Download CHAPTER 7 CONSERVATION OF ENERGY • Conservation of

CONSERVATION OF MECHANICAL ENERGY
Consider the motion of an object in one dimension. In the
previous chapter we found that the work done by a force
moving the object from point 1 to point 2 is
2
CHAPTER 7
CONSERVATION OF ENERGY
• Conservation of mechanical energy
• Conservation of total energy of a system
! various examples
• Origin of friction
W12 = ∫ F.dx = ΔK = K2 − K1,
1
where K is the kinetic energy. We also found that if the
force is conservative, then
2
∫ F.dx = −ΔU = −(U2 − U1 ),
1
where U is the potential energy.
∴ΔK = −ΔU = K2 − K1 = −(U2 − U1 ),
K2 + U2 = K1 + U1.
If define the total energy as E = K + U,
i.e.,
then
ΔE = (K 2 − K1 ) + (U2 − U1 ) = 0,
i.e., the total energy remains unchanged. This is a
statement of the conservation of energy. Let’s see how this
works with a specific example.
Consider a system consisting of an object and the Earth.
We release the object from a height h above the ground.
v1 = 0
We take the ground as the zero of potential energy.
h
U1 = mgh : K1 = 0
v1 = 0
h
U = mgy : K =
1
mv 2
2
y
1
U2 = 0 : K 2 = mv 2 2
2
v2
0
At the bottom: (v =
v !2
2
+ 2aΔy) ⇒ v 2 = 2gh
1
∴K2 = mv 22 = mgh .
2
At any position y: v 2 = 2g(h − y)
1
∴K = mv 2 = mg(h − y).
2
U = mgy : K =
1
mv 2 = mg(h − y)
2
y
v2
v
2
v
U1 = mgh : K1 = 0
U2 = 0 : K 2 =
1
mv 2 2 = mgh
2
But we note that
K1 + U1 = K2 + U2 = K + U,
i.e., K + U = constant ⇒ E mech .
We define E mech as the
total mechanical energy of the system
(i.e., the sum of the potential and kinetic energies).
In this case: E mech = mgh , the initial energy of the
system and the total mechanical energy remains constant
with this value while the object falls.
See how the individual parts of the mechanical energy
vary with height as the objects falls ...
m
E mech = K + U
Energy
K=
h
1
mv 2
2
2
1
h
U = mgy
h
t
Question 7.1: Consider two identical ice cubes sliding
down frictionless ramps with the same overall heights.
Since E mech = K + U ⇒ constant (in time)
ΔE mech = ΔK + ΔU = 0
∴ΔK = −ΔU.
i.e., the increase in kinetic energy equals the decrease in
potential energy. This is known as the ...
conservation of mechanical energy
Note that the mechanical energy is conserved only if
conservative forces are involved.
The ice cubes are released from the top of the ramps at
the same time and they travel identical distances down
the ramps. What can we say about their relative speeds
at the ends of the ramps?
M
1
h
h
2M
2
h
h
U= 0
What if the two ice cubes have different masses, e.g., M
Set the gravitational potential energy U = 0 at the bottom
and 2M? What can we say about their speeds at the
of the ramps. Conservation of energy tells us that:
bottom of the ramps?
for each ice cube. But, in each case,
If they start from rest we found K bottom = Utop .
(K bottom + Ubottom ) = (K top + Utop )
1
But K bottom = mv 2 and Utop = mgh ,
2
K top = 0 and Ubottom = 0,
∴K bottom = Utop .
But Utop = mgh for both ice cubes. So, their kinetic
energies at the bottom,
1
K bottom = mv 2 = mgh ,
2
are the same also. Since the cubes are identical,
m1 = m2 .
∴v1 = v 2 ,
i.e., they have the same speeds at the bottom.
∴v 2 = 2gh ,
which does not depend on mass! So, again, v1 = v 2 ,
even though one ice cube has twice the mass of the
other!
The reason is that although the work done on mass M by
the gravitational force is one-half that done on mass 2M,
the inertia of mass M is one-half that of mass 2M and so
the accelerations are the same (cf. Newton’s 2nd Law).
v!
