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Quick Question
Energy States
Left to their own devices, systems always seek out the
lowest energy state available to them.
Systems want to be at rest or in a constant state of
motion.
You have to do work on
the Rock to roll it back up
the hill. This will give the
Rock Energy – the
potential of rolling back
down – Potential Energy.
Potential Energy
The energy an object has due to its position in a force field.
For example: gravity or electricity
The Potential Energy is relative to a ‘ground’ that is defined.
(Potential Energy: U or PE)
Gravitational Potential Energy
PE = mgh
Force
Distance
It takes work to move the object and that gives it energy!
Same change in
height!
PE = mgh
The “ground”: h = 0
IMPORTANT!
Either path
gives the
same
potential
energy!
WHY?
Work Up an Incline
The block of ice weighs 500 Newtons. How much work does it take
to push it up the incline compared to lifting it straight up?
Ignore friction.
Work Up an Incline
Work = Force x Distance
W = Fd = 500 N ⋅ 3m = 1500 J
3
F = ? F = mg sin θ = 500 N ⋅ = 250 N
6
W = Fd = 250 N ⋅ 6m = 1500 J
Straight up:
Push up:
What is the PE at the top?
mg = 500N
An incline is a simple machine!
1500J
Simple Machines
Force Multipliers
Same Work, Different Force, Different Distance
Chapter 8: Conservation of Energy
Energy can neither be created nor destroyed.
It may change in form or be transferred from
one system to another.
The total amount of energy in the Universe
is constant and can never change.
Ei = E f
Except for VERY brief amounts of time according to the Heisenberg Uncertainty
Principle.
Total Mechanical Energy
The total mechanical energy of a system is defined as the sum
of the kinetic and potential energies:
Emech = K + U
If only conservative forces act, the total mechanical energy is
conserved.
K f + U f = Ki + U i
Work Done by a Conservative Force
Conservative forces do work on a system such that Energy is
exactly transferred between kinetic energy and potential energy,
there is no energy transferred (lost) to friction or heat. Thus
Conservation of Mechanical Energy gives:
Ei = E f
Ki + U i = K f + U f
−(U f − U i ) = K f − K i
−ΔU = ΔK
By the Work-Energy Theorem:
Wc = ΔK
The work done by a conservative force:
Wc = −ΔU
Electrical Potential Energy
The Electric Field does
work on the charge to move
it from A to B, converting
Potential Energy into
Kinetic Energy, same as
with the gravitational field.
The field does work ON the
object as it loses potential
energy and gains Kinetic
Energy:
-
WAB = ΔKE = −ΔPE
ΔPE < 0
-
Conservation of Mechanical Energy
If there are no frictional forces, PE is converted into KE.
Total Energy: 10,000J
Total Energy: 10,000J
Total Energy: 10,000J
Total Energy: 10,000J
Total Energy: 10,000J
Conservative Force of Gravity
Frictionless Ramp
Ei = E f
50J
0J
KEi + PEi = KE f + PE f
25J
25J
0J
50J
Conservation of Mechanical Energy
Potential Energy of a Spring
Kinetic
Kinetic & Potential
Potential
Kinetic
Conservative Forces and
Potential Energy
• Define a potential energy function, U, such
that the work done by a conservative force
equals the decrease in the potential energy
of the system
dU
Fx = −
dx
• The work done by such a force, F, is
xf
WC = ∫ Fx dx = −ΔU
xi
Conservative Forces and
Potential Energy – Check
dU
Fx = −
dx
• Look at the case of a deformed spring
dU s
d ⎛1 2⎞
Fs = −
= − ⎜ kx ⎟ = − kx
dx
dx ⎝ 2
⎠
– This is Hooke’s Law
– Gravitational Potential & Force:
Fg = −
dU g
dx
=−
d
( mgy ) = −mg
dx
HO Problem
1. A single conservative force Fx = (6.0x – 12) N (x is
in m) acts on a particle moving along the x axis. The
potential energy associated with this force is
assigned a value of +20 J at x = 0. What is the
potential energy at x = 3.0 m?
a.
+11 J
b.
+29 J
c.
+9.0 J
xf
WC =
Fx dx = −ΔU
d.
–9.0 J
xi
e.
