Note Chapter4 SF017

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PHYSICS CHAPTER 4

CHAPTER 4: CHAPTER 4: Work, Energy and PowerWork, Energy and Power

(3 Hours)(3 Hours)

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PHYSICS CHAPTER 4

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and useDefine and use work done work done byby a force. a force.

CalculateCalculate work done from the force-displacement work done from the force-displacement graph.graph.

DiscussDiscuss the area under graph. the area under graph. State and explainState and explain the relationship between work and the relationship between work and

change in energy.change in energy.

Learning Outcome:

4.1 Work and energy (1 hour)

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sFW

PHYSICS CHAPTER 4

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4.1 Work and energy4.1.1 Work, W Work done by a constant forceWork done by a constant force is defined as the product of the component of the force the product of the component of the force

parallel to the displacement times the displacement of a parallel to the displacement times the displacement of a bodybody.

OR

is defined as the scalar (dot) product between force and the scalar (dot) product between force and displacement of a bodydisplacement of a body.

Equation :

θFssθFW coscos

sFW

force of magnitude:F

sFθ

and between angle the:

body theofnt displaceme : s

where

PHYSICS CHAPTER 4

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It is a scalarscalar quantity. Dimension :

The S.I. unit of work is kg mkg m22 s s22 or joule (J)joule (J). The joule joule (1 J) is defined as the work done by a force of 1 N the work done by a force of 1 N

which results in a displacement of 1 m in the direction of which results in a displacement of 1 m in the direction of the forcethe force.

Work done by a variable forceWork done by a variable force Figure 4.1 shows a force, F whose magnitude changes with the

displacement, s.

For a small displacement, s1 the force remains almost

constant at F1 and work done therefore becomes W1=F1 s1 .

sFW 22TML W

22 s m kg 1m N 1J 1

PHYSICS CHAPTER 4

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To find the total work done by a variable force, W when the

displacement changes from s=s1 to s=s2, we can divide the displacement into N small successive displacements :

s1 , s2 , s3 , …, sN Thus

FN

F4

s4 sN

s1 s2

F/N

s0

F1

s1

W1

NN2211 sFsFsFW ...

Figure 4.1Figure 4.1

PHYSICS CHAPTER 4

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When N , s 0, therefore

2

1

s

sFdsW

graphnt displaceme-force under the area theW

F/N

s/ms1 s20

Work = Area

PHYSICS CHAPTER 4

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4.1.2 Applications of work’s equation

Case 1 : Case 1 : Work done by a horizontal force, F on an object (Figure 4.2).

Case 2 : Case 2 : Work done by a vertical force, F on an object (Figure 4.3).

0θF

s


θFsW cosFsW

and

90θθFsW cosJ 0W

andF

s


PHYSICS CHAPTER 4

8

Case 3 : Case 3 :

Work done by a horizontal forces, F1 and F2 on an object

(Figure 4.4).

Case 4 : Case 4 : Work done by a force, F and frictional force, f on an object

(Figure 4.5).

0cos sFW 11 0cos sFW 22

sFWW nettnett

1F

2F

s

Figure 4.4Figure 4.4 sFsFWWW 2121 21nett FFF sFFW 21 and

cos mafθFFnett sFW nettnett sfFWnett cos masWnett

f F

Figure 4.5Figure 4.5 s

and

OR

PHYSICS CHAPTER 4

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Caution : Work doneWork done on an object is zerozero when FF = = 00 or ss = = 00 and

= = 9090.

PHYSICS CHAPTER 4

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Sign for work.

If 0< <90 (acute angleacute angle) then cos > 0 (positive value)therefore

WW > 0 (positive) > 0 (positive) work done on the system ( by work done on the system ( by the external force) the external force)

where energy where energy is transferred to the is transferred to the system.system.

If 90< <180 (obtuse angleobtuse angle) then cos <0 (negative value) therefore

WW < 0 (negative) < 0 (negative) work done by the system work done by the system where energy is where energy is

transferred transferred from the from the system.system.

cosFsW

PHYSICS CHAPTER 4

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You push your physics reference book 1.50 m along a horizontal table with a horizontal force of 5.00 N. The frictional force is 1.60 N. Calculate

a. the work done by the 5.00 N force,

b. the work done by the frictional force,

c. the total work done on the book.

Solution :Solution :

a. Use work’s equation of constant force,

Example 1 :

m 1.50s

N 5.00FN 1.60f

cosθFsWF

J 7.50FW

0θand

0cos1.505.00 FW

PHYSICS CHAPTER 4

12


b.

c.

