3. types of energy

7/29/2019 3. Types of Energy

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UNIT 1: CONSERVATION LAWS

Lesson 3:

Types of Energy

CENTRE HIGH: PHYSICS 30


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Recommended Reading:

Ladner pp. 6 - 8p. 10

Heath pp. 287 - 296


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Recall: Work is a bridge between dynamics and energetics

WorkW = Fd cos W = E

Force Energy

We will now see how work was used to derive each form of

energy.


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B3. Work and Energy

Work:If work is done on an object, its energy will change

W = E


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Energy

- energy is the ability to do work

- the more energy an object has, the more work it can do


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C. MECHANICAL ENERGY (ME)

- there are three types of mechanical energy:

1. Kinetic energy (KE)

- due to speed

2. Gravitational Potential Energy (PEg)

- due to height

3. Elastic Potential Energy (Ee)

- due to stretching / compressing

- all are scalars (they have no direction)


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C1. Kinetic Energy (KE or Ek)

- energy due to speed or motion

- if an object has speed, it can do work on another object


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Consider a car moving towards a large crate

e.g.

How do we know that the car has energy?


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When the car hits the crate, it does work on the crate

(i.e. it moves the crate over a distance)

Thus, it must have energy due to its motion


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Equation:

KE = 0.5 mv2

where m is the mass of the object (in kg)

v is the speed of the object (in m/s)

Units: Joules (J)

Based on the equation above, what is an

equivalent unit for a Joule?


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Equation:

KE = 0.5 mv2


v is the speed of the object (in m/s)

Units: Joules (J)

1 J = 1 kg m 2 = 1 kg m2

s s2

Note: KE is always positive (it can never be negative)


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Derivation:

To see how the equation KE = 0.5 mv2

was derivedusing the concept of work, you can read:

Ladner p. 10

Heath p. 287

You are not responsible for this, though.


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Ex. 1 A 560 kg car is travelling at 40.0 km/h.

a) Find its kinetic energy.

b) If it then loses 21.1 kJ, then find its final speed (in m/s).


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a)

KEi = 0.5 m vi2

Remember: The speed must be in m/s


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a) vi = 40.0 x 1000 m = 11.1111 m/s

3600 s

KEi = 0.5 mvi2

= 0.5 (560 kg) (11.111 m/s)2

= 34568 J

KEi = 3.46 x 104 J


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b) Since the car loses 21.1 kJ,

KEf = KEi - 21.1

103

J= (34568 J) - (21.1 103 J)

= 13468 J


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b)

KEf = 0.5 mvf2

vf2 = KEf

0.5 m

vf = KEf = 13468 J

0.5 m 0.5 (560 kg)

= 6.94 m/s


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Practice Problems:

Try:Ladner p.11 #1 - 6


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C2. Gravitational Potential Energy (PEg or Epg)

- stored energy due to an object's height

- if an object has height, it can do work on another object


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Consider an object located a height above a nail

Object

NailHow do we know the object

has energy?


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When the block is dropped,

it does work on the nail

(applies a force over a distance)

Thus, it must have energy


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Equation:

PEg = m g h


g is the magnitude of gravity (in N/kg)

h is the height of an object

Units:

Joules (J)


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Derivation:

To see how the equation PEg = m g h was derivedusing the concept of work, you can read:

Heath p. 298 - 299



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Determining heights:

Height is always measured relative to some reference height

At ref height, h = 0

Above ref height, h is positive (h > 0)

Below ref height, h is negative (h < 0)


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e.g. Consider an object moving on the following path:

B

15 m

A

12 m

C


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If the reference height is at A:

B

15 m

A Ref h

12 m

hA = 0 C


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hB = + 15 m

(above ref height) B

15 m

A Ref h

12 m

C


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hC = 12 m

B (below ref height)

15 m

A Ref h

12 m

C


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But, if the reference height is at C:

B

15 m

A

12 m

C Ref h


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hA = + 12 m

(above ref height)

B

15 m

A

12 m

C Ref h


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hB = + 27 m

(above ref height) B

15 m

A

12 m

C Ref h


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hC = 0

B (at ref height)

15 m

A

12 m

Ref h C


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Why is PEg relative to reference height?

