Download Power is the time rate at which work W is done by a force •Average

Power
Power is the time rate at which work W is done by a force
•Average power
Pavg = W/∆t (energy per time)
•Instantaneous power
P = dW/dt = (Fcosφ dx)/dt = F v cosφ= F . v
Unit: watt
1 watt = 1 W = 1 J/s
1 horsepower = 1 hp = 550 ft lb/s = 746 W
kilowatt-hour is a unit for energy or work:
1 kW h = 3.6 M J
Problem 07-51
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
A) Work done on the particle
B) Average power
C) Angle between di and df
Problem 07-51
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
A) Work done on the particle
F is constant in this problem, so
(
v v v
W = F ⋅ df − di
)
= (3.0î + 7.0 ĵ + 7.0k̂ ) ⋅ ((−5.0 − 3.0)î + ((4.0 + 2.0) ĵ + (7.0 − 5.0)k̂ )
= (3.0î + 7.0 ĵ + 7.0k̂ ) ⋅ (−8.0î + 6.0 ĵ + 2.0k̂ )
= −24.0 + 42.0 + 14.0 = 32.0J
Problem 07-51
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0 ĵ + 7.0k̂ )
B) Average power
Pave = W/∆t = 32.0J/4s = 8.0W
Problem 07-51
v
Force F = (3.0î + 7.0 ĵ + 7.0k̂ )
Initial
Displacement
v
d i = (3.0î − 2.0 ĵ + 5.0k̂ )
v
d i = 38
in ∆t = 4.0s
Final Displacement
v
d f = (−5.0î + 4.0ˆj + 7.0k̂ )
v
d f = 90
C) Angle between di and df
di and df are two vectors, so take the dot product
v v
d f ⋅ d i = d f d i cos φ ⇒
cos φ = (−5.0î + 4.0 ĵ + 7.0k̂ ) ⋅ (3.0î − 2.0 ĵ + 5.0k̂ ) / ( d f d i )
(−15.0 − 8.0 + 35.0k̂ ) / ((9.49)(6.16) ) = 12.0 / 58.5 = 0.205
⇒ φ = cos −1 0.205 = 78.2o
Chapter 8: Potential Energy and
Conservation of Energy
Work and kinetic energy are energies of motion.
rf
v v
1
1
2
2
∆K = K f − K i = mv f − mvi = Wnet = ∫ F ⋅ d r
2
2
ri
Consider a vertical spring oscillating with mass m
attached to one end. At the extreme ends of travel
the kinetic energy is zero, but something caused it
to accelerate back to the equilibrium point.
We need to introduce an energy that depends on
location or position. This energy is called potential
energy.
Chapter 8:
Potential Energy and Conservation
of Energy
Work done by gravitation for a ball thrown
upward then it fell back down
b
Wab+ Wba = − mgd + mg d = 0
The gravitational force is said to be a
conservative force.
A force is a conservative force if the net
work it does on a particle moving around
every closed path is zero.
a
Conservative forces
Wab,1 + Wba,2 = 0
Wab,2 + Wba,2 = 0
therefore: Wab,1 = Wab,2
i.e. The work done by a conservative force on a
particle moving between two points does not
depend on the path taken by the particle.
So, ... choose the easiest path!!
Conservative & Non-conservative forces
Conservative Forces: (path independent)
gravitational
spring
electrostatic
Non-conservative forces: (dependent on path)
friction
air resistance
electrical resistance
These forces will convert mechanical energy into
heat and/or deformation.
Gravitation Potential Energy
Potential energy is associated with the configuration of a
system in which a conservative force acts: ∆U = −W
For a general conservative force F = F(x)
xf v
∆U = − W = − ∫ F( x ) ⋅ dx̂
xi
Gravitational potential energy:
∆U = − ∫ (−mg)dy = mg(y f − y i )
yf
yi
assume Ui = 0 at yi = 0 (reference point)
U(y) = mgy
(gravitational potential energy)
only depends on vertical position
Nature only considers changes in Potential energy
important. Thus, we will always measure ∆U.
So, . . . We need to set a reference point!
The potential at that point could be defined to be
zero (i.e., Uref. point = 0) in which case we drop the
“∆” for convenience.
BUT, it is always understood to be there!
• Potential energy and reference point
A 0.5 kg physics book falls from a table 1.0 m to the ground.
What is U and ∆U if we take reference point y = 0 (assume
U = 0 at y = 0) at the ground?
y
∆U = Uf − Ui = −(mgh)
Physics
y=h
0
The book lost potential energy.
Actually, it was converted to
kinetic energy
h
y=0
• Potential energy and reference point
A 0.5 kg physics book falls from a table 1.0 m to the ground.
What is U and ∆U if we take reference point y = 0 (assume
U = 0 at y = 0) at the table?
y
∆U = Uf − Ui = mg(−h)
Physics
y=0
0
The book still the same lost
potential energy.
h
y = −h
Sample problem 8-1: A block of cheese, m = 2 kg, slides
along frictionless track from a to b, the cheese traveled a
total distance of 2 m along the track, and a net vertical
distance of 0.8 m. How much is Wg?
Easier (or another way)?
b
Wnet
v v
= ∫ Fg ⋅ d r
a
But, ... We don’t know the
angle between F and dr
Sample problem 8-1: A block of cheese, m = 2 kg, slides
along frictionless track from a to b, the cheese traveled a
total distance of 2 m along the track, and a net vertical
distance of 0.8 m. How much is Wg?
c
Split the problem into two parts
since gravity is conservative
}=d
c
Wnet
b
v
v
= ∫ Fg ⋅ dx̂ + ∫ Fg ⋅ dŷ
a
0
c
mg(b−c) = mgd
Elastic Potential Energy
Spring force is also a conservative force
F = −kx
xf
∆U = − W = ∫ (−kx )dx
xi
Uf – Ui = ½ kxf2 – ½ kxi2
Choose the free end of the relaxed spring as the
reference point:
that is: Ui = 0 at xi = 0
U = ½ kx2 Elastic potential energy
Conservation of Mechanical Energy
• Mechanical energy
Emec = K + U
• For an isolated system (no external forces), if there are
only conservative forces causing energy transfer within
the system….
