What speed does it give the same pebble when it is
stretched to a distance 4d?
A)
B)
C)
D)
E)
16 v
4v
4 sqrt(4) v
64 v
sqrt(4) v
Lecture 17
Purdue University, Physics 149
4
Energy and Work
• Work: Transfer of Energy by Force
WF = |F| |s| cosθ
θ
• Kinetic Energy (Energy of Motion)
K = 1/2 mv2
• Work-Energy Theorem:
ΣWNC = ΔK + ΔU
• Gravitational Potential Energy:
gy
m1m2
Ugrav = mgy = -Wgrav
U = −G
Lecture 17
Purdue University, Physics 149
r
5
Hooke’s Law
• The deformation–change in size or shape–of the object is
proportional to the magnitude of the force that causes the
deformation.
– Magnitude:
g
k|x|
| |
– Direction: Whether the spring is compressed or stretched, the
Hooke’s force always points toward its relaxation position (x=0).
– Spring constant k
• A characteristic of a spring
• Unit: N/m
Lecture 17
Purdue University, Physics 149
x=0
|Fx|=k|x|
to left
6
Work by Variable Force
• W = Fx Δx
Force
– Work is area under F vs x plot
Work
– Spring: F = k x
Distance
i
• Area = ½ k x2 =Wspring
Work is the area under the F vs x p
plot
Force
Work
Distance
Lecture 17
Purdue University, Physics 149
7
Work and Potential Energy of a Spring
• The force of a spring
• The work
F = -kx
1 2 1 2
W = − kx f + kx i
2
2
xi = 0 x f = x
1 2
W = − kx
k
2
• The potential energy
ΔU = −W spring
Lecture 17
1 2 1 2
= kx f − kx i
2
2
Purdue University, Physics 149
xi = 0 x f = x
1 2
U = kx
2
Lecture 15 page 8
ILQ
•
You compress a spring by a distance x and
store 10 J of energy
energy. How much energy is
stored if you compress the spring a distance 2x?
a)
b)
c)
d)
5J
20 J
30 J
40 J
Lecture 17
Purdue University, Physics 149
9
Work Done by an Ideal Spring
•
The work done by a spring as its movable end moves from
equilibrium (xi=0) to the final position xf is
Wspring = ∫
xf
=∫
xf
xi = 0
xi = 0
G
G
FHooke ⋅ ds
(−kx) dx
1
= − kx 2f
2
•
•
beyond the scope
of this class
Å independent of path !
The work done by the spring is the
(
(negative)
ti ) area under
d th
the Fx(x)
( ) graph.
h
The work done by an ideal spring
depends only on the initial and final
positions of the moveable end
(independent of the path taken).
Æ Hooke
Hooke’s
s force is a conservative force.
force
Æ Potential energy may be defined.
Lecture 17
Purdue University, Physics 149
Area = ½⋅base⋅height
10
Elastic Potential Energy
•
The work done by an ideal spring is independent of path (see
previous slide) Æ It means that the spring force is a conservative
f
force
Æ Thus,
Th
we may define
d fi potential
t ti l energy for
f the
th spring
i fforce.
ΔU = U ( x f ) − U ( xi ) ≡ −WHooke
xf G
G
=
−
F
⋅
d
s
beyond the scope
∫ Hooke
of this class
xi
xf
= − ∫ (−kx) dx
xi
=
Lecture 17
1
1
U ( x f ) = U ( xi ) + kx 2f − kxi2
2
2
1 2 1 2
kx f − kxi
2
2
arbitrary choice
If we let U ( xi ) ≡ 0 at xi = 0,
U (x f ) =
Purdue University, Physics 149
1 2
kx f
2
11
ILQ
• All springs and masses are identical. (Gravity
acts down).
)
– Which of the systems below has the most potential
energy stored in its spring(s), relative to the relaxed
position?
iti ?
A) 1
B) 2
C) same
(1)
Lecture 17
(2)
Purdue University, Physics 149
12
ILQ: Solution
• The displacement of (1) from equilibrium will be
half of that of (2) (each spring exerts half of the
force needed to balance mg)
0
d
2d
(1)
Lecture 17
(2)
Purdue University, Physics 149
13
ILQ: Solution
• The potential energy stored in (1) is 2 ⋅ 1 k d 2 = k d 2
2
• The potential energy stored in (2) is 1 k(2d)2 = 2kd2
2
The spring
Th
i P
P.E.
E iis
twice as big in (2) !
0
d
2d
(1)
Lecture 17
Purdue University, Physics 149
(2)
14
Vertical Springs
• A spring is hung vertically. Its
relaxed position is at y = 0 (a).
