Transcript Energy - Ohio Wesleyan University

Energy
• Concept of energy important due to principle of
conservation of energy
– Energy can be converted from one form to another, but
cannot be created or destroyed
• There are many different kinds of energy (thermal,
chemical, electromagnetic, etc.)  we’ll focus on
mechanical energy, specifically energy of motion
(kinetic energy)
• Using energy considerations, you can solve
problems where you can’t use constantacceleration formulas because the applied forces
aren’t constant
• Definition of work: A quantity that depends on the
product of an applied force and the displacement of
a body while it experiences the force
Work
• Mathematically:
– Work = (component of force in direction of displacement)
 (magnitude of displacement) OR
– Work = (magnitude of force)  (component of
displacement in direction of force)
• Can be summarized by: W  Fx cos 
• Work is a scalar quantity with units of Nm = J
(Joule) in SI units
• Work can be positive,
or zero depending
 negative,

on the direction of F and x
Sign of W

0°  90°
90°  180°
+
–
90°
0
Work
• Pushing an eraser along a table:
F
f

x
N
W
• What is the sign of the work done during the
displacement …
– by force F on eraser: +
– by friction on eraser: –
– by gravitational force on eraser: 0
– by eraser on hand pushing eraser: –
– Now lift the eraser. What is the sign of the work done by
the gravitational force on the eraser: –
• Be careful to designate work done by what force
acting on what object
Example Problem #5.5
Starting from rest, a 5.00-kg block slides 2.50 m down a rough
30.0° incline. The coefficient of kinetic friction between the
block and the incline is mk = 0.436. Determine (a) the work
done by the force of gravity, (b) the work done by the friction
force between block and incline, and (c) the work done by
the normal force. (d) Qualitatively, how would the answers
change if a shorter ramp at a steeper angle were used to
span the same vertical height?
Partial solution (details given in class):
(a) 61.3 J
(b) – 46.3 J
(c) 0
Total Work
• When multiple forces are acting on a body, calculate
the total work done on the body by:
1. Wtot = Sum of individual works due to each force acting
on the body
2. Calculate
net force (from vector sum) on body and

use Fnet in the definition of work shown previously
• Method #1 tends to be easier
• Total work done on a body is related to changes in
the speed of the body:
– Wtot > 0: Object speeds up
– Wtot < 0: Object slows down
– Wtot = 0: Object maintains speed
• Let’s find a quantitative relationship between Wtot
and speed
Total Work and Kinetic Energy
• For a constant net force directed along direction of
Fnet  ma
motion: v2  v02  2ax
• Putting these 2 equations
together to eliminate a:
Wtot
1 2 1 2
Fnet x  mv  mv 0
2
2
K2
K1
• ½ mv2 has units of energy (J); we define it to be the
kinetic energy K of a body (energy due to motion)
– Never negative, 0 only when v = 0
• From above: Wtot  K2  K1  K (Work–Energy Theorem)
– Wtot > 0  K2 > K1  v increases
– Wtot < 0  K2 < K1  v decreases
– Wtot = 0  K2 = K1  v unchanged
CQ1: Interactive Conceptual Example:
Battle of the Kinetic Energies
Which block (m1 or m2) has the larger kinetic
energy at the finish line?
A) m1
B) m2
C) Their kinetic energies are the same at the
finish line.
D) Neither block has kinetic energy at the
finish line.
PHYSLET Exercise 7.1.3, Prentice Hall (2001)
Example Problem #4.50
A car is traveling at 50.0 km/h on a flat highway.
(a) If the coefficient of friction between road and tires
on a rainy day is 0.100, what is the minimum
distance in which the car will stop?
(b) What is the stopping distance when the surface is
dry and the coefficient of friction is 0.600?
Solution is presented on the next slide.
v0
Example Problem #4.50

