CONSERVATION OF MECHANICAL ENERGY previous chapter we ...cosweb1.fau.edu/~jordanrg/phy2048/chapter_7/notes_7.pdf · CHAPTER 7 CONSERVATION OF ENERGY • Conservation of mechanical

CHAPTER 7

CONSERVATION OF ENERGY

• Conservation of mechanical energy

• Conservation of total energy of a system ! various examples

• Origin of friction

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

W12 = F.dx

1

2∫ = ΔK = K2 − K1,

where K is the kinetic energy. We also found that if the force is conservative, then

F.dx

1

2∫ = −ΔU = −(U2 − U1),

where U is the potential energy.

∴ΔK = −ΔU = K2 − K1 = −(U2 − U1),i.e., K2 + U2 = K1 + U1.

If define the total energy as E = K + U, then ΔE = (K2 − 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. We take the ground as the zero of potential energy.

At the bottom: (v2 = v!2 + 2aΔy)⇒ v2

2 = 2gh

∴K2 =

12

mv22 = mgh .

At any position y: v2 = 2g(h − y)

∴K =

12

mv2 = mg(h − y).

0

U2 = 0: K2 = 1

2mv2

2

U = mgy: K = 1

2mv2

U1 = mgh: K1 = 0 v1 = 0

v2

v h

yBut we note that

K1 + U1 = K2 + U2 = K + U,

i.e., K + U = constant ⇒ Emech.

We define Emech as the

total mechanical energy of the system(i.e., the sum of the potential and kinetic energies).

In this case: Emech = mgh , the initial energy of the

system and the total mechanical energy remains constant with this value while the object falls.

U2 = 0: K2 = 1

2mv2

2 = mgh

U = mgy: K = 1

2mv2 = mg(h − y)

U1 = mgh: K1 = 0 v1 = 0

v2

v h

y

See how the individual parts of the mechanical energy vary with height as the objects falls ...

Since Emech = K + U ⇒ constant (in time)

ΔEmech = 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 energyNote that the mechanical energy is conserved only if conservative forces are involved.

m

h

Energy Emech = K + U

K = 1

2mv2

U = mgy

tQuestion 7.1: Consider two identical ice cubes sliding down frictionless ramps with the same overall heights. 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?

h h

1 2

Set the gravitational potential energy U = 0 at the bottom of the ramps. Conservation of energy tells us that:

Kbottom + Ubottom( ) = Ktop + Utop( )for each ice cube. But, in each case,

Ktop = 0 and Ubottom = 0,

∴Kbottom = Utop.

But Utop = mgh for both ice cubes. So, their kinetic

energies at the bottom,

Kbottom =

12

mv2 = mgh ,

are the same also. Since the cubes are identical,

m1 = m2.

∴v1 = v2,

i.e., they have the same speeds at the bottom.

h h

1 2

U = 0What if the two ice cubes have different masses, e.g., M and 2M? What can we say about their speeds at the bottom of the ramps?

If they start from rest we found Kbottom = Utop.

But Kbottom =

12

mv2 and Utop = mgh ,

∴v2 = 2gh ,

which does not depend on mass! So, again, v1 = v2,

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).

h h

M 2M

Question 7.2: Two stones are thrown from the top of a building 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 ...

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.

v! v!

This is a conservation of energy problem. The change in gravitational potential energy ( ΔU = −mgh ) is converted to kinetic energy. Therefore, for each stone ...

Kf + Uf = Ki + Ui.

When they hit the ground, where Uf = 0,

12

mv2 =12

mv!2 + mgh ,

i.e., v2 = v!2 + 2gh .

So their final speeds v = v!2 + 2gh( ) are the same since

they started from the same height. But the time they take depends on the initial value of vy, 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!

v! v!

h

Other examples of conservation of energy ...

Simple pendulum

A B C h

Emech = K + UG ⇒ constant (= mgh )

Continuous conversion of potential energy to kinetic energy and vice versa.

UG

K

Energy

t A → B → C → B → A

Emech = K + UG = mgh

HMM12VD1.MOV

Oscillating spring

Consider a mass (M) resting on a frictionless surface and attached to a spring. We put the position of the mass at

x = 0 when the spring is relaxed. If the spring is compressed a distance x!, and then released, it oscillates

back-and-forth. There are two contributions to the mechanical energy, the kinetic energy of the mass,

K = 12( )Mv2,

and the elastic energy associated with the spring,

UE = 12( )kx2,

where k is the spring constant and x is the instantaneous position measured from the equilibrium position ( x = 0).