This is a conservation of energy
v!
problem. The change in
v!
v!
gravitational potential energy
Question 7.2: Two stones are
thrown from the top of a building
( ΔU = −mgh ) is converted to
h
kinetic energy. Therefore, for
at the same instant with the same
initial speed; one is thrown
horizontally and the other is
thrown vertically upward. Which
one of the following statements
best describes what happens. The stones will strike the
ground ...
each stone ...
K f + Uf = Ki + Ui.
When they hit the ground, where Uf = 0,
1 2 1
mv = mv !2 + mgh ,
2
2
i.e.,
(
v 2 = v !2 + 2gh .
)
So their final speeds v = v !2 + 2gh are the same since
A: at the same time with equal speeds.
B: at different times with equal speeds.
C: at the same time with different speeds.
D: at different times with different speeds.
they started from the same height. But the time they
take depends on the initial value of v y, which is always
greater for the stone thrown upwards. Therefore, the
time it takes before hitting the ground is longer. So, the
answer is B - different times with equal speeds.
Note, the result is independent of the masses!
Other examples of conservation of energy ...
Oscillating spring
Simple pendulum
x=0
Relaxed position
x!
B
A
C
h
HMM12VD1.MOV
Energy
E mech = K + UG = mgh
t
B
→
C
→
B
→
A
E mech = K + UG ⇒ constant (= mgh )
Continuous conversion of potential energy to kinetic
energy and vice versa.
Consider a mass (M) resting on a frictionless surface and
attached to a spring. We put the position of the mass at
compressed a distance x !, and then released, it oscillates
back-and-forth. There are two contributions to the
K
→
1
kx 2 : K = 0
2 !
x = 0 when the spring is relaxed. If the spring is
UG
A
UE =
mechanical energy, the kinetic energy of the mass,
K = 12 Mv 2 ,
( )
and the elastic energy associated with the spring,
UE = 12 kx 2 ,
( )
where k is the spring constant and x is the instantaneous
position measured from the equilibrium position ( x = 0).
x=0
x!
A
UE =
1
kx 2 : K = 0
2 !
UE = 0 : K =
B
x!
C
UE =
1
kx 2 : K = 0
2 !
D
UE = 0 : K =
A
UE =
Energy
1
mv 2
2
Conservation of mechanical energy gives us
another way to determine velocities,
displacements, etc., as in these next problems.
1
mv 2
2
1
kx 2 : K = 0
2 !
1
E total = kx! 2
2
Question 7.3: An object, of mass 2.5 kg, is confined to
move along the x-axis. If the potential energy function is
UE
U(x) = 3x 2 − 2x 3,
where U(x) is in Joules and x is in meters and the
K
A
B
C
D
A
B
C
D
time
A
( )
The total energy is the initial energy, i.e., E total = 12 kx !2 ,
where x ! is the initial displacement from the equilibrium
position. Note that K + UE = constant = E total .
velocity of the object is 2.0 m/s at x = −1 m, what is its
velocity at x = 1 m?
Since a conservative force is involved, the total mechanical
energy of the object is conserved. At x = −1 m the kinetic
energy is
1
1
K1 = mv 12 = (2.5 kg)(2.0 m/s) 2 = 5.0 J,
2
2
and the potential energy is
2
3
U1 = [3(−1) − 2(−1) ] J = 5.0 J.
So, the total energy at x = −1 m is
E mech = 10.0 J.
The kinetic energy at x = 1 m is
K2 = E mech − U2
2
3
= 10.0 J − [3(1) − 2(1) ] J = 9.0 J.
∴v 2 =
2K 2
2(9.0 J)
=
= 2.68 m/s.
m
2.5 kg
Question 7.4: A pendulum consists of a 2 kg mass
hanging vertically from a string, of negligible mass, with
a length of 3 m. The mass is struck horizontally so that it
has an initial horizontal velocity of 4.5 m/s. At the point
where the string makes an angle of 30! with the vertical,
what is (a) the speed of the mass, (b) the potential energy
of the system, and (c) the tension in the string? (d) What
is the maximum angle achieved by the string?
ℓ
T
T
h = ℓ − ℓ cos 30"
v = 4.5 m/s
2
mg
ℓ
30" ℓ cos30"
1
mg
Then
at 1 ...
at 2 ...
1
K1 = mv 12 : U1 = 0.
2
1
K2 = mv 22 : U2 = mgh .
