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Unformatted text preview: CONSERVATION OF MECHANICAL ENERGY
Consider a system consisting of an object and the Earth.
We release the object from a height h above the ground.
We take the ground as the zero of potential energy. CHAPTER 7
CONSERVATION OF ENERGY
• Conservation of mechanical energy
• Conservation of total energy of a system
o various examples U1 = mgh : K1 = 0 v1 = 0 v 1
U = mgy : K = mv 2
2 v2 h 1
U2 = 0: K 2 = mv 2 2
2 y • Origin of friction 0
2 2 At the bottom: ( v = v o + 2a∆y ) ⇒ v 22 = 2 gh
1
∴ K2 = mv 22 = mgh .
2
At any position y: v 2 = 2 g( h − y )
1
∴ K = mv 2 = mg( h − y ).
2 1 See how the individual parts of the mechanical energy
vary with time as the objects falls ...
v1 = 0 U1 = mgh : K1 = 0 m
h h v 1
U = mgy : K = mv 2 = mg( h − y)
2 y
v2 1
U2 = 0: K 2 = mv 2 2 = mgh
2 E mech = K + U h→ 1
K = mv 2
2 But we note that
U = mgy K1 + U1 = K2 + U2 = K + U , t 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). Since E mech = K + U ⇒ constant (in time) ∆E mech = ∆K + ∆U = 0
∴ ∆K = − ∆ U . In this case: E mech = mgh , the initial energy of the i.e., the increase in kinetic energy equals the decrease in system. potential energy.
Conservation of mechanical energy 2 Note that we can derive the conservation law in another
way ...
The work done by the Other examples of conservation of energy ...
Simple pendulum gravitational force on the
v1 = 0 h object is
W1→2 = mgh . But, the change in potential
v B A C h energy of the Earthobject
system is v2 ∆U1→2 = U2 − U1 HMM12VD1.MOV Energy E mech = K + UG = mgh = 0 − mgh = − mgh , i.e., a decrease. UG ∴ W1→2 = − ∆U1→2. K Using the workkinetic energy theorem we get
W1→2 = ∆K1→2 = − ∆U1→2, i.e., the increase in kinetic energy from 1 → 2 is equal to
the decrease in potential energy, as before.
Note that mechanical energy (PE + KE) is
conserved only if conservative forces are involved. t
A → B → C → B → A E mech = K + UG ⇒ constant ( = mgh ) Continuous conversion of potential energy to kinetic
energy and vice versa. 3 Oscillating spring
When a mass (M)
C is hung from a
ys = kMg ys y=0 spring, the spring M extends to an y=0
yo oscillates upanddown (with simple harmonic motion). D A equilibrium position, which we define at y = 0 . If the
spring is further extended, to y = − y o , and released, it UE = 0 B A Energy 1
E total = kyo 2
2 E total UE C y=0
yo B UE = 0 K D A t A A There are two contributions to the mechanical energy,
the kinetic energy of the mass, K = 12 Mv 2 , and the
elastic energy associated with the spring, UE = 12 ky 2 , () () where k is the spring constant and y is the instantaneous
position measured from the equilibrium position. B C D A B C D A The total energy is the initial energy, i.e.,
1
E total = ky o2,
2
where y o is the initial displacement from the equilibrium
position. Note that
K + UE = constant = E total . 4 Conservation of mechanical energy
gives us another way to determine
velocities, displacements, etc. l T 30o l cos 30o
h = l − l cos 30o
v = 4.5m/s 2
1 mg mg Question 1: A pendulum consists of a 2kg mass attached
to a string, of negligible mass, with a length of 3m. The The total mechanical energy at any point is:
E = K + U. mass is struck horizontally so that it has an initial
horizontal velocity of 4.5m/s. At the point where the Take the zero of potential energy at 1 , i.e., put U1 = 0. string makes an angle of 30o with the vertical, what is (a) Then the speed of the mass, (b) the potential energy of the at 1 ... 1
K1 = mv12 : U1 = 0 .
2 at 2 ... 1
K2 = mv 22 : U2 = mgh .
2 system, and (c) the tension in the string? (d) What is the
maximum angle achieved by the string? But energy is conserved ... there are no other external
forces, so E1 = E2. 5 T l T l 30o l cos 30o
h = l − l cos 30
v = 4.5m/s 2
mg T 1 o mg (a) h = l − l cos 30o
v = 4.5m/s 1
mg (c) Identify forces acting on the bob at 30o ∴ K1 + U1 = K2 + U2
1
1
i.e., mv12 = mv 22 + mgh .
