Physics for Scientists
and Engineers
Chapter 10
Energy
Spring, 2008
Ho Jung Paik
Introduction to Energy
„
„
„
„
Energy is one of the most important concepts in
science although it is not easily defined
Every physical process that occurs in the Universe
involves energy and energy transfers
The energy approach to describing motion is
particularly useful when the force is not constant
An approach will involve Conservation of Energy
„
This could be extended to biological organisms,
technological systems and engineering situations
31-Mar-08
Paik
p. 2
Energy of Falling Object
„
2
2
v
=
v
Kinematic eq. f
i + 2 a y ( y f − yi )
„
„
Gravity is the only force acting
Multiplying m & putting a y = g ,
1
2
1 2 1 2
mv f = mvi − mg ( y f − yi )
2
2
1 2
1 2
⇒ mv f + mgy f = mvi + mgyi
2
2
„
„
Define K = mv : kinetic energy
U g = mgy : gravitational potential energy
1
2
2
Then, K f + U gf = K i + U gi : Conservation of Energy
31-Mar-08
Paik
p. 3
Kinetic Energy
„
„
Kinetic energy is the energy associated with the
motion of an object
Kinetic energy is always given by K = ½mv 2,
where v is the speed of the object
„
„
„
Kinetic energy is a scalar, independent of direction of v
Kinetic energy cannot be negative
Units of energy: 1 kg m2/s2 ≡ 1 Joule
„
Energy has dimensions of [M L2 T−2]
31-Mar-08
Paik
p. 4
Potential Energy
„
Potential energy is the energy associated with the
relative position of objects that exert forces on
each other
„
„
„
A potential energy can only be associated with specific
types of forces, i.e. conservative forces
There are many forms of potential energy:
gravitational, electromagnetic, chemical, nuclear
Gravitational potential energy is associated with
the distance above Earth’s surface: Ug = mgy
31-Mar-08
Paik
p. 5
Total Mechanical Energy
„
„
The sum of kinetic and potential energies is the
mechanical energy of the system: E = K + ∑Ui
When conservative forces act in an isolated system,
kinetic energy gained (or lost) by the system as its
members change their relative positions is balanced
by an equal loss (or gain) in potential energy
„
„
This is the Conservation of Mechanical Energy
If there is friction or air drag, the total mechanical energy
is not conserved, i.e. constant
31-Mar-08
Paik
p. 6
Gravitational Potential Energy
„
„
„
„
The system consists of Earth and a book
Initially, the book is at rest (K = 0) at y =
yb (Ug = mgyb)
When the book falls and reaches ya,
Ka + mgya = 0 + mgyb
⇒ Ka = mg (yb - ya) >0
The kinetic energy that was converted from potential
energy depends only on the relative displacement yb - ya
„ It doesn’t matter where we take y = 0
31-Mar-08
Paik
p. 7
Example 1: Launching a Pebble
Bob uses a slingshot to shoot a 20.0 g
pebble straight up with a speed of 25.0 m/s.
How high does a pebble go?
1 2
Before : K b = mv = 6.25 J, U gb = 0
2
After : K a= 0 , U ga = mgy = 0.196 ya
Using energy conservation, K a + U ga = K b + U gb
0 + 0.196 ya = 6.25 + 0 , or ya = 31.9 m
Kinetic energy was converted into gravitational
potential energy by 6.25 J.
31-Mar-08
Paik
p. 8
Energy Conservation in 2-D
We will now analyze 2 - D motion
along a frictionless incline.
dvs
∑ Fs = mas = m ,
dt
dv
dv ds
− mg sin θ = m s
= mvs s
ds
ds dt
Multiplying both sides by ds, − mg sin θds = mvs dvs
But sin θds = dy. Integrating from initial to final positions,
1 2
1 2
mv f + mgy f = mvi + mgyi
2
2
This is the same equation as in a 1 - D motion.
31-Mar-08
Paik
p. 9
Example 2: Downhill Sledding
Christine runs forward with her
sled at 2.0 m/s. She hops onto
the sled at the top of a 5.0 m
high, very slippery slope.
What is her speed at the bottom?
Using the conservation of energy,
1 mv 2 + mgy = 1 mv 2 + mgy
0
1
2 0
2 1
v1 = v02 + 2 g ( y0 − y1 ) = 10 m/s
Notice that (1) the mass cancelled out and (2) only Δy is needed.