2
Energy is conserved as there are no other external forces.
∴E1 = E2 .
1
mg
∴K1 + U1 = K2 + U2
1
1
i.e., mv12 = mv 22 + mgh .
2
2
(a)
E = K + U.
Take the zero of potential energy at 1 , i.e., put U1 = 0.
h = ℓ − ℓ cos 30"
v = 4.5 m/s
2
mg
The total mechanical energy at any point is:
30" ℓ cos30"
Re-arranging, we get v 22 = v12 − 2gh .
∴v 22 = (4.5 m/s) 2 − 2(9.81 m/s2 )(3 m)(1 − cos30" )
i.e., v 2 = 3.52 m/s.
(b) U2 = mgh
= (2 kg) × (9.81 m/s 2 ) × (3 m)(1− cos30" )
= 7.89 J.
T
ℓ
T
30" ℓ cos30"
30"
h = ℓ − ℓ cos 30"
2
mg
mg
v = 4.5 m/s
1
mg
(c) Identify forces acting on the mass at 30" at 2 :
mv 22
T − mg cos30" =
ℓ
Question 7.5: (Alternative solution to question 3.6.) A
skier starts from rest at A and slides down the slope to B
and then to C. If the track is frictionless, (a) what is the
skier’s speed at B? (b) What is the skier’s speed at C?
Assume the skier turns all the corners smoothly with no
change in speed.
∴T = 25.3 N.
(d) At maximum height: K = 0, so U = K1,
1
i.e., mgh max = mgℓ(1− cosθmax ) = mv12 .
2
v12
(4.5 m/s)2
∴cosθmax = 1−
=1−
= 0.656,
2gℓ
2 × (9.81 m/s 2 ) × (3 m)
i.e., θ max = cos −1(0.656) = 49.0" .
A
10 m
C
45!
B
30!
5m
A
10 m
C
45!
5m
30!
B
(a) Since the track is frictionless, mechanical energy is
conserved, i.e., (K B + UB ) = (K A + UA ) . We define
UB = 0, and note that K A = 0.
1
∴K B = mv B2 = UA = mgh A ,
2
(
)
so v B = 2gh A = 2 × 9.81 m/s 2 × (10 m) = 14.0 m/s.
(b) By conservation of energy (KC + UC ) = (K B + UB ) ,
1
1
but UB = 0, therefore KC = mv C2 = mv B2 − mgh C.
2
2
∴v C2 = v B2 − 2gh C
2
(
2
)
= (14.0 m/s) − 2 × 9.81 m/s × (5 m) ,
i.e., v C = 9.89 m/s.
These answers are the same as in chapter 3. Note they
do not depend on the angles only on the heights!
5m
k = 400 N/m
x
Question 7.6: A 3 kg object is released from rest at a
height of 5 m on a curved frictionless ramp, as shown
above. At the foot of the ramp, there is a spring with a
force constant of k = 400 N/m. The object slides down
the ramp into the spring, compressing it a distance x
before coming momentarily to a stop. (a) What is the
distance x? (b) What happens to the object after it
comes to a stop?
k = 400 N/m
5m
(a) At the top:
x
UG = 0
UG = mgh = (3 kg)(9.81 m/s 2 )(5 m) = 147.2 J,
K = 0 and UE = 0.
∴E top = K + UG + UE = 147.2 J.
At the bottom, when the object is at rest against the spring,
1
UG = 0, K = 0 and UE = kx 2 .
2
2
∴E bottom = K + UG + UE = UE = 200x J.
But mechanical energy is conserved ...
∴E bottom = E top , i.e., 200x 2 = 147.2 J,
∴x = 0.858 m.
(b) The object will retrace its path back up to the start
position at 5 m (energy is conserved). However, with
friction on the track then mechanical energy is NOT
conserved; some will be “lost” (i.e., converted to heat) as
the frictional force is non-conservative, so x < 0.858 m.
h
h
A
B
Question 7.7: Two identical balls roll along similar
tracks; the only difference is that track B has a small
depression in it but the overall distances the balls travel
from start to finish are the same. Since the balls “fall”
the same overall heights (h), conservation of mechanical
energy tells us that their speeds are the same at the end
points if we ignore friction. But, if it’s a race, which one
(if either) will reach the finish line first?
h
h
A
Question 7.8: A physics student, with mass 80 kg, does a
×
bungee jump from a platform 100 m above the ground.