Relaxed position

x = 0

x! UE = 1

2kx!2 : K = 0

Energy

time

UE

K

Etotal =

12

kx!2

C D B AA C D B A

C

A

D

B

A

x = 0

x! UE = 1

2kx!2 : K = 0

UE = 0 : K = 1

2mv2

UE = 1

2kx!2 : K = 0

UE = 0 : K = 1

2mv2

UE = 1

2kx!2 : K = 0

x!

The total energy is the initial energy, i.e., Etotal = 12( )kx!2,

where x! is the initial displacement from the equilibrium

position. Note that K + UE = constant = Etotal.

Question 7.3: An object, of mass 2.5 kg, is confined to move along the x-axis. If the potential energy function is

U(x) = 3x2 − 2x3,

where U(x) is in Joules and x is in meters and the velocity of the object is 2.0 m/s at x = −1 m, what is its velocity at x = 1 m?

Conservation of mechanical energy gives us another way to determine velocities, displacements, etc., as in these next problems.

Since a conservative force is involved, the total mechanical energy of the object is conserved. At x = −1 m the kinetic energy is

K1 =

12

mv12 =

12(2.5 kg)(2.0 m/s)2 = 5.0 J,

and the potential energy is

U1 = [3(−1)2 − 2(−1)3] J = 5.0 J.

So, the total energy at x = −1 m is

Emech =10.0 J.

The kinetic energy at x = 1 m is

K2 = Emech − U2

= 10.0 J − [3(1)2 − 2(1)3] J = 9.0 J.

∴v2 =

2K2m

=2(9.0 J)2.5 kg

= 2.68 m/s.

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?

The total mechanical energy at any point is:

E = K + U.

Take the zero of potential energy at 1 , i.e., put U1 = 0.

Then

at 1 ... K1 =

12

mv12: U1 = 0.

at 2 ... K2 =

12

mv22: U2 = mgh .

Energy is conserved as there are no other external forces.

∴E1 = E2.

1 2

ℓ ℓ cos30"

h = ℓ − ℓcos30"

30"

v = 4.5 m/s

T

mg mg 1

2

ℓ ℓ cos30"

h = ℓ − ℓcos30"

30"

v = 4.5 m/s

T

mg mg

(a) ∴K1 + U1 = K2 + U2

i.e., 12

mv12 =

12

mv22 + mgh .

Re-arranging, we get v22 = v1

2 − 2gh .

∴v22 = (4.5 m/s)2 − 2(9.81 m/s2 )(3 m)(1− cos30")

i.e., v2 = 3.52 m/s.

(b) U2 = mgh

= (2 kg) × (9.81 m/s2) × (3 m)(1− cos30")

= 7.89 J.

1 2

ℓ ℓ cos30"

h = ℓ − ℓcos30"

30"

v = 4.5 m/s

T

mg mg

(c) Identify forces acting on the mass at 30" at 2 :

T − mg cos30" =

mv22

ℓ

∴T = 25.3 N.

(d) At maximum height: K = 0, so U = K1,

i.e., mghmax = mgℓ(1− cosθmax ) =

12

mv12.

∴cosθmax = 1−

v12

2gℓ= 1−

(4.5 m/s)2

2 × (9.81 m/s2) × (3 m)= 0.656,

i.e., θmax = cos−1(0.656) = 49.0".

30"

T

mgQuestion 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.

5 m 45! 30! 10 m

A

B

C

(a) Since the track is frictionless, mechanical energy is conserved, i.e., KB + UB( ) = KA + UA( ). We define

UB = 0, and note that KA = 0.

∴KB =

12

mv B2 = UA = mgh A,

so vB = 2gh A = 2 × 9.81 m/s2( ) × 10 m( ) = 14.0 m/s.

(b) By conservation of energy KC + UC( ) = KB + UB( ),

but UB = 0, therefore KC =

12

mvC2 =

12

mv B2 − mghC.

∴vC2 = vB

2 − 2ghC

= 14.0 m/s( )2 − 2 × 9.81 m/s2( ) × 5 m( ),

i.e., vC = 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!

5 m 45! 30! 10 m

A

B

C

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

x

5 m

(a) At the top:

UG = mgh = (3 kg)(9.81 m/s2)(5 m) = 147.2 J,

K = 0 and UE = 0.