When a mass m is hung from
its end, the new equilibrium
position is ye (b).
z
(b)
(a)
j
k
Recall that the force of a spring is
Fs = -kx. In case (b)
Fs = mg and x = ye:
-kyye - mg
g=0
y=0
(ye < 0))
mg = -kye
m
mg
y = ye
-kye
(ok since ye is a negative number)
Lecture 17
Purdue University, Physics 149
15
Vertical Springs
• Th
The potential
t ti l energy off the
th
spring-mass system is:
(b)
(a)
j
1 2
U = ky + mgy + C
2
k
but mg = -kye
y=0
1 2
U = ky − kye y + C
2
m
choose C to make U=0 at y = ye:
1 2
2
0 = kye − kye +C
2
Lecture 17
1
2
C = ky
k e
2
Purdue University, Physics 149
mg
y = ye
-kye
16
Vertical Springs
• So:
1 2
1 2
U = ky
k − kkyey + kkye
2
2
1 2
= k y + ye2 − 2yey
2
(
)
(b)
(a)
j
k
y=0
which can be written:
1
2
U = k ( y − ye )
2
Lecture 17
Purdue University, Physics 149
m
mg
y = ye
-kye
17
Vertical Springs
1
2
U = k(y − ye )
2
• So if we define a new y′
coordinate system such that y′ =
0 is at the equilibrium position, (
y′ = y - ye ) then we get the
simple result:
(b)
(a)
j
k
m
y′ = 0
1 ′2
U = kyy
2
Lecture 17
Purdue University, Physics 149
18
Vertical Springs
• If we choose y = 0 to be at the
equilibrium
q
p
position of the mass
hanging on the spring, we can define
the potential in the simple form.
((b))
((a))
j
k
1
U = ky 2
2
• Notice that g does not appear in this
expression!!
–B
By choosing
h
i our coordinates
di t and
d constants
t t
cleverly, we can hide the effects of gravity.
Lecture 17
Purdue University, Physics 149
y=0
m
19
ILQ: Energy Conservation
• In (1) a mass is hanging from a spring.
In (2) an identical mass is held at the height of the
end
d off the
th same spring
i iin it
its relaxed
l
d position.
iti
– Which correctly describes the relation of the potential
energies of the two cases?
(a) U1 > U2
(b) U1 < U2
(c) U1 = U2
case 2
case 1
d
Lecture 17
Purdue University, Physics 149
20
ILQ: Solution
• IIn case 1,
1 it is
i simplest
i l t tto choose
h
th
the mass tto
have zero total potential energy (sum of spring
and gravitational potential energies) at its
equilibrium position.
1
• In case 2 the total potential energy is then U 2 = kd 2
2
relaxed
y=d
d
y = 0, U1 = 0
The answer is (b) U1 < U2.
Lecture 17
Purdue University, Physics 149
21
Power (Rate of Work)
• P = W / Δt
– Units: Joules/Second = Watt
•
ΔW
P=
Δt
W = F.Δrr = F Δr cosθ
F
Δrr
= F (v Δt) cosθ
v
• P = F v cosθ
• How much power does it take for a (70 kg) student to run
up the
stairs (5 meters) in 7 seconds?
P=W/t
= mgh /t
= (70 kg) (9
(9.8
8 m/s2) (5 m) / 7 s
= 490 J/s
Lecture 17
or 490 Watts
Purdue University, Physics 149
22
Example
• Lars, of mass 82.4 kg, has been working out and can do
work for about 2.0 min at the rate of 746 W. How long will
it take him to climb three flights of stairs, a vertical height
of 12.0m?
– As Lars climbs the stairs, he increases his gravitational
potential energy.
– The rate of potential energy increase must be equal to
the rate he does work.
ΔE ΔU mgΔy
=
=
Δt
Δt
Δt
mgΔy (82.4kg )(9.80m / s 2 )(12.0m)
⇒ Δt =
=
= 13.0 s
Pav
746W
Pav =
Lecture 17
Purdue University, Physics 149
23
Power
• A 2000 kg trolley is pulled up
a 30 degree hill at 20 mi/hr
b a winch
by
i h att th
the ttop off th
the
hill. How much power is the
winch providing?
.
.
y
x
v
T
winch
θ
mg
• The power is P = F v = T v
• Since the trolley is not accelerating, the
net force on it must be zero. In the x
direction:
– T - mg sin θ = 0
– T = mg sin θ
Lecture 17
Purdue University, Physics 149
24
Power
y
.
• P = T v = Tv
parallel to v
since T is p
x
v
T
winch
• So P = mgv
g sin θ
θ
v = 20 mi/hr = 8.94 m/s
mg
g = 9.81 m/s2
m = 2000 kg
sin θ = sin(30o) = 0.5
and P = (2000 kg)(9.81 m/s2)(8.94 m/s)(0.5)
= 87,700 W
Lecture 17
Purdue University, Physics 149
25
Power
•
Power is the rate at which energy is transferred, or equivalently, the
rate at which work is done (that is, work per a time interval).