B
×
x
When the balls reach the bottom of the slope (at ×), they
both have the same speed ( v = 2gh ). On track A the
speed of the ball remains constant the rest of the way,
since the path is horizontal and there’s no friction.
However, on track B, the ball gains additional speed
when it falls a further distance x; so its speed over the
red region increases to
2g(h + x). When it climbs the
depression its speed drops back to 2gh . So, between
the bottom of the ramp and the finish line the ball on
track B has the greater average speed ... therefore, it
reaches the finish first!
If the rubber rope has an unstretched length of 50 m and
a spring constant k = 200 N/m, (a) how far above the
ground will the student be at his lowest point? (b) How
far above the ground will he achieve his greatest speed?
(c) What is his greatest speed?
(a) Put the zero of gravitational potential energy at the
ground.
K = 0 : UE = 0 : UG = mgh = 78, 480 J
ℓ = 50 m
The two solutions are: y ! = 66.27 m and y ! = 25.89 m.
Clearly, the appropriate solution is y ! = 25.89 m.
(b) To determine the position where the jumper’s speed
h = 100 m
ℓ
K = 0 : UE =
lowest point
y"
1 2
kℓ : UG = mgy"
2
UG = 0
is greatest, we need an expression for the kinetic energy
dK
(K) in terms of the vertical position (y) and set
= 0,
dy
i.e., where K is a maximum.
K = 0 : UE = 0 : UG = mgh = 78, 480 J
The total energy is 78,480 J, which is conserved
throughout the jump. At his lowest point
1
K + kℓ 2 + mgy " = 78,480 J,
2
but K = 0 and ℓ = 50 − y " .
2
∴100(50 − y " ) + 80 × 9.81y " = 78,480,
i.e., 100y "2 − (10,000 − 784.8)y "
+ (250,000 − 78,480) = 0.
2
∴y " − 92.15y " + 1,715.2 = 0,
i.e., y " =
92.15 ± (92.15)2 − 4 × 1,715.2
.
2
ℓ = 50 m
h = 100 m
ℓ′
fastest point
y
1
K : UE = kℓ ′ 2 : UG = mgy
2
UG = 0
If y is the height above the ground where their speed is
greatest and ℓ′ is the amount the rope is stretched, then,
as energy is conserved, at the fastest point:
1
K + kℓ′ 2 + mgy = 78,480 J.
2
Substituting for ℓ′ = 50 − y, we find
K + 100(50 − y)2 + 784.8y = 78,480,
i.e., K = 78,480 − 100(50 − y)2 − 784.8y
= −171,520 + 9,215.2y − 100y 2 ,
dK
so
= 9,215.2 − 200y.
dy
dK
The extremum occurs when
= 0,
dy
i.e., at y =
Check that this is a maximum:
dy 2
former, but what does the latter solution correspond to?
Note that both solutions correspond to positions where
the kinetic energy of the jumper is zero. Consider a
scenario in which the rope is a vertical spring and the
jumper is a mass attached to the spring.
9,215.2
= 46.08 m.
200
d 2K
In part (a) we found two values for the lowest point y !,
i.e., y ! = 25.89 m and y ! = 66.27 m. We chose the
= −200, i.e., < 0.
“highest” point
66.27 m
lowest point
25.89 m
mid point of
oscillation
46.08 m
When the spring is stretched and released, the mass will
(c) Since Kmax = 78,480 − 100(50 − y)2 − 784.8y, when
y = 46.08m, we have
1
K max = mv max 2 = 40,780 J,
2
i.e., v max =
2 × 40,780 J
= 31.9 m/s.
80 kg
oscillate up-and-down . The solution y ! = 25.89 m
corresponds to the lowest point of the oscillation; the
solution y ! = 66.27 m corresponds to the highest point
of the oscillation! The mid-point is at y = 46.08 m,
where the kinetic energy is greatest.