x
N
v=0
f
mg
Wtot = K2 – K1 = ½ m(v22 – v12) = –½ mv02 (since v2 = 0)
 Wtot = Wf + WN + Wmg = Wf
 Wf = fx cos180° = –fx = – (mN)x = –mmgx
 –mmgx = –½ mv02  x = v02 / 2mg
(a) For v0 = 50.0 km/h = 13.9 m/s, m = 0.100:
x = 98.6 m
(b) For m = 0.600:
x = 16.4 m
Potential Energy
• A body gains or loses kinetic energy (energy of
motion) because of work done by external forces
acting on it:
1
K  mv22  v12   Wtot  Fx cos 
2
• In some situations, it seems as if this energy were
stored somewhere in a “hidden” location
• Consider motion on a swing:
v=0
K=0
(max
height)
v = vmax
K = Kmax
v=0
K=0
(max
height)
Potential Energy
• It seems as though the kinetic energy is converted
to some other form as you swing toward the highest
points in the motion
– Points to idea of energy associated with position of a
body
– This kind of energy is a measure of the potential (or
possibility) of work to be done
• Potential Energy = stored energy that
can be converted to other forms of
energy
• Consider the fall of a book
toward the floor:
Gravitational Potential Energy
• Once the book begins to fall, gravity does work on it,
accelerating it toward the ground
• Work done by gravity on the book:
Wg  Fg y cos  Fg y  mgy
• More generally: Wg  mgyi  y f 
• Product of weight mg and vertical height y is defined
as the gravitational potential energy: U  mgy
g
– Has units of kgm2/s2 = J
• Work done by gravity can thus be interpreted as a
change in the gravitational potential energy:
Wg  mg y1  y2   U1 U2
Path independent – all that matters is change in vertical position
Gravitational Potential Energy
• When a body moves up:
– Vertical position increases relative to ground
– Work done by gravity on the body is negative
– Gravitational potential energy increases
• When a body moves down:
– Vertical position decreases relative to ground
– Work done by gravity on the body is positive
– Gravitational potential energy decreases
• Going back to the Work – Energy Theorem:
Wtot  K 2  K1
• If gravity is the only force acting on the object:
Wtot  Wg  U1  U 2
Total Mechanical Energy
• Putting these 2 equations together, we get:
K2 – K1 = U1 – U2
• We can rewrite this as: K1 + U1 = K2 + U2 or
1 2
1 2
mv1  mgy 1  mv 2  mgy 2
2
2
(if only gravity does work)
• Both sides of this equation must equal a constant
since equation must hold for 2 arbitrary vertical points
y1 (object having velocity v1) and y2 (object having
velocity v2)
• This constant is defined as the total mechanical
energy of the system = E = K + U
– “System” means the body of mass m + the earth since
gravitational potential energy is a shared property of both
Conservation of Mechanical Energy
• Since E = constant, we say that total energy is
conserved, and this is our first example of the
conservation of mechanical energy
• Note that conservation of total mechanical
energy depends on the change in gravitational
potential energy
– It’s the difference in gravitational potential energy
that is physically significant
– Thus it doesn’t matter what height we choose to
be y = 0, the origin of the vertical coordinate
– We can define Ug = 0 at whatever height we
choose
CQ2: A ball is thrown multiple times from the top of a
building with the same initial height and with the same
initial speed, but with a varying launch angle relative to
the horizontal. How does the speed of the ball when it
hits the ground depend on the launch angle, neglecting
air resistance?
Projectile Launch Interactive
A) It increases with increasing launch angle.
B) It decreases with increasing launch angle.
C) It will increase only if the ball is thrown at an
angle below the horizontal.
D) It does not depend on launch angle.
CQ3: A 4-kg ball is thrown straight into the
air at 6 m/s. How high does it travel?
A)
B)
C)
D)
1.0 m
1.6 m
1.8 m
2.0 m
Design of a Loop–the–Loop Roller Coaster
Suppose we wish to design the following Loop–the–
1
Loop roller coaster:
2
y
0
y1 = H
R
y2 = 2R
What is the speed of the roller coaster car at the top of
the loop (pt. 2)? (Assuming cars fall under influence of
gravity only.)
Conservation of mechanical energy:
½ mv12 + mgy1 = ½ mv22 + mgy2
Assume that roller coaster starts from rest at top of
hill. Then we have: mgH = ½ mv22 + mg(2R)
v22 = 2mg(H – 2R) / m = 2g(H – 2R)
Design of a Loop–the–Loop Roller Coaster
v2 = [2g(H – 2R)]1/2
(check the next time you’re at Cedar Point or King’s
Island!)
Work done by Varying Forces
• Examples of forces that vary in magnitude with
distance:
– Restoring force of spring
– Gravity
• Example: Force directed along
line of motion but varies in
strength:
Fx

x
W ≈ Fx ave,1x + Fx ave,2 x + …
W = area under Fx vs. x curve
Springs
–x
F x=0
(Spring compressed)
x=0
(Spring relaxed)
x
x=0 F
(Spring stretched)
• The spring exerts a restoring force on the block:
Fs  kx (Hooke’s Law) Hooke's Law Interactive
– k = spring constant (measures stiffness of spring)
– “Slinky” has k = 1 N/m; auto suspensions have k = 105 N/m
• Work done on spring to stretch:
1 2
W  kx
2
• Work done by spring on block while
stretching:
W = –½ kx2