∴Etop = K + UG + UE = 147.2 J.

At the bottom, when the object is at rest against the spring,

UG = 0, K = 0 and UE =

12

kx2.

∴Ebottom = K + UG + UE = UE = 200x2 J.

But mechanical energy is conserved ...

∴Ebottom = Etop, i.e., 200x2 = 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.

k = 400 N/m

x

5 m UG = 0

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

A

B

h

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!

A

B

x

×

×

h

h

Question 7.8: A physics student, with mass 80 kg, does a bungee jump from a platform 100 m above the ground. 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.

The total energy is 78,480 J, which is conserved throughout the jump. At his lowest point

K +

12

kℓ2 + mgy " = 78,480 J,

but K = 0 and ℓ = 50 − y".

∴100(50 − y")2 + 80 × 9.81y" = 78,480,

i.e., 100y"2 − (10,000 − 784.8)y" + (250,000 − 78,480) = 0.

∴y"2 − 92.15y" +1,715.2 = 0,

i.e., y" =

92.15 ± (92.15)2 − 4 ×1,715.22

.

lowest point

K = 0 : UE = 0 : UG = mgh = 78,480 J

UG = 0

ℓ = 50 m

ℓ

y" K = 0 : UE = 1

2kℓ2 : UG = mgy"

h = 100 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 is greatest, we need an expression for the kinetic energy

(K) in terms of the vertical position (y) and set

dKdy

= 0,

i.e., where K is a maximum.

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:

K +

12

k ′ ℓ 2 + mgy = 78,480 J.

fastest point

K = 0 : UE = 0 : UG = mgh = 78,480 J

UG = 0

ℓ = 50 m

′ ℓ

K : UE = 1

2k ′ ℓ 2 : UG = mgy

h = 100 m

y

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 −100y2,

so

dKdy

= 9,215.2 − 200y.

The extremum occurs when

dKdy

= 0,

i.e., at y =

9,215.2200

= 46.08 m.

Check that this is a maximum:

d2Kdy2 = −200, i.e., < 0.

(c) Since Kmax = 78,480 −100(50 − y)2 − 784.8y, when

y = 46.08m, we have

Kmax =

12

mvmax2 = 40,780 J,

i.e., vmax =

2 × 40,780 J80 kg

= 31.9 m/s.

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

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.

When the spring is stretched and released, the mass will 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.

lowest point 25.89 m

“highest” point 66.27 m mid point of

oscillation 46.08 m

When non-conservative forces are involved, the situation is more complicated; we need a more general statement of energy conservation, which involves the total energy of a system (e.g., bike + Earth + rider ...), i.e.,

Esystem = Ei

i∑ = Emech + Echem +…

If there is no input to the system from any other external sources (e.g., wind, a push), 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 stored by rider).

If an external source does work ( Wext) on a system then:

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.

Let’s look at some examples:• with chemical changes• with frictional forces

Question 7.9: A physics student, with mass 60 kg, climbs a 120 m high hill. (a) What is the increase in his

gravitational potential energy? (b) Where does this energy come

from? (c) If the student’s body is 20% efficient, i.e., for every

100 J of chemical energy used only 20 J are converted to mechanical energy with the remainder ( 80 J) resulting in

thermal energy, how much chemical energy is used by the student during the climb?

(a) ΔU = U2 − U1 = mgh

= (60 kg)(9.81 m/s2)(120 m) = 7.06 ×104J.

The total mechanical energy change is 70.6 kJ.

(b) The system is the climber and the Earth. With no external forces

Esys = Emech + Eclimber,

where Eclimber is the energy stored within the climber.

Since ΔEsys = 0, ΔEmech = −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 Emech + 80 J of Etherm

∴ 1 J of Emech 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.

Why do you have to keep pedaling when traveling on a level road, even though you’re not changing your speed (i.e., with no change in ΔK)? Isn’t

the work-energy theorem ( W⇒ ΔK) violated? No! From earlier we have

Esys = Ei

i∑ = Emech + Etherm + Echem.

ΔEsys = ΔEmech + ΔEtherm + ΔEchem = 0.

If you did no work (i.e., ΔEchem = 0), the work against

friction, air drag, etc., ( ΔEtherm) would have to come

from somewhere ... it would come from a loss of mechanical energy (kinetic energy), since

ΔEtherm = −ΔEmech,

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 = −ΔEthermal.