We could have taken the bottom of the hill as y = 10.0 m and the top as
y = 15.0 m, and gotten the same answer.
31-Mar-08
Paik
p. 10
Energy Conservation: Pendulum
„
„
„
As the pendulum swings, there is a
continuous exchange between
potential and kinetic energies
At A, all the energy is potential
At B, all of the potential energy at A
is transformed into kinetic energy
„ Let zero potential energy be at B
ƒ At C, the kinetic energy has been transformed back into
potential energy
31-Mar-08
Paik
p. 11
Example 3: Ballistic Pendulum
A 10 g bullet is fired into a 1200 g wood block hanging from a
150 cm long string. The bullet embeds itself into the block,
and the block then swings out to an angle of 40°.
What was the speed of the bullet?
31-Mar-08
Paik
p. 12
Example 3, cont
This problem has two parts: (a) the bullet and the block make a perfectly
inelastic collision, in which momentum is conserved, and (b) the total mass
as a system undergoes a pendulum motion, in which energy is conserved.
(a) Momentum conservation : (mw + mb )v1 = mw v w 0 + mb vb 0
Since vw 0 = 0, vb 0 =
mw + mb
v1
mb
1
1
mtot v22 + mtot gy2 = mtot v12 + mtot gy1
2
2
Since v2 = 0 and we can choose y1 = 0, v1 = 2 gy2 = 2 gL(1 − cos θ )
(b) Energy conservation :
mw + mb
Therefore, vb 0 =
mb
=
31-Mar-08
2 gL(1 − cos θ )
1.210 kg
2(9.80 m/s 2 )(1.50 m )(1 − 0.766 ) = 320 m/s
0.010 kg
Paik
p. 13
Example 4: Roller Coaster
A roller-coaster car is
released from rest at
the top of the first rise
and then moves freely
with negligible friction.
The roller coaster has a
circular loop of radius R
in a vertical plane.
(a) Suppose first that the car barely makes it around the loop:
at the top of the loop the riders are upside down and feel
weightless, i.e. n = 0. Find the height of the release point
above the bottom of the loop, in terms of R so that n = 0.
31-Mar-08
Paik
p. 14
Example 4, cont
(a) At the top of the loop, the car and riders
are in free fall. From Newton' s 2nd law,
v2
− mg = −m
⇒ v = gR
R
Energy is conserved between the release
and the top of the loop :
K i + U gi = K f + U gf
1 2
0 + mgh = mv + mg (2 R)
2
1
gh = gR + 2 gR ⇒ h = 2.5R
2
31-Mar-08
Paik
p. 15
Example 4, cont
(b) Show that the normal force on the car at the bottom of the
loop exceeds the normal force at the top of the loop by six
times the weight of the car.
1 2
mvb
2
vb2 = 2 gh
(b) Energy conservation : (bottom) mgh =
⇒
1 2
mvt + mg (2 R )
2
vt2 = 2 gh − 4 gR
(top) mgh =
vb2
vt2
Newton' s 2nd law : (bottom) nb − mg = + m
(top) − nt − mg = − m
R
R
Substituting vb2 and vt2 ,
nb = mg (1 + 2h / R )
nt = mg (− 5 + 2h / R )
⇒
nb − nt = 6mg
The normal force on each rider follows the same rule. Such a large normal
force is very uncomfortable for the riders. Roller coasters are therefore
built helical so that some of the gravitational energy goes into axial motion.
31-Mar-08
Paik
p. 16
Hooke’s Law
„
„
Force exerted by a spring is
always directed opposite to the
displacement from equilibrium
Hooke’s Law: Fs = −kΔs
„
„
„
„
Δs is the displacement from the
equilibrium position
k is the spring constant and
measures the stiffness of the spring
F is called the restoring force
If it is released, it will oscillate back and forth
31-Mar-08
Paik
p. 17
Measuring the Force Constant
∑ F = F − mg = ma
y
s
y
At equilibrium,
a y = 0, − kΔs − mg = 0
Therefore,
mg
(N/m )
k=
Δs
Measure m and Δs to determine k .
31-Mar-08
Paik
p. 18
Energy of Object on a Spring
dvs
∑ Fs = mas , − k ( s − se ) = m
dt
Use the chain rule,
dvs ds
dvs
− k ( s − se ) = m
= mvs
ds dt
ds
or − k ( s − se )ds = mvs dvs
Define u = s − se = Δs, hence du = ds.
