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UNIT 1: CONSERVATION LAWS

Lesson 4:

Conservation of Mechanical Energy

CENTRE HIGH: PHYSICS 30

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D. Conservation of Energy

Recommended Reading:

Ladner pp. 16 - 17

Heath pp. 291 - 292, 300

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D. Conservation of Energy

Recall:

There are two major advantages to the energetics perspective:

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D. Conservation of Energy

Recall:

There are two major advantages to the energetics perspective:

1. Energy is a scalar

- does not have direction

- no need for vector math

2. It is conserved

We will now analyze situations where total mechanical energyis conserved.

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D1. Total Mechanical Energy (MET)

MET = sum of all mechanical energies= KE + PEg (+ Ee)

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Ex. 1 mcar = 400 kg

20 m/s

Aconstant speed

16 m 20 m/s

B

If the car's PEg is zero at A, then find MET at B (in kJ).

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Solution

A Ref h (h = 0)

-16 m 20 m/s

B

Since the height is zero at A,

then location B is below the reference height

So, the height at B is negative

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Solution

A Ref h (h = 0)

-16 m 20 m/s

B

PEgB = m g h

= (400 kg) (9.81 N/kg) (-16 m) = -62,784 J

KEB = 0.5 mv2

= 0.5 (400 kg) (20 m/s)2 = 80,000 J

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MET = PEgB + KEB

= -62,784 J + 80,000 J

= 17, 216 JUnit you want

MET = 17, 216 J x 1 kJ

1000 J

Unit to cancel

= 17 kJ

Check:

2 sig digs; units

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D2. Energy Conversions

Recall W =

E

- there are two ways that energy can change:

1. Energy Transformation

2. Energy Transfer

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1. Energy Transformations

- when energy changes from one form (or type)

to another

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e.g. Dropping a ball:

Rest

Ref h

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Since the object is losing height,

it is losing PEg All PEg Rest

Where does this energy go?

No PEg Ref h

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Since the object is losing height,

it is losing PEg All PEg Rest

No KEAt the same time, the object is

gaining speed.

Thus, it gains KE.

No PEg Ref h

All KE

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Since the object is losing height,

it is losing PEg All PEg Rest

No KEAt the same time, the object is

gaining speed.

Thus, it gains KE.

No PEg Ref h

All KE

We say that the PEg has transformed

into KE.

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2. Energy Transfer

When energy goes from one object to another.

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e.g. Car colliding with another stationary car:

moving car stationary car

(KE) (no KE)

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KE No KE

Both have KE

After the collision, the blue car has gained kinetic energy.

Where did the energy come from?

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KE No KE

Both have KE

Some kinetic energy has transferred from the first car

to the second

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D3. Conservation of Mechanical Energy

- total mechanical energy stays constant (conserved) when:* the system is closed

- no objects are added / lost

* no friction

* only mechanical forms of energy change

- only Fg and Fs do work on the object

- mechanical energy can convert only to another

form of mechanical energy

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Equations:

If mechanical energy is conserved, then you can use two

equations:

1. MET remains constant

METi = METfPEgi + KEi = PEgf + KEf

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2. No overall change in energy

PEg +

KE = 0

PEg = - KE

That is,

if PEg increases by 100 J,

then KE decreases by 100 J (no overall change)

So, one form of mechanical energy transforms intoanother form of mechanical energy

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e.g. Consider an object thrown downward (No air resistance)

PEgi = 600 J

KEi = 200 J

PEgf Ref h (h = 0)

KEf

What is the total mechanical energy at the start?

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METi PEgi = 600 J

800 J KEi = 200 J

PEgf Ref h (h = 0)

KEf

What is the total mechanical energy at the end?

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METi PEgi = 600 J

800 J KEi = 200 J

METi = METf

METf PEgf Ref h (h = 0)

800 J KEf

So, what is its kinetic energy at the end?

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METi PEgi = 600 J

800 J KEi = 200 J

METf PEgf= 0 Ref h (h = 0)

800 J KEf = 800 J

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PEgi = 600 J

KEi = 200 J

Calculate and interpret:

- the change in PEg

- the change in KE

PEgf = 0

KEf = 800 J

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PEgi = 600 J

KEi = 200 J PEg = PEgf - PEgi

= 0 - 600 J= - 600 J

object loses 600 J of PEgPEgf = 0 (since it loses height)

KEf = 800 J

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PEgi = 600 J

KEi = 200 J KE = KEf - KEi

= 800 J - 200 J= + 600 J

object gains 600 J of KEPEgf = 0 (since it gains speed)

KEf = 800 J

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PEgi = 600 J

KEi = 200 J

The object loses 600 J of PEg,but it gains 600 J of KE

We say that the PEg has

transformed into KE.PEgf = 0

KEf = 800 J

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Animations:

1. Dropped object:

http://departments.weber.edu/physics/amiri/director/dcrfiles/energy/FallingBallS.dcr

Can you explain what happens in the animation?

http://departments.weber.edu/physics/amiri/director/dcrfiles/energy/FallingBallS.dcrhttp://departments.weber.edu/physics/amiri/director/dcrfiles/energy/FallingBallS.dcrhttp://departments.weber.edu/physics/amiri/director/dcrfiles/energy/FallingBallS.dcrhttp://departments.weber.edu/physics/amiri/director/dcrfiles/energy/FallingBallS.dcr