s
F
F 
s
Springs
• This expression comes from finding the area under
the curve of F = kx (applied force on spring Fapp vs. x):
• Avoid pitfall of saying: W = Fx = (kx)(x) = kx2
• This is wrong because F is not constant in x
• Writing F = F(x) may help remind you
Elastic Potential Energy
• Is there potential energy stored in springs? Yes!
x1
x2
F
x=0
(Spring stretched to position x1)
x=0 F
(Spring stretched to position x2)
• The work done on the block by the spring is:
1 2 1 2
W  kx1  kx 2
2
2
• Similar to gravity, we can define an elastic potential
energy
1 2 such that W = U1,el – U2,el
U el 
2
kx
Elastic Potential Energy
Uel
• Graphically:
x<0
(spring compressed)
0
x
x>0
(spring extended)
• Note that:
– Uel is > 0 always
– Uel = 0 at x = 0 (relaxed position)
– We do not have freedom to pick x = 0 wherever we want,
in order to be consistent with U = ½ kx2
• If the elastic force is the only force that does work:
Wtot = Wel = K2 – K1 = U1,el – U2,el (from the Work–
Energy Theorem)
Conservative and Nonconservative Forces
• Rearranging terms: K1 + U1,el = K2 + U2,el
• Or: ½ mv12 + ½ kx12 = ½ mv22 + ½ kx22
• In other words, the total mechanical energy E = K +
U is conserved when the elastic force is the only
force that does work
Spring Energy Interactive
• Mechanical energy is always conserved if only
conservative forces are doing work (gravity, elastic
forces, electromagnetic forces)
– Work done by these forces are independent of path and
depend only on the starting and stopping points
– Can always write work done in terms of potential energy
• In general, nonconservative forces are doing work
as well (“Wnc”)
– In these cases, the process of converting kinetic to
potential energy is not reversible
Conservative and Nonconservative Forces
• Some nonconservative forces cause mechanical
energy to be lost (“dissipative” forces)
– Kinetic friction, fluid/air resistance
• Other nonconservative forces cause an increase in
mechanical energy
– Chemical reaction in firecracker causing fragments to
shoot off
• In either case, we cannot write a potential energy
function representative of these forces
• However, total energy is always conserved if we
consider the energy that changes form from one
type to another
• In general: K1  U1, g  U1,el  Wnc  K2  U2, g  U2,el
(from the Work – Energy Theorem)
CQ4: A 2-kg ball and an 8-kg ball are placed on
separate springs, each with the same spring
constant. The springs are compressed by the
same distance and released. Which of the
following is true about the maximum heights
reached by the balls?
A)
B)
C)
D)
The 2-kg ball will go four times as high.
The 2-kg ball will go twice as high.
The balls will reach equal maximum heights.
The 8-kg ball will go four times as high.
CQ5: Objects A and B are placed on the spring as
shown. Object A has twice as much mass as
object B. If the spring is depressed and released,
propelling the objects into the air, object A will:
A)
B)
C)
D)
rise one fourth as high as object B.
rise half as high as object B.
rise to the same height as object B.
rise twice as high as object B.
Example Problem #5.39
The launching mechanism of a toy gun consists of a spring of
unknown spring constant, as shown in the figure. If the
spring is compressed a distance of 0.120 m and the gun
fired vertically as shown, the gun can launch a 20.0-g
projectile from rest to a maximum height of 20.0 m above
the starting point of the projectile. Neglecting all resistive
forces, (a) describe the mechanical energy
transformations that occur from the time the gun is fired
until the projectile reaches its maximum height, (b)
determine the spring constant, and (c) find the speed of
the projectile as it moves through the equilibrium position
of the spring (where x = 0), as shown in the figure.
Partial solution (details given in class):
(b) 544 N/m
(c) 19.7 m/s
CQ6: Interactive Example Problem:
Inverse Bungee Jumper
What is the necessary spring constant for the
spring?
A)
B)
C)
D)
250 N/m
330 N/m
360 N/m
390 N/m
ActivPhysics Online Problem 5.4
Example Problem #5.48
In a circus performance, a monkey is strapped
to a sled and both are given an initial speed
of 4.0 m/s up a 20° inclined track. The
combined mass of monkey and sled is 20 kg,
and the coefficient of kinetic friction between
sled and incline is 0.20. How far up the
incline do the monkey and sled move?
Solution (details given in class):
1.5 m
Example Problem #5.71
Two objects are connected by a light string
passing over a light, frictionless pulley as
shown in the figure. The 5.00-kg object
is released from rest at a point 4.00 m
above the floor.
(a) Determine the speed of each object
when the two pass each other.
(b) Determine the speed of each object at
the moment the 5.00-kg object hits the
floor.
(c) How much higher does the 3.00-kg
object travel after the 5.00-kg object hits
the floor?
Solution (details given in class):
(a) 3.13 m/s
(b) 4.43 m/s
(c) 1.00 m
Power
• Definition of work says nothing about the passage
of time, but sometimes we want to know how
quickly work is done
• Power = time rate at which work is done (scalar
quantity)
• Average work done per unit time = Average power
= Pave = W / t (W = work done during time
interval t)
• Can also write Pave = Fvave
• Units of power: 1 J/s = 1 Watt (W)
Elevator Power Demo
CQ7: A winch is used to lift heavy objects to the top of a
building under construction. A winch with a power of 50
kW was replaced with a new winch with a power of 100
kW. Which of the following statements about the new
winch is NOT true?
A) The new winch can do twice as much work in the
same time as the old winch.
B) The new winch takes twice as much time to do the
same work as the old winch.
C) The new winch can raise objects with twice as
much mass at the same speed as the old winch.
D) The new winch can raise objects with the same
mass at twice the speed of the old winch.