θfsW f cos

fF WWW

OR

sFW nett

sfFW

180θand

180cos1.501.60fW

J 2.40fW

2.407.50W

J 5.10W

1.501.605.00W

J 5.10W

PHYSICS CHAPTER 4

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A box of mass 20 kg moves up a rough plane which is inclined to

the horizontal at 25.0. It is pulled by a horizontal force F of magnitude 250 N. The coefficient of kinetic friction between the box and the plane is 0.300.

a. If the box travels 3.80 m along the plane, determine

i. the work done on the box by the force F,

ii. the work done on the box by the gravitational force,

iii. the work done on the box by the reaction force,

iv. the work done on the box by the frictional force,

v. the total work done on the box.

b. If the speed of the box is zero at the bottom of the plane,

calculate its speed when it is travelled 3.80 m.

(Given g = 9.81 m s2)

Example 2 :

PHYSICS CHAPTER 4

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25


a. Consider the work done along inclined plane, thus

i.

25

kf

N

F

s

gmW

yF

25cosmg

xF

25sinmg

m 3.800.300; ;N 250 ;kg 20 sμFm k

a

25

θsFW xF cos 0θand

0cos3.8025cos250FW

J 861FW

xy

PHYSICS CHAPTER 4

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a. ii.

iii.

iv.

θsmgWg cos25sin 180θand

180cos3.8025sin9.8120gWJ 315gWθNsWN cos 90θand

J 0NW

θsfW kf cos 180θand

180cossNμW kf

J 323fW

smgFμW kf 25cos25sin

3.8025cos9.812025sin2500.300 fW

PHYSICS CHAPTER 4

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a. v.

b. Given

By using equation of work for nett force,

Hence by using the equation of linear motion,

fNgF WWWWW

J 223W

3230315861W

masW 3.8020223 a

2s m 2.93 a

asuv 22 2

0u

1s m 4.72 v 3.802.9320 2v

PHYSICS CHAPTER 4

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A horizontal force F is applied to a 2.0 kg radio-controlled car as it moves along a straight track. The force varies with the displacement of the car as shown in figure 4.6. Calculate the work

done by the force F when the car moves from 0 to 7 m.


Example 3 :

5

47

053 6

(N)F

5 (m)s


graph under the area sFW

4672

15356

2

1W

J 18W

PHYSICS CHAPTER 4

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Exercise 4.1 :1. A block of mass 2.50 kg is pushed 2.20 m along a frictionless

horizontal table by a constant 16.0 N force directed 25.0 below the horizontal. Determine the work done on the block by

a. the applied force,

b. the normal force exerted by the table, and

c. the gravitational force.

d. Determine the total work on the block.

(Given g = 9.81 m s2)

ANS. : 31.9 J; (b) & (c) U think; 31.9 JANS. : 31.9 J; (b) & (c) U think; 31.9 J

2. A trolley is rolling across a parking lot of a supermarket. You apply a constant force to the trolley as it undergoes a displacement . Calculate

a. the work done on the trolley by the force F,

b. the angle between the force and the displacement of the

trolley.

ANS. : ANS. : 150 J; 108150 J; 108

N j40i30 F

m j3.0i9.0 s

PHYSICS CHAPTER 4

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Exercise 4.1 :3.

Figure 4.7 shows an overhead view of three horizontal forces acting on a cargo that was initially stationary but that now moves across a frictionless floor. The force magnitudes are

F1 = 3.00 N, F2 = 4.00 N and F3 = 10.0 N. Determine the total

work done on the cargo by the three forces during the first 4.00 m of displacement.

ANS. : 15.3 JANS. : 15.3 J

3F

1F

2F

y

x

35

50


PHYSICS CHAPTER 4

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4.1.3 Energy is defined as the system’s ability to do workthe system’s ability to do work. The S.I. unit for energy is same to the unit of work (joule, Jjoule, J). The dimension of energy ,

is a scalar quantityscalar quantity. Table 4.1 summarises some common types of energy.

22 TMLWorknergyE

Forms of Energy

Description

ChemicalEnergy released when chemical bonds between atoms and molecules are broken.

Electrical Energy that is associated with the flow of electrical charge.

Heat Energy that flows from one place to another as a result of a temperature difference.

InternalTotal of kinetic and potential energy of atoms or molecules within a body.

PHYSICS CHAPTER 4

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Forms of Energy

Description

Table 4.1Table 4.1

Nuclear Energy released by the splitting of heavy nuclei.