2
2
1
1
∴ × 2 × (4.5)2 = × 2 × v 22
2
2
+2 × 9.81 × 3(1 − cos 30o ), i.e., v 2 = 3.52 m/s.
o (b) U2 = mgh = 2 × 9.81 × 3(1 − cos 30 ) = 7.89J . mg 2
mg 30o 30o l cos 30o T − mg cos 30o = mv 22
l ∴ T = 25.3N.
(d) At maximum height: K = 0, so U = K1,
1
i.e., mgh max = mgl(1 − cos θ max ) = mv12.
2 v2
(4.5)2
∴ cos θ max = 1 − 1 = 1 −
= 0.656 ,
2 gL
2 × 9.81 × 3 i.e., θ max = 49.0o. 6 k = 400N/m 5m x (a) At the top
5m k = 400N/m
x Question 2: A 3kg object is released from rest at a height
of 5m on a curved frictionless ramp, as shown above. At
the foot of the ramp, there is a spring with a force
constant of k = 400N/m. The object slides down the
ramp into the spring, compressing it a distance x before
coming momentarily to rest. (a) What is the distance x?
(b) What happens to the object after it comes to rest? UG = 0 UG = mgh = 3 × 9.81 × 5 = 147.2J ,
K = 0 and UE = 0.
∴ E top = K + UG + UE = 147.2J. At the bottom, when the object is at rest against the spring,
1
UG = 0 , K = 0 and UE = kx 2.
2 ∴ E bottom = K + UG + UE = UE = 200 x 2J .
But mechanical energy is conserved ...
∴ E bottom = E top , i.e., 200 x 2 = 147.2,
∴ x = 0.858 m. (b) The object will retrace its path back up to the start
position at 5m (energy is conserved).
Note: with friction on the track then mechanical energy will
be “lost” (converted to heat), so x < 0.858 m, as the
frictional force is nonconservative. 7 (a) Put the zero of gravitational potential energy at the
ground.
K = 0 : UE = 0 : UG = mgh = 78480J Question 3: A physics student, with mass 80kg, does a
l = 50 m bungee jump from a platform 100m above the ground. If
the rubber rope has an unstretched length of 50m and a h = 100m
∆l spring constant k = 200N/m, (a) how far above the
ground will the student be at his lowest point? (b) How lowest point
yo far above the ground will he achieve his greatest speed?
(c) What is his greatest speed? 1
K = 0 : UE = k( ∆l) 2 : UG = mgyo
2 UG = 0 The total energy is 78480J, which is conserved
throughout the jump. At his lowest point
1
K + k ( ∆l)2 + mgy o = 78480,
2
but K = 0 and ∆l = h − l − y o = 50 − y o. ∴100(50 − y o )2 + 80 × 9.81y o = 78480 ,
i.e., 100 y o2 − (10000 − 784.8) y o + (250000 − 78480) = 0. ∴ y o2 − 92.15y o + 1715.2 = 0 ,
i.e., y o = 92.15 ± (92.15)2 − 4 × 1715.2
.
2 8 The two solutions are: y o = 66.27 m and y o = 25.89 m.
Clearly, the appropriate solution is y o = 25.89 m. Substituting for l′ = h − 50 − y = 50 − y , we find K + 100(50 − y )2 + 784.8 y = 78480 ,
i.e., K = 78480 − 100(50 − y )2 − 784.8 y We will return to the other solution in a moment.
(b) To determine the position where the jumper’s speed = −171520 + 9215.2 y − 100 y 2 , so dK
= 9215.2 − 200 y .
dy is greatest, we will get an expression for the kinetic
energy (K) in terms of the vertical position (y) and set
dK
= 0, i.e., where K is a maximum.
dy
K = 0 : UE = 0 : UG = mgh = 78480J
l = 50m h = 100m max speed l′ y The extremum occurs when dK
= 0,
dy i.e., at y = 46.08 m.
Check that this is a maximum: d 2K
= −200 , i.e., < 0.
dy 2 We will find the result by another method in a moment.
1
K : UE = kl ′ 2 : UG = mgy
2 UG = 0 Since energy is conserved
1
K + kl′2 + mgy = 78480.
