PHY082 (PHYSICS II)

CHAPTER 5: MATTER

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Lecture Outline

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5.1 Elasticity

Concept and Definition

5.2 Hooke’s Law

5.2.1 Definition

5.2.2 Force – Extension Graph

5.2.3 Work done & elastic potential energy

5.3 Buoyancy in fluid

5.3.1 Archimedes’ Principle

5.3.2 Problem solving involving buoyant force

5.3.3 Bernoulli’s Principle and applications

5.1 Elasticity

• Every body will change its shape or size when external force acting on it – deformation.

• If the body returns to original shape or size when the external force is removed, then it is said to be elastic.

• Example is a spring.• Objects which are elastic obeys Hooke’s law.

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5.2 Hooke’s Law

5.2.1 Definition:• An object exerted by force, such as pulling force

will cause the length of the change.• If the amount of elongation is small compare to

the length of the object, Hooke’s Law state that:

The change in length is proportional to the applied force.

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lkF ∆=

• F = force (or weight) pulling on the object,

∀ ∆ L = increase in length,• k = proportionality constant

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lkF ∆= ….(Eq. 5.1)

• First noted by Robert Hooke (1635 – 1703)

• Valid for almost any solid material.• But applicable for only to a certain

point – force is not too strong.• Up to a point – Hooke’s Law is not

valid.

Figure 5.1

Example 5.1• A 0.7 kg wooden block is used to stretch a spring as

shown in figure below. When a force of 100 N is exerted on the spring, it is stretch 0.2 m to the right direction. Calculate the spring constant. (PHY082 - APR2011).

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lkF ∆=

kxF =

N/m 500m 0.2

N 100 ===x

Fk

Hooke’s Law

Hooke’s Law for Spring

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• This proportionality, holds until the force reaches the proportional limit.

• Beyond that, the object will still return to its original shape up to the elastic limit.

• Beyond the elastic limit, the material is permanently deformed.

• The maximum elongation is reached at the breaking point.

• Ultimate Strength of the material – Maximum force that can be applied without breaking.

Figure 5.2

5.2.2 Force – Extension Graph

Stress and Strain

• Stress is defined as the force per unit area.

• Strain is defined as the ratio of the change in length to the original length. (how much an object is deform).

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A

F==area

force Stress,σ [Unit: N/m2](Eq. 5.2)…

0length original

lengthin change Strain,

l

l∆==ε(Eq. 5.3)…

• Therefore, the elastic modulus is equal to the stress divided by the strain:

• E = Young’s modulus

• Unit: Newton/ meter2 (N/m2)

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[Source: Giancoli]

Example 5.2

• Tension in a piano wire (Giancoli).

A 1.60 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.30 cm when tightened?

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N1200

)m101.3)(m60.1

m0030.0)(N/m100.2( 26211

0

=

××=

∆=

−

Al

lEF

5.2.3 Work done & Elastic potential energy

Work done, W

• Review:

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[Source: Giancoli]

sFW .=

∫∫

=

=x

x

p

dxkx

dxxFW

0

0

)(

2

2

1kxW =

kxF =

….(Eq. 5.4)

13[Source: Giancoli]

Work done = Area under the

curve

2

2

1))((

2

1kxkxxW ==

[Unit: Joule (J)]

Potential Energy, U

Notes:• From principle of conservation energy:

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2

2

1kxWU == ….(Eq. 5.5)

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2

1

2

1mvkx =

Potential Energy = Kinetic Energy

Example 5.3• Work done on a spring. (Giancoli)

A person pulls on the spring in Figure below, stretching it 3.0 cm, which requires a maximum force of 75 N. How much work does the person do?

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N/m105.2

m030.0

N75

max

max

3×=

==x

Fk

J 1.1

)m030.0)(N/m105.2(2

1

2

1

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2

=

×=

= kxW

Example 5.4

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• A 0.7 kg wooden block is used to stretch a spring as shown in figure below. When a force of 100 N is exerted on the spring, it is stretch 0.2 m to the right direction. Calculate the elastic potential energy stored in the spring. Given the k = 500 N/m. (PHY082 - APR2011).

J 10

)2.0)(N/m500(2

1

2

1

2

2

=

=

= kxU

5.3 Buoyancy in Fluid

• Buoyant force, FB: The force acting perpendicularly upward on a body when a part or the whole body is immersed in a fluid.

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Figure 5.6

….(Eq. 5.6)

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5.3.1 Archimedes’ Principle:

The buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by that object.

Figure 5.7

Example 5.5

• A block of dimensions 5 m x 2 m x 20 m floats in water with half its volume immersed in water. If density of water 1000 kgms-3, calculate the buoyant force acting on the block? Assume g = 10ms-2.

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N 100.1

))2025(2

1)(10)(1000(

forceBuoyant

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W

×=

××=

= gVρ

Example 5.6

• A metal block is hung against a Newton scale. When in air, the scale reads 5.5 N but when fully submerged in water the reading becomes 4.8 N. What is the factor that contributes to the difference? Calculate the buoyant force that acts on the block.

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N 0.7

4.8-5.5

==

Buoyant force = Actual weight in air – weight in water

Solution:

Example 5.7

• A hot air balloon floats in air. If the total mass of the balloon is 270 kg, what is the buoyant force on the balloon? Assume g = 10 ms-2.

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N 2700

)ms kg)(10 (270

g 2-

===m

Buoyant force = Weight of object

Solution:

5.3.3 Bernoulli’s Principle & applications

• Bernoulli’s Principle states that where the velocity of a fluid is high, the pressure is low, and where the velocity is low, the pressure is high.

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↓↑ , fluid Pv ↑↓ , fluid Pv

22221

211 2

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1 gyvPgyvP ρρρρ ++=++ ….(Eq. 5.7)

constant21 2 =++ gyvP ρρ

Applications:

1. Torricelli’s (fig. 13-25)

2. Atomizer (a)

3. Ping-pong ball in jet of air (b)

4. Airplane wing (c)

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