– Average Power: the average rate of energy conversion
– Instantaneous Power: the instantaneous rate at which a force F
does work when the object it acts on moves with velocity v
P=
•
•
W FΔr cos θ
=
= Fv cos θ
Δt
Δt
Power is a scalar quantity.
Units: W, J/s, etc.
– Unit conversion: 1 W = 1 J/s
– Note that kWh (kilowatt-hour) is a unit of energy, not power.
•
Power is denoted by P.
Lecture 17
Purdue University, Physics 149
26
ILQ
•
What power must an engine have if it is to be
used to raise a 25 kg load 10 m in 4 seconds?
a)
b)
c)
d)
25 W
625 W
1000 W
2500 W
Lecture 17
Purdue University, Physics 149
27
Example: The Dart Gun
In this case, there are three forces acting on the dart.
But, the directions of gravity and normal forces are perpendicular
to the displacement of the dart, so the work done by the two forces
are zero. Hooke force (a conservative force) is the only force
which does work, so the mechanical energy is conserved.
Emech = Ki + Ui = Kf + Uf = const (b/c Wnc = 0)
Ki = 0 (b/c vi = 0)
Ui = ½kxi2
Kf = ½mvf2
Uf = 0 (b/c xf = 0)
Æ 0 + ½kxi2 = ½mvf2 + 0
Thus, vf = sqrt(k/m)⋅|xi| = 11 m/s
Lecture 17
Purdue University, Physics 149
28
Key Ideas
• Work-Energy
–
–
–
–
ΣF=ma
multiply both sides by d
ΣFd=mad
(note: a d = ½ Δv2)
Σ F d = ½ m Δv2
Σ W = ΔK
K
D fi W
Define
Work
k and
d Ki
Kinetic
ti E
Energy
• Impulse-Momentum
p
–
–
–
–
Lecture 17
ΣF=ma
multiply both sides by Δt
Σ F Δt = m a Δt (note: a Δt = Δv)
Σ F Δt = m Δv
Σ I = Δp
Define Impulse and Momentum
Purdue University, Physics 149
29
Momentum is Conserved
• Momentum is “Conserved” meaning it can not be
created nor destroyed
– Can be transferred
• Total Momentum does not change with time
• Momentum is a VECTOR
3 Conservation Laws in one!
Lecture 17
Purdue University, Physics 149
30
A Vector Conservation Law
• When a vector quantity is conserved in an
interaction both its magnitude and direction are
interaction,
unchanged (or equivalently, all components are
unchanged).
g )
Lecture 17
Purdue University, Physics 149
31
Example: Momentum
• What is the momentum of an automobile (weight = 9800 N)
g at 35 m/s to the south?
when it is moving
– W = mg
Æ m = W/g = (9800 N) / (9.8 m/s2) = 1000 kg
– p = mv = (1000 kkg)) ⋅ (35 m/s
/ south)
th)
= 35,000 kg⋅m/s south
Don’t forget that momentum p is a vector! (We need
both its magnitude and direction.)
Lecture 17
Purdue University, Physics 149
32
Pushing Off…
• Fred and Jane are on skates facing each other.
Jane then pushes Fred with force F
– N2L Fred: FJF = mFred a
a = Δv/Δt
ΔvFred = a Δt
= (FJF/mFred
F d) Δt
mFred ΔvFred = FJF Δt
– N2L Jane: FFJ = mJane a
ΔvJane = a Δt
= (F/mJane) Δt
mJane ΔvJane = FJF Δt
– N3L: For every action, there is an equal and opposite
reaction. FFJ=-FJF
mFred ΔvFred = -mJane ΔvJane
Lecture 17
Purdue University, Physics 149
33
Pushing Off…
Fred (75 kg) and Jane (50 kg) are on skates facing each other. Jane then
pushes Fred with a constant force F = 45 N for a time Δt = 3 seconds.
Who will be moving fastest at the end of the push?
A) Fred
B) Same
C) Jane
Fred
Jane
F = +45 N (positive direct.)
F = -45 N Newton’s 3rd law
I = +45 × 3 Ns = 135 Ns
I = -45 × 3 Ns = -135 Ns
I = Δp
I = Δp
= mvf – mvi
= mvf – mvi
I/m = vf - vi
I/m = vf - vi
vf = 135 N-s / 75 kg
vf = -135 N-s / 50 kg
= 1.8
1 8 m/s
/
= -2.7
2 7 m/s
/
Note: Pfred + Pjane = (1.8) 75 + (-2.7) 50 = 0!
Lecture 17
Purdue University, Physics 149
34