To understand this, answer the

following:

How much work can this object do

if it is dropped?


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If dropped forward:

Big PEg

Lots of work

Ref h


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If dropped forward:

Big PEg

Lots of work

Ref h

But what if it was dropped backwards?


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Small PEg

Little work

Ref h

If dropped backward, it has far less PEg


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So, how much PEg does this

object have?

It depends on the ref height chosen

Thus, PEg is not a measurable valueIt is relative.


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But as we will discover in the next example,

Change in PEg (PEg) is independent of ref height

and thus, it is measurable


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Ex. 2 An 8.00 N object is lifted from position A to position B.

B

150 cm

A

60 cm

ground

Find the change in gravitational potential energy

if the ref height is:

a) the ground b) position A


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For both questions, notice:

PEg

= PEgf

- PEgi

= m g hf - m g hi

= m g (hf - hi)

Also, the weight of the object is 8.00 N

i.e.

Weight = m g = 8.00 N


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a) DLESC B

hf= +150 cm

A

hi =+60 cm

Ref height

(h = 0)


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a) DLESC B

hf= +150 cm

A

hi =+60 cm

Ref height

(h = 0)PEg = m g (hf - hi)

= (8.00 N) (1.50 m - 0.60 m)

= 7.2 J

Positive answer

PEg is increasing


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b) B

90 cm

150 cm

A Ref height

60 cm (h = 0)

When the ref height is A,

hi = 0 (since it starts at the reference height)

hf = +0.90 m


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b) B

90 cm

150 cm

A Ref height

60 cm (h = 0)

PEg = m g (hf - hi)

= (8.00 N) (0.90 m - 0 m)

= 7.2 J

Same answer as (a)


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From this example, you can see that

the change in gravitational potential energy (PEg)

is independent of reference height.

That is, we got the same value for PEg

with two different reference heights

Why is this valuable?

It means that PEg is measurable


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Practice Problems:

Try:

Ladner p.8 #1, 3 - 5


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C3. Elastic Potential Energy (Ee or Es)

- energy due to stretching / compressing an elastic material

- if you compress a spring, it can do work on an object

e.g.

Compressed spring

The spring did work on the object.


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Equation:

Ee

= 0.5 kx2

where k is the stiffness or spring constant (in N/m)

x is the amount the spring is stretched / compressed

from its rest length (in m)Ee is the energy stored within a spring (in J)


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Derivation:

To see how the equation Ee

= 0.5 kx2 was derived

using the concept of work, you can read:

Heath p. 293 - 295



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Ex. 3 A spring has a rest length of 24.0 cm. If the spring is

stretched to a length of 35.0 cm and stores 2.8 kJ of

energy, then what is its spring constant (in N/m)?

Show why the units will become N/m.


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Ee = 0.5 kx2

We know that Ee = 2.8 103 J,

but what is x?


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x is the length the spring has been stretched from its rest length

i.e.

24.0 cm (rest length)

x = 11.0 cm

35.0 cm


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Ee = 0.5 kx2

k = Ee

= 2.8 x 103 J

0.5 x2 0.5 (0.110 m)2

= 4.6 x 105 N/m


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We need to show that 1 J = 1 N

m2 m


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We need to show that 1 J = 1 N

m2 m

Based on the equation W = F d, we know that 1 J = 1 Nm

So, J = N m = Nm2 m2 m


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Practice Problems:

Try:

Heath p.297 #1 - 3


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SUMMARY

1. When work is done on an object, its energy __________

Energy is the ability to do _____________

2. KE is the energy due to ____________

PEg is the energy due to ______________

Ee is the energy due to ________________

3. Which is the only energy that can be negative? When?

4. PEg is relative to ___________________

But __________________ is not.

5. What does it mean when E is positive? negative?


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SUMMARY

1. When work is done on an object, its energy changes

Energy is the ability to do work

2. KE is the energy due to speed (motion)

PEg is the energy due to height

Ee is the energy due to stretching / compressing


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SUMMARY

3. Which is the only energy that can be negative? When?

Only PEg can be negative

When the object is below the reference height

4. PEg is relative to reference height

But change in PEg is not.

5. IfE is positive, then E is increasing

If it is negative, it is decreasing

3. types of energy

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