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 the bar. The mug slides down the ramp and onto the bar, 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 mug and the surfaces of the ramp and the bar is 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?

θ 0.50 m

2.0 m There are two contributions to the total energy of the system; Emech and Etherm (assuming that the work done

against friction produces heat).

∴Esys = Emech + Etherm.

If there are no external forces, then Esys is constant, i.e.,

ΔEsys = ΔEmech + ΔEtherm = 0.

Now, ΔEmech = ΔK + ΔU and ΔEtherm = fk.Δs, where

fk is the frictional force and Δs is the distance through

which the frictional force is active.

(a) In this problem, ΔEtherm comprises two components:

(1) down the ramp and (2) along the bar.

(1) Down the ramp: Δs1 =

0.50 msin θ

and

fk1 = µkN = µkmg cosθ.

θ 0.50 m

2.0 m

mg

N

θ 0.50 m

2.0 m

mg

N

∴ΔEtherm1 = (µkmg cosθ)Δs1 = (µkmg cosθ)

0.50 msin θ

=

0.15 × (1.5 kg) × (9.81 m/s2) × (0.50 m)tanθ

=1.104tanθ

J.

(2) Along the bar: Δs2 = 2.0 m and fk2 = µkmg

∴ΔEtherm2 = (µkmg)Δs2

= 0.15 × (1.5 kg) × (9.81 m/s2) × (2.0 m) = 4.415 J.

∴ΔEtherm = ΔEtherm1 + ΔEtherm2 =

1.104tanθ

+ 4.415 J.

Using conservation of energy ΔK + ΔUg + ΔEtherm = 0,

but ΔK = 0, i.e., ΔEtherm = −ΔUg.

ΔUg = −mgh = −(1.5 kg) × (9.81 m/s2 ) × (0.50 m)

= −7.358 J.

∴

1.104tanθ

+ 4.415 = 7.358 J,

i.e., θ = tan−1 1.104

7.358 − 4.415⎛ ⎝

⎞ ⎠ = 20.6!.

(b) On the ramp: ΔEsys = ΔEmech + ΔEtherm1 = 0,

i.e., ΔK1 + ΔUg + ΔEtherm1 = 0.

If v is the velocity at the bottom of the ramp

ΔK1 =

12

mv2 = −ΔUg − ΔEtherm1

= mgh −

1.104 Jtanθ

= 7.385 J − 2.937 J = 4.448 J.

∴v2 =

2ΔK1m

=2 × 4.448 J

1.50 kg,

i.e., v = 2.44 m/s.

θ 0.50 m

2.0 m mg

N

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) Put Ug = 0 at position 2 where the kinetic energy is

a maximum. Then conservation of energy gives:

K1 + Ug1 + Ue1 = K2 + Ug2 + Ue2,

i.e., mg (5.0 m) + Δy( ) = K2 +

12

k(Δy)2.

∴K2 = mg (5.0 m) + Δy( ) − 1

2k(Δy)2.

For maximum kinetic energy:

dK2d(Δy)

= mg − kΔy = 0,

i.e., Δy =

mgk

=(2.4 kg) × (9.81 m/s2)

400 N/m= 0.059 m.

1

2

3

start position

max kinetic energy

max compression

5.0 m

Δy y!

But K2 = mg (5.0 m) + Δy( ) − 1

2k(Δy)2

= (2.4 kg) × (9.81 m/s2) × (5.059 m)

−

12(400 N/m) × (0.059 m)2 = 118.4 J.

(b) Now we put Ug = 0 at position 3 .

Conservation of energy gives:

K1 + Ug1 + Ue1 = K3 + Ug3 + Ue3,

i.e., mg (5.0 m) + y!( ) = 1

2ky!2.

∴200y!2 − 23.54y! −117.7 = 0

1

2

3

start position

max kinetic energy

max compression

5.0 m

Δy y!

i.e., y! =

23.54 ± (23.54)2 + 4 × 200 ×117.72 × 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 compressed a distance Δℓ (from the top of the spring), so the net force acting on the mass is zero, i.e., when

kΔℓ = mg .

∴Δℓ =

mgk

=(2.4 m) × (9.81 m/s2)

400 N/m= 0.059 m.

Note that this is the position where the kinetic energy was a maximum in part (a)!

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.