When non-conservative forces are involved, the situation
is more complicated; we need a more general statement
Let’s look at some examples:
of energy conservation, which involves the total energy
of a system (e.g., bike + Earth + rider ...), i.e.,
Esystem = ∑ Ei = E mech + Echem + …
•
with chemical changes
•
with frictional forces
i
If there is no input to the
system from any other external
Question 7.9: A physics student, with mass 60 kg,
sources (e.g., wind, a push),
climbs a 120 m high hill. (a) What is the increase in his
then Esystem = constant
i.e., ΔEsystem = 0.
So an increase in one form of energy (e.g., mechanical) is
compensated by a decrease in another form (e.g., energy
gravitational potential energy?
(b)
Where does this energy come
from? (c) If the student’s body is
20% efficient, i.e., for every
stored by rider).
100 J of chemical energy used
If an external source does work ( Wext) on a system then:
mechanical energy with the
only 20 J are converted to
ΔEsystem = Wext .
This is the more general form of the work-energy
theorem. An example might be the action of a wind
acting on the rider.
remainder ( 80 J) resulting in
thermal energy, how much chemical energy is used by
the student during the climb?
Why do you have to keep
(a) ΔU = U2 − U1 = mgh
2
pedaling when traveling on a
4
= (60 kg)(9.81 m/s )(120 m) = 7.06 × 10 J.
level road, even though you’re
The total mechanical energy change is 70.6 kJ.
not changing your speed (i.e.,
with no change in ΔK)? Isn’t
(b) The system is the climber and the Earth. With no
the work-energy theorem ( W ⇒ ΔK) violated? No!
external forces
From earlier we have
Esys = E mech + Eclimber ,
where Eclimber is the energy stored within the climber.
Since ΔEsys = 0, ΔE mech = −ΔEclimber ,
i.e., the increase in mechanical energy comes from a
decrease in (chemical) energy stored by the climber,
which he got from metabolizing food.
(c) 100 J of Echem → 20 J of E mech + 80 J of E therm
∴ 1 J of E mech requires 5 J of Echem
i.e., an increase of 70.6 kJ in mechanical energy requires
5 × (70.6 kJ) = 353 kJ of chemical energy. The balance
(353 kJ − 70.6 kJ = 282.4 kJ) appears as thermal energy.
Esys = ∑ Ei = E mech + E therm + Echem.
i
ΔEsys = ΔE mech + ΔE therm + ΔEchem = 0.
If you did no work (i.e., ΔEchem = 0), the work against
friction, air drag, etc., ( ΔE therm) would have to come
from somewhere ... it would come from a loss of
mechanical energy (kinetic energy), since
ΔE therm = −ΔE mech,
and so you would slow down. To overcome these losses
and maintain constant speed, and satisfy ΔEsys = 0, you
have to do an amount of work equal to the energy lost
through friction, air resistance, etc., i.e.,
ΔEchem = −ΔE thermal .
N
θ
0.50 m
2.0 m
Question 7.10: A physics student has been given the task
of designing a system to deliver a glass of beer along a bar
to a designated spot. The beer mug, of mass 1.5 kg, is
released from the top of a ramp, which is 0.50 m above
mg
θ
0.50 m
2.0 m
There are two contributions to the total energy of the
system; E mech and E therm (assuming that the work done
against friction produces heat).
∴Esys = E mech + E therm.
If there are no external forces, then Esys is constant, i.e.,
the bar. The mug slides down the ramp and onto the bar,
ΔEsys = ΔE mech + ΔE therm = 0.
where it has to come to a stop 2.0 m from the bottom of
the ramp. If the coefficient of kinetic friction between the
Now, ΔE mech = ΔK + ΔU and ΔE therm = f k .Δs, where
f k is the frictional force and Δs is the distance through
mug and the surfaces of the ramp and the bar is 0.15,
which the frictional force is active.
(a) what is the required value of the angle θ?
(a) In this problem, ΔE therm comprises two components:
(b) What is the speed of the mug at the bottom of the
(1) down the ramp and (2) along the bar.
0.50 m
(1) Down the ramp: Δs1 =
and
sin θ
ramp?
f k1 = µ k N = µ k mg cosθ.
N
θ
mg
0.50 m
i.e.,
1.104
⎞ = 20.6! .