+20 J
∫
Energy Diagrams and Stable
Equilibrium: Mass on a Spring
dU
Fx = −
dx
• The x = 0 position is one
of stable equilibrium
• Configurations of stable
equilibrium correspond to
those for which U(x) is a
minimum.
• x=xmax and x=-xmax are
called the turning points
Energy Diagrams and Unstable
dU
Equilibrium
F =−
x
• Fx = 0 at x = 0, so the
particle is in equilibrium
• For any other value of x,
the particle moves away
from the equilibrium
position
• This is an example of
unstable equilibrium
• Configurations of unstable
equilibrium correspond to
those for which U(x) is a
maximum. Ex: A pencil
standing on its end.
dx
dU
Fx = −
dx
P7.47
For the potential energy curve shown, (a) determine whether the force Fx is positive, negative, or zero at the five points indicated. (b) Indicate points of stable, unstable, and neutral equilibrium. (c) Sketch the curve for Fx
versus x from x = 0 to x = 9.5 m.
a) Fx is zero at points A, C and E;
Fx is positive at point B and
negative at point D.
Fx
B
b) A and E are unstable, and C is stable.
A
C
E
D
x (m)
Ski Hill Problem
If the skier has an initial velocity of 12m/s,
what is his final velocity at the top of the
ramp?
Ignore Friction.
= 12m / s
Ski Hill Problem
Take the ground to be the ground: U = 0.
Ki + Ui = K f + U f
1
1
2
2
mvi + 0 = mv f + mgh
2
2
v f = v − 2 gh
2
i
9.75m / s
= 12m / s
PE=0
HO Problem
2. A pendulum is made by letting a 2.0-kg object
swing at the end of a string that has a length of 1.5
m. The maximum angle the string makes with the
vertical as the pendulum swings is 30°. What is the
speed of the object at the lowest point in its
trajectory?
a.
2.0 m/s
b.
2.2 m/s
c.
2.5 m/s
d.
2.7 m/s
e.
3.1 m/s
You Try: HO Problem
3.
A certain pendulum consists of a 1.5-kg mass
swinging at the end of a string (length = 2.0 m). At
the lowest point in the swing the tension in the
string is equal to 20 N. To what maximum height
above this lowest point will the mass rise during its
oscillation?
a.
77 cm
b.
50 cm
c.
63 cm
d.
36 cm
e.
95 cm
Nonconservative Forces
Nonconservative forces do work on a system such that
Mechanical Energy is ‘lost’ or transformed into internal energy
(heat) or can’t be directly transformed back into KE or PE.
Conservation of Energy gives:
Energy Bank
Ei = E f
Expenditures
K i + U i = K f + U f − Wnc
Examples: Air resistance, Friction, Applied Forces
(ex : W f = − f ⋅ Δr )
Nonconservative Forces
The change in mechanical energy of a system is due
to the nonconservative forces acting on it.
Ei = E f
K i + U i = K f + U f − Wnc
ΔU + ΔK = Wnc
Ex: Change in energy bank spent on friction:
ΔU + ΔK = − fd
Ski Hill Problem
If the skier has an initial velocity of 12m/s, what
is his final velocity at the top of the ramp? The
coefficient of kinetic friction between the skies
and the hill is 0.13. The mass of the skier is
80kg.
= 12m / s
PE=0
Ski Hill Problem
Ki + Ui = K f + U f −W f
1
1
2
2
mvi + 0 = mv f + mgh + fd
2
2
Wf = − f ⋅ d
v f = v − 2 gh − 2 fd / m
2
i
What is f and d?
= 12m / s
d
μ = .13
m=80kg
PE=0
Ski Hill Problem
v f = v − 2 gh − 2 fd / m
2
i
f = μ mg cos θ , d = h / sin θ
v f = v − 2 gh − μ 2 g cos θ (h / sin θ )
2
i
v f = v − 2 gh(1 + μ cot θ )
2
i
v f = 9.27m / s
(< 9.75m / s )
Does not depend on the mass of the skier!!
HO Problem
4. A 12-kg projectile is launched with an initial vertical
speed of 20 m/s. It rises to a maximum height of 18 m
above the launch point. How much work is done by the
dissipative (air) resistive force on the projectile during
this ascent?
a.
–0.64 kJ
b.
–0.40 kJ
c.
