PHYS 20 LESSONS Unit 4: Energetics Lesson 4: Conservation of Mechanical Energy.

PHYS 20 LESSONS

Unit 4: Energetics

Lesson 4: Conservation of MechanicalEnergy

Reading Segment:

Conservation ofMechanical Energy

To prepare for this section, please read:

Unit 4: p. 14

D. Conservation of Energy

Recall:

There are two major advantages to the energetics perspective:

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 energy is conserved.

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

Equations:

If mechanical energy is conserved, then you can use two equations:

1. EmT remains constant

EmTi = EmTf

Epgi + Eki = Epgf + Ekf

2. No overall change in energy

Epg + Ek = 0

Epg = - Ek

That is,

if Epg increases by 100 J,

then Ek decreases by 100 J (no overall change)

So, one form of mechanical energy transforms into

another form of mechanical energy

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

Epgi = 600 J

Eki = 200 J

Epgf Ref h (h = 0)

Ekf

What is the total mechanical energy at the start?

EmTi Epgi = 600 J

800 J Eki = 200 J

Epgf Ref h (h = 0)

Ekf

What is the total mechanical energy at the end?

EmTi Epgi = 600 J

800 J Eki = 200 J

EmTi = EmTf

EmTf Epgf Ref h (h = 0)

800 J Ekf

So, what is its kinetic energy at the end?

EmTi Epgi = 600 J

800 J Eki = 200 J

EmTf Epgf = 0 Ref h (h = 0)

800 J Ekf = 800 J

Epgi = 600 J

Eki = 200 J

Calculate and interpret:

- the change in Epg

- the change in Ek

Epgf = 0

Ekf = 800 J

Epgi = 600 J

Eki = 200 J Epg = Epgf - Epgi

= 0 - 600 J

= - 600 J

object loses 600 J of Epg

Epgf = 0 (since it loses height)

Ekf = 800 J

Epgi = 600 J

Eki = 200 J Ek = Ekf - Eki

= 800 J - 200 J

= + 600 J

object gains 600 J of Ek

Epgf = 0 (since it gains speed)

Ekf = 800 J

Epgi = 600 J

Eki = 200 J

The object loses 600 J of Epg,

but it gains 600 J of Ek

We say that the Epg has

transformed into Ek.

Epgf = 0

Ekf = 800 J

Animations:

1. Dropped object:

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

Can you explain what happens in the animation?

You should have noticed:

The Epg goes down by 1078 J,

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

That is, the Epg has transformed entirely into Ek

Thus, the total mechanical energy will remain the same

throughout the entire motion

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

Ex. 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 Epg vs t, Ek vs t, and a EmT vs t graph

a) EmTi = EmTf Rest

Epgi + Eki = Epgf + Ekf

Ref h (h = 0)

Be certain to state 7.0 m/s

where the height is zero

a) EmTi = EmTf Rest

Epgi + Eki = Epgf + Ekf

Ref h (h = 0)

7.0 m/s

Since it starts at rest, Eki = 0

Since it has no height at the end, Epgf = 0

EmTi = EmTf

Epgi + Eki = Epgf + Ekf

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

EmTi = EmTf

Epgi + Eki = Epgf + Ekf

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

b) Epg Ek

t t

EmT

t

b) Epg Ek

height speed

decreases increases

t t

EmT constant

t

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

A 6.70 m/s

v ?

C

8.70 m B

4.20 m

1.90 m

D

Find its speed at C.

A 6.70 m/s

v ?

C

8.70 m B

4.20 m

1.90 m

D (h = 0)

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

the lowest location in the diagram

A 6.70 m/s

v ?

C

8.70 m B

4.20 m

1.90 m

D (h = 0)

Total mechanical energy remains the same.

So, EmTA = EmTC

EpgA + EkA = EpgC + EkC

Ref h = D

EpgA = mghA

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

EkA = 0.5mvA2

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

EpgC = mghC

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

Find EkC:

EmTA = EmTB

EpgA + EkA = EpgC + EkC

102.42 J + 26.93 J = 49.44 J + EkC

EkC = 79.908 J

Find speed at C:

EkC = 0.5 mv2

v2 = EkC

0.5 m

v = EkC = 79.908 J = 11.5 m/s

0.5 m 0.5 (1.20 kg)

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 4 p. 15 #1 - 3

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 Ek, find the object's new height.

Assume a conservative system.

Solution

The total mechanical energy must remain the same

So, if the object loses 92.0 J of Ek,

it must gain 92.0 J of Epg

That is, the Ek converts (transforms) into Epg

Find Epgi:

Epgi = mghi

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

= 103.0 J

Find Epgf:

Since Epg must increase by 92.0 J,

Epgf = Epgi + 92.0 J

= 103.0 J + 92.0 J

= 195.0 J

Find the final height:

Epgf = mghf

hf = Epgf = 195.0 J = 1.42 m

mg (14.0 kg) (9.81 N/kg)

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 4 p. 15 #4 - 8