2 (c) Since K max = 78480 − 100(50 − y )2 − 784.8 y when y = 46.08 m, we have
1
mv max2 = 40780J ,
2
i.e., v max = 2 × 40780
= 31.9m/s.
80 9 DISCUSSION PROBLEM [7.1]:
K = 0 : UE = 0 : UG = mgh = 78480J
l = 50 m h = 100m DISCUSSION PROBLEM [7.2]:
∆l lowest point
yo 1
K = 0 : UE = k( ∆l) 2 : UG = mgyo
2
UG = 0 Can you think of an alternative method for determining
the position where the jumper’s speed is greatest? In part (a) of the previous problem we found two values
for the lowest point y o, i.e., y o = 25.89 m and
y o = 66.27 m. Clearly, the appropriate solution is the former. What does the latter solution correspond to? 10 When nonconservative forces are involved, the situation
is a little more complicated: FTR is the result of a (downward) force F acting on the pedal(s) that is transmitted by a chain to a geared cog
wheel attached to the axle of the wheel, which causes it
to rotate. Case Study ...
What makes a bicycle move when you pedal it? θ Motion
F Force of tire on road FTR F Force of road on tire FRT By Newton’s 3rd Law FRT = FTR .
If there is no slip (only static friction) the point of contact
of tire does not move relative to the road. Therefore
neither FTR nor FRT do any work! But it is FRT that
“propels” the bike forward (⇒ ma ). If FRT does no work
how can the kinetic energy change ... ∆K = W ?
Let’s look at the origin of FTR . FTR FRT Since the force F  produced by musclepower  moves
through a displacement it does work. It is this work that
propels the bicycle. Where does the force (work)
originate?
... from energy supplied by the rider’s muscles
(i.e., from food!) ... and it is a nonconservative force. It
is the work done by the rider that results in the motion,
i.e., Wrider ⇒ ∆K .
So the rider has to be part of the system. 11 Thus, a more general
statement of energy Let’s look at some examples: conservation must involve • with chemical changes the total energy of a system • with frictional forces (e.g., bike + Earth + rider ...)
Esystem = ∑ Ei = E mech + Echem + K
i If there is no input to the system from any other external
sources (e.g., wind, a push, etc.), then Esystem = constant
i.e., ∆Esystem = 0 . Question 4: A physics student, with mass 60kg, climbs a
120m high hill. (a) What is the increase in his
gravitational potential
energy? (b) Where does
this energy come from? So an increase in one form of energy (e.g., mechanical) is (c) The student’s body is compensated by a decrease in another form (e.g., energy 20% efficient, i.e., for stored by rider. every 100J of chemical
energy expended only 20J If an external source does work ( Wext ) on a system then: Wext = ∆Esystem . are converted to
mechanical energy, with the remainder (80J) resulting in This is the more general form of the workenergy thermal energy. How much chemical energy is theorem. An example might be the action of a wind expended by the student during the climb? acting on the rider. 12 (a) ∆U = U2 − U1 = mgh = 80 × 9.81 × 1200 = 94.2 kJ .
The total mechanical energy change ⇒ 94.2 kJ . DISCUSSION PROBLEM [7.3]:
(b) The system is the climber and the Earth. With no
external forces Esys = E mech + Eclimber , where Eclimber
is the energy stored within the climber. Since ∆Esys = 0,
∆E mech = − ∆Eclimber ,
i.e., the energy comes from a decrease in (chemical)
energy stored by the climber, which he got from
metabolizing food.
(c) 100J (Echem) ⇒ 20J ( E mech ) + 80J ( E therm)
∴ 1J ( E mech) requires 5J ( Echem) Why does a cyclist need to keep pedaling when traveling
on a level road, even though she’s not changing her
speed (i.e., with no change in ∆K )? Isn’t the workenergy theorem ( W ⇒ ∆K ) violated? Explain. i.e., 94.2kJ ( E mech) requires 5 × 94.2 = 471kJ of
chemical energy. 13 N θ mg 0.50m 2.0 m
0.50 m θ
2.0m Question 5: 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 1.5kg is
released from the top of a ramp, which is 0.50m above 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.,
∆Esys = ∆E mech + ∆E therm = 0. where it should come to a stop 2.0m from the bottom of Now, ∆E mech = ∆K + ∆U and ∆E therm = f k .∆s, where
f k is the frictional force and ∆s is the distance through the ramp. If the coefficient of kinetic friction between which the frictional force is active. the bar. The mug slides down the ramp and onto the bar, the mug and the surfaces of the ramp and bar in 0.15, (a)
what is the required value of the angle θ ? (b) What is
the speed of the mug at the bottom of the ramp? (a) In this problem, ∆E therm comprises two components:
(1) down the ramp and (2) along the bar.