Integrating,
Δs f
sf
1 2 Δs f
1
1
2
2
− ∫ k ( s − se )ds = − ∫ kudu = − ku Δs = − k (Δs f ) + k (Δsi )
i
2
2
2
Δsi
si
vf
1 2 1 2
1 2 1
1 2 1
2
2
(
)
(
)
mv
dv
=
mv
−
mv
⇒
mv
+
k
Δ
s
=
mv
+
k
Δ
s
i
i
f
f
∫ s s 2 f 2 i
2
2
2
2
vi
31-Mar-08
Paik
p. 19
Elastic Potential Energy
„
Potential energy of an object attached to a
spring is
Us =½ k(Δs)2
„
„
The elastic potential energy depends on the
square of the displacement from the spring’s
equilibrium position
The same amount of potential energy can be
converted to kinetic energy if the spring is
extended or contracted by the same displacement
31-Mar-08
Paik
p. 20
Example 5: Spring Gun Problem
A spring-loaded toy gun launches a 10 g
plastic ball. The spring, with spring
constant 10 N/m, is compressed by 10
cm as the ball is pushed into the barrel.
When the trigger is pulled, the spring is
released and shoots the ball out.
What is the ball’s speed as it leaves the
barrel? Assume friction is negligible.
1 2 1
1
1
1
1
2
2
2
mvi + k ( Δxi ) = mv 2f + k (Δx f ) , 0 + k (x1 − xe ) = mv 2f + 0
2
2
2
2
2
2
k ( x1 − xe ) 2
(10 N/m)(−0.10 m − 0 m) 2
vf =
=
= 3.16 m/s
m
0.010 kg
31-Mar-08
Paik
p. 21
Example 6: Pushing Apart
A spring with spring constant 2000 N/m is sandwiched
between a 1.0 kg block and a 2.0 kg block on a frictionless
table. The blocks are pushed together to compress the spring
by 10 cm, then released.
What are the velocities of the blocks as they fly apart?
31-Mar-08
Paik
p. 22
Example 6, cont
From energy conservation,
1
2
mv12,i + 12 mv22,i + 12 k (Δxi ) 2 = 12 mv12,f + 12 mv22,f + 12 k (Δx f ) 2
0 + 0 + 12 k (Δxi ) 2 = 12 mv12,f + 12 mv22,f + 0
From momentum conservation,
m2
m1v1,i + m2 v2 ,i = m1v1,f + m2 v2 ,f , 0 = m1v1,f + m2 v2 ,f ⇒ v1,f = − v2 ,f
m1
Substituting this into the energy equation,
2
1
2
⎛ m2
⎞ 1
⎛ m2 ⎞ 2
2
2
1
⎟⎟v2 ,f = k (Δx1 ) 2
m1 ⎜⎜ v2 ,f ⎟⎟ + 2 m2 v2 ,f = 2 k (Δx1 ) , m2 ⎜⎜1 +
⎝ m1
⎠
⎝ m1 ⎠
v2 ,f
k (Δx1 ) 2
m
=
= 1.8 m/s, v1,f = − 2 v2 ,f = −3.6 m/s
m2 (1 + m2 m1 )
m1
31-Mar-08
Paik
p. 23
Example 7: Spring and Gravity
A 10 kg box slides 4.0 m down
the frictionless ramp shown in
the figure, then collides with a
spring whose spring constant is
250 N/m.
(a) What is the maximum compression of the spring?
(b) At what compression of the spring does the box have its
maximum velocity?
31-Mar-08
Paik
p. 24
Example 7, cont
(a) Choose the origin of the
coordinate system at the
point of maximum
compression. Take the
coordinate along the
ramp to be s.
K 2 + U s 2 + U g 2 = K 1 + U s1 + U g 1
0 + 12 k (Δs ) + 0 = 0 + 0 + mg (4.0 m + Δs ) sin 30o
2
(125 N/m )(Δs ) − (49 kg m/s )Δs − 196 kg m /s
2
2
2
2
=0
Δs = 1.46 m, − 1.07 m (unphysical)
31-Mar-08
Paik
p. 25
Example 7, cont
(b) For this part of the problem, it
is easiest to take the origin at
the point where the spring is
at its equilibrium position.