We know: ∆K = W
(work-kinetic energy theorem)
Also:
∆U = −W
(definition of potential energy)
Therefore: ∆K + ∆U = 0 (Kf – Ki) + (Uf – Ui) = 0
therefore K1 + U1 = K2 + U2
Emec,1 = Emec,2
the mechanical energy is conserved
Conservation of mechanical energy
For an isolated system with only conservative forces
( F = mg, F = −kx ) act inside
Emec,1 = Emec,2
K1 + U1 = K2 + U2
½ mv12 + mgy1 + ½ kx12 = ½ mv22 + mgy2 + ½ kx22
The bowling ball pendulum demonstration
Mechanical
energy is
conserved if
there are only
conservative
forces acting
on the system.
Pendulum (1-D) vertically
Find the maximum speed of the mass.
θ
Isolated system with only gravity
(conservative force) acting on it.
L
1
3
y
mg
y=0
Let 1 be at maximum height (y = ymax), 2
2 be at minimum height (y = 0), and 3
y = L(1−cosθ)
be somewhere inbetween max and min.
Pendulum (1-D) vertically
Where does it have the max speed?
θ
L
1. 1: max. height
2. 2: min. height
3. 3: somewhere in between max
and min
4. none of the above
3
y=0
1
y
2
mg
θ
Lcosθ
Find the maximum speed of
the mass.
L
Pendulum (1-D) vertically
L
1
Isolated system with only gravity
(conservative force) acting on it.
0
K1 + U1 = K2 + U2 = Etotal
½
mv12
0
+ mgy1 = ½
mv22
+ mgy2
0
Kmax must be when Umin (= 0)
3
y
2
mg
y=0
In general,
y = L(1−cosθ)
Answer: 2
Pendulum (1-D) vertically
Find the maximum speed of the mass.
Let 1 be at maximum height (y = ymax)
and 2 be at minimum height (y = 0)
Emech,1 = Emech,2 = Etotal
θ
L
y = L(1−cosθ)
K1 + U1 = K2 + U2 = Etotal
y
0
0
mgymax = ½ mvmax2 = Etotal
y=0
⇒ v max = 2gy max = 2gL(1 − cos θ max )
mg
Pendulum (1-D) vertically
Find the speed of the mass as a
function of angle.
Let 1 be at maximum height (y = ymax)
and 2 be at some height (y)
Emech,1 = Emech,2 = Etotal
θ
L
y = L(1−cosθ)
K1 + U1 = K2 + U2 = Etotal
mgymax = ½ mv2 + mgy = Etotal
y
y=0
mg
v = 2g (y max − y ) = 2g(L(1 − cos θ max ) − L(1 − cos θ))
= 2gL(cos θ − cos θ max )
Pendulum (1-D) vertically
Find the speed of the mass as a
function of angle.
Let 1 be at maximum height (y = ymax)
and 2 be at some height (y)
θ
y
y=0
v = 2g (y max − y ) = 2g(L(1 − cos θ max ) − L(1 − cos θ))
= 2gL(cos θ − cos θ max )
U
K
0
−θmax 0
+θmax
0
−θmax
L
θ
mg
Pendulum (1-D) vertically
Find the speed of the mass as a
function of angle.
Let 1 be at maximum height (y = ymax)
and 2 be at some height (y)
θ
y
y=0
U
E
K
0
−θmax 0
+θmax
0
−θmax
L
θ
mg
Potential Energy Curve
• We know
∆U(x) = −W = −F(x) ∆x
Therefore F(x) = −dU(x)/dx
Work done by external force
• When no friction acts within the system, the net
work done by the external force equals to the
change in mechanical energy
W = ∆Emec = ∆K + ∆U
• Friction is a non-conservative force
• When a kinetic friction force acts within the
system, then the thermal energy of the system
changes:
∆Eth = fkd
Therefore
W = ∆Emec + ∆Eth
Work done by external force
• When there are non-conservative forces (like
friction) acting on the system, the net work done
by them equals to the change in mechanic energy
Wnet = ∆Emec = ∆K + ∆U
Wfriction = −fkd
Conservation of Energy
The total energy E of a system can change only by
amounts of energy that are transferred to or from the
system
W = ∆E = ∆Emec + ∆Eth + ∆Eint
If there is no change in internal energy, but friction acts
within the system: W = ∆Emec + ∆Eth
If there are only conservative forces act within the
system: W = ∆Emec
If we take only the single object as the system
W = ∆K
Law of Conservation of Energy
• For an isolated system (W = 0), the total energy E
of the system cannot change
∆Emec + ∆Eth + ∆Eint = 0
• For an isolated system with only conservative
forces, ∆Eth and ∆Eint are both zero. Therefore:
∆Emec = 0
Sample 8-7. In the figure, a 2.0kg package slides along a
floor with speed v1 = 4.0m/s. It then runs into and
compresses a spring, until the package momentarily stops.
Its path to the initially relaxed spring is frictionless, but as
it compresses the spring, a kinetic friction force from the
floor, of magnitude 15 N , act on it. The spring constant is
10,000 N/m. By what distance d is the spring compressed
when the package stops?