0.50
(1) down the ramp: ∆s1 =
and
sin θ
f k1 = µ mN = µ k mg cos θ . 14 N θ mg 0.50m 2.0m N 0.50 0.15 × 1.5 × 9.81 × 0.5
∴ ∆E therm1 = µ k mg cos θ.
=
sin θ
tan θ
1.104
=
J.
tan θ (2) Along the bar: ∆s2 = 2.0 m and f k2 = µ k mg
∴ ∆E therm2 = µ k mg × 2.0 = 0.15 × 1.5 × 9.81 × 2.0 = 4.415J.
1.104
∴ ∆E therm =
+ 4.415J .
tan θ
Using conservation of energy ∆K + ∆Ug + ∆E therm = 0,
but ∆K = 0 , so ∆E therm = − ∆Ug.
∆Ug = − mgh = −1.5 × 9.81 × 0.5 = −7.358J , θ mg 0.50m 2.0m (b) On the ramp: ∆Esys = ∆E mech + ∆E therm1 = 0,
i.e., ∆K1 + ∆Ug + ∆E therm1 = 0 .
1
1.104
∴ ∆K1 = mv 2 = − ∆Ug − ∆E therm1 = mgh −
2
tan θ = 7.385 − 2.937 = 4.448J.
2 × 4.448
∴ v2 =
= 5.93 ⇒ v = 2.44 m/s.
1.50 1.104
+ 4.415 = 7.385,
tan θ
1.104 = 20.6o .
i.e., θ = tan −1 7.358 − 4.415 15 Hey .. before going any further ...
what’s the origin of friction? DISCUSSION PROBLEM [7.4]: Magnum XL200, Cedar Point Park, Ohio Friction results from attractive forces between atoms on
the two surfaces. When sliding occurs work is done to • What forces are involved in a roller coaster? stretch the “bonds”; they break and the participating
• Are the forces conservative or nonconservative ? atoms vibrate.
~ Then why do the surfaces get hot ? ~ ∆K ⇒ ∆ T (kinetic theory). • What changes of energy are involved? Increased vibration (velocity) ⇒ increased temperature. 16 1 start position
5.0m
∆y 2 max kinetic energy yo 3 max compression (a) Let Ug = 0 at position 2 . Then conservation of
Question 6: A block of mass 2.4kg is dropped onto a energy gives: spring, with a spring constant of 400N/m, from a height K1 + Ug1 + Ue1 = K2 + Ug2 + Ue2, of 5.0m above the top of the spring. (a) What is the 1
i.e., mg(5 + ∆y ) = K2 + k ( ∆y )2 .
2
1
∴ K2 = mg(5 + ∆y ) − k ( ∆y )2 .
2 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). For maximum kinetic energy: dK2
= 0,
d ( ∆y ) i.e., mg − k∆y = 0 ⇒ ∆y = mg 2.4 × 9.81
=
k
400 = 0.059 m. 17 1
But K2 = mg(5 + ∆y ) − k ( ∆y )2
2
1
= 2.4 × 9.81 × 5.059 − × 400 × (0.059)2
2 = 118.4J . ( v max = 9.93m/s.) (b) Now we let Ug = 0 at position 3 . i.e., y o = 23.54 ± (23.54)2 + 4 × 200 × 117.7
2 × 200 ∴ y o = 0.828 m or y o = −0.711m.
Clearly, the former is the physically meaningful root and represents the maximum compression of the spring. 1 start position
5.0 m ∆y
yo 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
3 max compression compressed a distance ∆l (from the top of the spring), so
the net force acting on the mass is zero, i.e., when k∆l = mg . Then conservation of energy gives: K1 + Ug1 + Ue1 = K3 + Ug3 + Ue3,
1
i.e., mg(5 + y o ) = ky o2.
2 ∴ ∆l = mg 2.4 × 9.81
=
= 0.059 m.
400
k Note that this is the position where the kinetic energy
was a maximum in part (a). ∴ 200 y o2 − 23.54 y o − 117.7 = 0 18 ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.
 Spring '08
 Guzman
 Physics, Conservation Of Energy, Energy, Potential Energy

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