K 2 + U s 2 + U g 2 = K1 + U s1 + U g 1
1
2
1
2
mv + mgy + k (Δs ) = 0 + mgy1 + 0
2
2
1
2
k (Δs ) − (mg sin 30o )(Δs ) + 12 mv 2 − mg (4.0 m sin 30o ) = 0
2
To find vmax wrt Δs, take the derivative of this equation and put
mg sin 30o
dv
= 0, Δs =
= 0.196 m
kΔs − mg sin 30 + mv
k
dΔs
dv
= 0.
dΔ s
o
31-Mar-08
Paik
p. 26
Elastic Collisions
„
Both momentum and kinetic
energy are conserved (U = 0)
m1 v1i + m2 v 2i = m1 v1f + m2 v 2f
1
1
1
1
2
2
2
m1v1i + m2 v2 i = m1v1f + m2 v22f
2
2
2
2
„
Typically, there are two unknowns (v2f and v1f), so
you need both equations: pi = pf and Ki = Kf
„
„
The kinetic energy equation can be difficult to use
With some algebraic manipulation, the two equations
can be used to solve for the two unknowns
31-Mar-08
Paik
p. 27
1-D Elastic Collisions
1
1
1
1
2
2
2
Kf = Ki :
m1v1f + m2 v2 f = m1v1i + m2 v22i
2
2
2
2
⇒ m1 (v1f − v1i )(v1f + v1i ) = − m2 (v2f − v2i )(v2f + v2i )
pf = pi :
⇒
m1v1 f + m2 v2 f = m1v1i + m2 v2i
m1 (v1f − v1i ) = −m2 (v2f − v2i )
Combine the two equations, (v1f + v1i ) = (v2f + v2i )
Solve for v1f or v2f and substituting into the momentum equation,
⎛ m1 − m2 ⎞
⎛ 2m2 ⎞
⎛ 2m1 ⎞
⎛ m2 − m1 ⎞
⎟⎟v1i + ⎜⎜
⎟⎟v2i , v2 f = ⎜⎜
⎟⎟v1i + ⎜⎜
⎟⎟v2i
v1 f = ⎜⎜
⎝ m1 + m2 ⎠
⎝ m1 + m2 ⎠
⎝ m1 + m2 ⎠
⎝ m1 + m2 ⎠
31-Mar-08
Paik
p. 28
1-D Elastic Collisions, cont
(1) m1 = m2 :
v1f = (0)v1i + (1)v2i = v2i , v2f = (1)v1i + (0)v2i = v1i
⇒ Particles exchange velocities.
(2) m1 << m2 , v2 i = 0 :
v1f ≈ (−1)v1i + (2)0 ≈ −v1i , v2f ≈ (0)v1i + (1)0 = 0
⇒ m1 bounces back while m2 remains stationary
(3) m1 >> m2 , v2 i = 0 :
v1f ≈ (1)v1i + (0)0 = v1i , v2f ≈ (2)v1i + (−1)0 = 2v1i
⇒ m1 continues to move at v1i while m2 moves at 2v1i
31-Mar-08
Paik
p. 29
2-D Elastic Collisions
„
Energy conservation:
½ m1v1i2 = ½ m1v1f2 + ½ m2v2f2
„
Momentum conservation:
x: m1v1i = m1v1f cosθ +m2v2f cosφ
y: 0 = m1v1f sinθ +m2v2f sinφ
„
Note that we have 3 equations
and 4 unknowns
„
We cannot solve for the unknowns
unless we have one final angle or
final velocity given
31-Mar-08
Paik
p. 30
Example 8: Billiard Ball
A player wishes to sink target ball
in the corner pocket. The angle
to the corner pocket is 35°.
At what angle is the cue ball
deflected?
Energy conservation :
1
1
1
2
2
m1v1i = m1v1 f + m2 v22 f
2
2
2
Momentum conservation :
m1v1i = m1v1 f cosθ + m2 v2 f cos 35°
0 = − m1v1 f sinθ + m2 v2 f sin 35°
31-Mar-08
Paik
p. 31
Example 8, cont
Since m1 = m2 ,
v12i = v12f + v22 f , v1i = v1 f cosθ + v2 f cos 35°, 0 = − v1 f sinθ + v2 f sin 35°
sin 35°
From the third equation, v1 f = v2 f
.
sin θ
Substituting this into the first and second equations,
2
sin
35° ⎞ 2
⎛
⎛ sin 35°
⎞
2
+ 1⎟ v2 f , v1i = ⎜
cosθ + cos 35° ⎟v2 f
v1i = ⎜
2
⎝ sin θ
⎠
⎠
⎝ sin θ
2
2
⎛ sin 35°
⎞ sin 35°
From these, ⎜
cosθ + cos 35° ⎟ =
+ 1.