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You should have noticed:

The PEg goes down by 1078 J,

but at the same time, the KE goes up by 1078 J

That is, the PEg has transformed entirely into KE

Thus, the total mechanical energy will remain the samethroughout the entire motion

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Other animations showing conservation of energy:

2. Pendulum:

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=27

3. Rollercoaster:

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/

coastwin.html

http://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=27http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/coastwin.htmlhttp://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/coastwin.htmlhttp://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/coastwin.htmlhttp://www.glenbrook.k12.il.us/gbssci/phys/mmedia/qt/energy/coastwin.htmlhttp://www.phy.ntnu.edu.tw/ntnujava/viewtopic.php?t=27

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Ex. 2 A ball is dropped from a height H and it lands with a

speed of 7.0 m/s. If the system is conservative,

a) find H

b) sketch a PEg vs t, KE vs t, and a MET vs t graph

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a) METi = METf Rest

PEgi+ KE

i= PEg

f+ KE

f

Ref h (h = 0)

Be certain to state 7.0 m/s

where the height is zero

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a) METi = METf Rest

PEgi+ KE

i= PEg

f+ KE

f

Ref h (h = 0)

7.0 m/s

Since it starts at rest, KEi = 0Since it has no height at the end, PEgf = 0

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METi = METf

PEgi + KEi = PEgf+ KEf

mghi = 0.5 mvf2

ghi = 0.5 vf2

If you divide both sides of the equation by m,

the mass cancels

Thus, the answer does not depend on mass.

i.e. It is true for any mass

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METi = METf

PEgi + KEi = PEgf+ KEf

mghi = 0.5 mvf2

ghi = 0.5 vf2

hi = 0.5 vf2 = 0.5 (7.0 m/s)2

g 9.81 m/s2

H = 2.5 m

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b) PEg KE

t t

MET

t

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b) PEg KE

height speed

decreases increases

t t

MET constant

t

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Ex. 3 A 1.20 kg car travels the following path. No friction.

A 6.70 m/s

v ?

C

8.70 m B4.20 m

1.90 m

D

Find its speed at C.

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A 6.70 m/s

v ?

C

8.70 m B

4.20 m

1.90 mD (h = 0)

The reference height (h = 0) is at D,

the lowest location in the diagram

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A 6.70 m/s

v ?

C

8.70 m B

4.20 m

1.90 mD (h = 0)

Total mechanical energy remains the same.

So, META

= METC

PEgA + KEA = PEgC + KEC

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Ref h = D

PEgA = mghA

= (1.20 kg) (9.81 N/kg) (8.70 m) = 102.42 J

KEA = 0.5mvA2

= 0.5 (1.20 kg) (6.70 m/s)2 = 26.93 J

PEgC = mghC

= (1.20 kg) (9.81 N/kg) (4.20 m) = 49.44 J

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Find KEC:

META = METB

PEgA + KEA = PEgC + KEC

102.42 J + 26.93 J = 49.44 J + KEC

KEC = 79.908 J

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Find speed at C:

KEC

= 0.5 mv2

v2 = KEC

0.5 m

v = KEC = 79.908 J = 11.5 m/s0.5 m 0.5 (1.20 kg)

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Ex. 4

A 14.0 kg object is currently moving at a height of 75.0 cm. If

it then loses 92.0 J of KE, find the object's new height.

Assume a conservative system.

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Solution

The total mechanical energy must remain the same

So, if the object loses 92.0 J of KE,

it must gain 92.0 J of PEg

That is, the KE converts (transforms) into PEg

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Find PEgi:

PEgi

= mghi

= (14.0 kg) (9.81 N/kg) (0.750 m)

= 103.0 J

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Find PEgf:

Since PEg

must increase by 92.0 J,

PEgf = PEgi + 92.0 J

= 103.0 J + 92.0 J

= 195.0 J

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Find the final height:

PEgf

= mghf

hf = PEgf = 195.0 J = 1.42 m

mg (14.0 kg) (9.81 N/kg)

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Practice Problems:

Try:

Ladner p.19 #1 - 10, 13

Heath p.304 #3, 4

Challenge:

Ladner p.19 #11, 12

Heath p. 344 #38

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SUMMARY

1. Identify the 3 conditions for MET to remain constant.

2. An object experiences a gain of 700 J of PEg. If the system is

conservative, describe what happens to:

a) its KE b) MET

3. A ball is thrown vertically upward to a maximum height. If

there is no AR, sketch the following graphs:

a) PEg vs t b) KE vs t c) MET vs t

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SUMMARY

1. Identify the 3 conditions for MET to remain constant.

Closed system

- no objects lost, added

No friction

Only mechanical energy can change

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2. An object experiences a gain of 700 J of PEg. If the system is

conservative, describe what happens to:

a) KE b) MET

a) KE must decrease by 700 J

KE = - PEg

b) MET remains constant

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3. A ball is thrown vertically upward to a maximum height. If

there is no AR, sketch the following graphs:

a) PEg vs t b) KE vs t c) MET vs t

a) PEg b) KE c) MET

t t theight speed