θ = tan −1 ⎛
⎝ 7.358 − 4.415⎠
N
2.0 m
∴ΔE therm1 = (µ k mg cosθ)Δs1 = (µ k mg cosθ)
0.50 m
sin θ
0.15 × (1.5 kg) × (9.81 m/s2 ) × (0.50 m) 1.104
=
=
J.
tan θ
tan θ
(2) Along the bar: Δs2 = 2.0 m and f k2 = µ k mg
∴ΔE therm2 = (µ k mg)Δs2
= 0.15 × (1.5 kg) × (9.81 m/s2 ) × (2.0 m) = 4.415 J.
θ
mg
0.50 m
2.0 m
(b) On the ramp: ΔEsys = ΔE mech + ΔE therm1 = 0,
i.e., ΔK1 + ΔUg + ΔE therm1 = 0.
ΔU g = −mgh = −(1.5 kg) × (9.81 m/s2 ) × (0.50 m)
If v is the velocity at the bottom of the ramp
1
ΔK1 = mv 2 = −ΔU g − ΔE therm1
2
1.104 J
= mgh −
= 7.385 J − 2.937 J = 4.448 J.
tan θ
2ΔK1 2 × 4.448 J
∴v 2 =
=
,
m
1.50 kg
= −7.358 J.
i.e., v = 2.44 m/s.
∴ΔE therm = ΔE therm1 + ΔE therm2 =
1.104
+ 4.415 J.
tan θ
Using conservation of energy ΔK + ΔUg + ΔE therm = 0,
but ΔK = 0, i.e., ΔE therm = −ΔU g .
∴
1.104
+ 4.415 = 7.358 J,
tan θ
1 start position
5.0 m
y!
Δy
2 max kinetic energy
3 max compression
(a) Put Ug = 0 at position 2 where the kinetic energy is
Question 7.11: A block of mass 2.4 kg is dropped onto a
spring, with a spring constant of 400 N/m, from a height
of 5.0 m above the top of the spring. (a) What is the
maximum kinetic energy of the block? (b) What is the
maximum compression of the spring?
This is an interesting problem as we will use different
positions for the zero of gravitational potential energy
in parts (a) and (b).
a maximum. Then conservation of energy gives:
K1 + Ug1 + Ue1 = K2 + Ug2 + Ue2,
1
i.e., mg ((5.0 m) + Δy ) = K2 + k(Δy)2 .
2
1
∴K2 = mg ((5.0 m) + Δy ) − k(Δy)2 .
2
dK 2
For maximum kinetic energy:
= mg − kΔy = 0,
d(Δy)
mg (2.4 kg) × (9.81 m/s 2 )
i.e., Δy =
=
= 0.059 m.
k
400 N/m
1
But K2 = mg ((5.0 m) + Δy ) − k(Δy)2
2
= (2.4 kg) × (9.81 m/s 2 ) × (5.059 m)
1
− (400 N/m) × (0.059 m)2 = 118.4 J.
2
(b) Now we put Ug = 0 at position 3 .
1 start position
5.0 m
y!
Δy
23.54 ± (23.54)2 + 4 × 200 × 117.7
i.e., y ! =
2 × 200
∴y ! = 0.828 m or y ! = −0.711 m.
Clearly, the former is the physically meaningful root and
represents the maximum compression of the spring.
If the mass were simply placed on the top of the spring,
the equilibrium position would occur when the spring is
2 max kinetic energy
compressed a distance Δℓ (from the top of the spring), so
3 max compression
the net force acting on the mass is zero, i.e., when
kΔℓ = mg .
Conservation of energy gives:
i.e.,
mg (2.4 m) × (9.81 m/s 2 )
∴Δℓ =
=
= 0.059 m.
k
400 N/m
K1 + Ug1 + Ue1 = K3 + Ug3 + Ue3,
Note that this is the position where the kinetic energy
1
mg ((5.0 m) + y ! ) = ky !2 .
2
was a maximum in part (a)!
∴200y !2 − 23.54y ! −117.7 = 0
Hey .. before finishing this chapter
...what’s the origin of friction?
Friction results because of attractive forces between atoms
on the two surfaces. When sliding occurs work is done to
stretch the ‘bonds’; they break and the participating atoms
vibrate about their equilibrium positions.
~ Then why do the surfaces get hot ? ~
ΔK ⇒ ΔT
(kinetic theory).
Increased vibration (velocity)
⇒ increased temperature.