2
sin θ
⎝ sin θ
⎠
This leads to cot θ = tan 35° or θ = 55°.
31-Mar-08
Paik
p. 32
Internal Energy
„
„
„
So far we have considered situations where
there is no friction
In the absence of friction, the mechanical
energy is conserved: E = K + U = constant
Where friction is present, some energy is
converted to internal energy, usually heat
„
Here K + E ≠ constant
31-Mar-08
Paik
p. 33
Example 9: Bullet Fired into Block
A 5.00-g bullet moving with an
initial speed of 400 m/s is fired
into and passes through a 1.00kg block. The block, initially at
rest on a frictionless, horizontal
surface, is connected to a spring
of force constant 900 N/m. The
block moves 5.00 cm to the
right after impact.
Find (a) the speed at which the bullet emerges from the block
and (b) the mechanical energy converted into internal energy
in the collision.
31-Mar-08
Paik
p. 34
Example 9, cont
(a) Energy conservation for the spring: 1 MVi 2 = 1 kx 2
2
2
kx
(900 N/m)(0.05 m)
Vi =
=
= 1.50 m/s
M
1.00 kg
2
2
Momentum conservation in the collision : mvi = MVi + mv
mvi − MVi (5.00 × 10 −3 kg)(400 m/s) − (1.00 kg)(1.5 m/s)
v=
=
= 100 m/s
−3
m
5.00 × 10 kg
[
(b) ΔE = ΔK + ΔU = 12 (5.00 ×10 −3 kg ) (100 m/s )2 − (400 m/s )2
[
]
]
+ 12 (900 N/m ) (5.00 ×10 − 2 m ) − (0 m ) = −374 J
2
2
The lost energy is due to the friction between the bullet
and block. The block heats up a little.
31-Mar-08
Paik
p. 35
Energy Diagrams
ƒ Particle under a free fall
31-Mar-08
Paik
p. 36
Energy Diagrams, cont
ƒ Particle attached to a spring
31-Mar-08
Paik
p. 37
General Energy Diagram
Equilibria are at points where
dU/dx = 0. At equilibria, F = 0.
Away from equilibria, dU/dx ≠ 0
and F ≠ 0.
31-Mar-08
Paik
p. 38
Molecular Bonding
„
Potential energy associated with the force
between two neutral atoms in a molecule can
be modeled by the Lennard-Jones function:
31-Mar-08
Paik
p. 39
Force Acting in a Molecule
„
„
The force is repulsive (positive) at small separations
The force is zero at the point of stable equilibrium
„
„
„
This is the most likely separation between the atoms in the molecule
(2.9 x 10-10 m at minimum energy)
The force is attractive (negative) at large separations
At great distances, the force approaches zero
31-Mar-08
Paik
p. 40
Example 10: Hemispherical Hill
A sled starts from rest at
the top of the frictionless,
hemispherical, snowcovered hill shown in the
figure.
(a) Find an expression for the sled's speed when it is at
angle φ.
(b) Use Newton's laws to find the maximum speed the sled
can have at angle φ without leaving the surface.
(c) At what angle φmax does the sled "fly off" the hill?
31-Mar-08
Paik
p. 41
Example 10, cont
(a) K 1 + U 1 = K 0 + U 0
1
2
mv12 + mgy 1 = 12 mv 02 + mgy 0
1
2
mv12 + mgR cos φ = 12 mv 02 + mgR
v1 = 2 gR (1 − cos φ )
mv 2
v2 ⎞
⎛
⇒ n = m ⎜ g cos φ − ⎟
(b) ∑ Fr = ma r , n − mg cos φ = −
R
R⎠
⎝
n decreases as v increases. When n = 0, the sled leaves the hill.
n ≥ 0 ⇒ v max =
gR cos φ max
(c) We have v for an arbitrary angle and vmax . Equating the two,
2
2
2 gR (1 − cos φ max ) = gR cos φ max , cos φ max = ⇒ φ max = cos −1 = 48 .2 o
3
3
31-Mar-08
Paik
p. 42