–0.52 kJ
d.
–0.28 kJ
e.
–0.76 kJ
A 2.00‐kg block situated on a rough incline is connected to a spring of negligible mass having a spring constant of 100 N/m. The pulley is frictionless. The block is released from rest when the spring is unstretched. The block moves 20.0 cm down the incline before coming to rest. Find the coefficient of kinetic friction between block and incline. The picture shows
the final state. Take it to be the ground where U=0.
P8.54
h
d
The gain in internal energy due to friction represents a loss in
mechanical energy that must be equal to the change in the
kinetic energy plus the change in the potential energy.
ΔU g + ΔU s + ΔK = Wnc
1 2
−mgh + kd + 0 = − fd
2
1
−mgd sin θ + kd 2 = − μ mg cos θ ⋅ d
2
1 2
kd − mgd sin θ
μ= 2
→
mg cos θ ⋅ d
μ = tan θ −
kd
2mg cos θ
μ = .115
Nonconservative Forces
System of Objects
Change in energy bank due to NC forces!!
Wnc = ΔU1 + ΔU 2 + ΔK1 + ΔK 2
m2
μ ≠0
Wnc < 0
m1
(ex : W f = − f ⋅ Δr )
A 20.0‐kg block is connected to a 30.0‐kg block by a string that passes over a light frictionless pulley. The 30.0‐kg block is connected to a spring that has negligible mass and a force constant of 250 N/m. The spring is unstretched when the system is as shown in the figure, and the incline is frictionless. The 20.0‐kg block is pulled 20.0 cm down the incline (so that the 30.0‐kg block is 40.0 cm above the floor) and released from rest. Find the speed of each block when the 30.0‐kg block is 20.0 cm above the floor (that is, when the spring is unstretched.)
P8.59
x =.2m
m1
m2
Take the equilibrium position shown as the ground for the m2. Take the initial
position the ground for m1. The blocks will have the same final speed.
0
0
0
0
( K i + U gi )1 + ( K i + U gi + U si ) 2 = ( K f + U fg )1 + ( K f + U gf + U sf ) 2
Energy Bank:
(U gi + U si ) 2 = ( K f + U fg )1 + K 2 f
:Expenditures
1 2 1
m2 gx + kx = (m1 + m2 )v 2 + m1 gx sin θ
2
2
v=
2 gx(m2 − m1 sin θ ) + kx 2
m1 + m2
v = 1.24 m s
Energy or Newton’s 2nd Law?
P8.63.A child slides without friction from a height h
along a curved water slide (Fig. P8.63). She is launched from a height h/5 into the pool. Determine her maximum airborne height y in terms of h and .
Energy or Newton’s 2nd Law?
P8.65 Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with man‐eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle with the vertical (Fig. P8.65). Taking D = 50.0 m, F
= 110 N, L = 40.0 m, and = 50.0°, (a) with what minimum speed must Jane begin her swing in order to just make it to the other side? (b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing? Assume that Tarzan has a mass of 80.0 kg.
HO Problem
5. A 1.5-kg block sliding on a rough horizontal surface is
attached to one end of a horizontal spring (k = 200 N/m)
which has its other end fixed. If this system is displaced
20 cm horizontally from the equilibrium position and
released from rest, the block first reaches the equilibrium
position with a speed of 2.0 m/s. What is the coefficient of
kinetic friction between the block and the horizontal
surface on which it slides?
a.
0.34
b.
0.24
c.
0.13
d.
0.44
e.
0.17
Power
• The time rate of energy transfer:
• The average power:
W
P=
dE
P=
dt
Δt
• The instantaneous power is the limiting value of
the average power as Δt approaches zero
P=
lim
Δt →0
W dW
=
dt
Δt
dW
dr
P=
= F ⋅ = F ⋅v
dt
dt
Units
Work
P=
time
Power has a unit of a Watt:
J
Watt =
s
Energy in terms of Power is
Energy = Power ⋅ Time
For example, Kilowatt-hour
You Try: Ch 7 HO 8
A 2.0-kg block slides down a plane (inclined at 40° with the
horizontal) at a constant speed of 5.0 m/s. At what rate is the
gravitational force on the block doing work?
a.
+98 W
b.
+63 W
c.
zero
d.
+75 W
e.
–75 W