Mass

Energy released when there is a loss of small amount of mass in a nuclear process. The amount of energy can be calculated from Einstein’s mass-energy

equation, E = mc2

Radiant Heat Energy associated with infra-red radiation.

SoundEnergy transmitted through the propagation of a series of compression and rarefaction in solid, liquid or gas.

Mechanicala. Kinetic

b. Gravitational

potentialc. Elastic

potential

Energy associated with the motion of a body.

Energy associated with the position of a body in a gravitational field.

Energy stored in a compressed or stretched spring.

PHYSICS CHAPTER 4

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and useDefine and use kinetic energy, kinetic energy,

Define and useDefine and use potential energy: potential energy:

i. gravitational potential energy, i. gravitational potential energy,

ii. elastic potential energy for spring, ii. elastic potential energy for spring,

State and useState and use the principle of conservation of energy. the principle of conservation of energy. ExplainExplain the work-energy theorem and the work-energy theorem and useuse the related the related

equation.equation.

Learning Outcome:

4.2 Conservation of energy (1 hour)

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2

2

1mvK

mghU

2

2

1kxU

PHYSICS CHAPTER 4

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4.2 Conservation of energy

4.2.1 Kinetic energy, K is defined as the energy of a body due to its motionthe energy of a body due to its motion. Equation :

Work-kinetic energy theoremWork-kinetic energy theorem Consider a block with mass, m moving along the horizontal

surface (frictionless) under the action of a constant nett force,

Fnett undergoes a displacement, s in figure 4.8.

2

2

1mvK

body a ofenergy kinetic:K

body a of speed : v

body a of mass : m

where

s

nettF

m


maFF nett (1)

PHYSICS CHAPTER 4

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By using an equation of linear motion:

By substituting equation (2) into (1), we arrive

Therefore

states “the work done by the nett force on a body equals the the work done by the nett force on a body equals the change in the body’s kinetic energychange in the body’s kinetic energy”.

as uv 222

s

uva

2

22 (2)

2

22

s

uvmFnett

ifnett KKmumvsF 22

2

1

2

1

KWnett

PHYSICS CHAPTER 4

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A stationary object of mass 3.0 kg is pulled upwards by a constant force of magnitude 50 N. Determine the speed of the object when it is travelled upwards through 4.0 m.

(Given g = 9.81 m s2)


The nett force acting on the object is given by

By applying the work-kinetic energy theorem, thus

Example 4 :

0 m; 4.0 ;N 50; kg 3.0 usFm

F

s

gm

F

gm

9.813.050 mgFFnett

N 20.6nettF

ifnett KKW

02

1 2 mvsFnett

23.02

14.020.6 v

1s m 7.41 v

PHYSICS CHAPTER 4

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A block of mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 below the horizontal. If the block starts from rest, calculate its final speed. You can ignore

the friction. (Given g = 9.81 m s2)


Example 5 :

s

36.9

0 m; 0.750 ; kg 2.00 usm

N

gm36.9

36.9sinmg36.9cosmg

a

x

y

PHYSICS CHAPTER 4

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Since the motion of the block along the incline surface thus nett force is given by

By using the work-kinetic energy theorem, thus

36.9sinmgFnett

N 11.8nettF

0 m; 0.750 ; kg 2.00 usm

36.9sin9.812.00nettF

ifnett KKW

02

1 2 mvsFnett

22.002

10.75011.8 v

1s m 2.98 v

PHYSICS CHAPTER 4

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An object of mass 2.0 kg moves along the x-axis and is acted on

by a force F. Figure 4.9 shows how F varies with distance

travelled, s. The speed of the object at s = 0 is 10 m s1.

Determine

a. the speed of the object at s = 10 m,

b. the kinetic energy of the object at s = 6.0 m.

Example 6 :

10

5

0 64 10

(N)F

7 (m)s


PHYSICS CHAPTER 4

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a.


1s m 10 kg; 2.0 umm 10 tom 0 fromgraph under the area sFW

57106102

11046

2

1W

J 32.5W

if KKW 22

2

1

2

1mumvW

22 102.02

12.0

2

132.5 v

1s m 11.5 v

PHYSICS CHAPTER 4

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b.


m 6 tom 0 fromgraph under the area sFW

10462

1W

J 50W

if KKW 2

2

1muKW f

2102.02

150 fK

J 150fK

PHYSICS CHAPTER 4

31

Exercise 4.2.1 :Use gravitational acceleration, g = 9.81 m s2

1. A bullet of mass 15 g moves horizontally at velocity of 250 m s1.It strikes a wooden block of mass 400 g placed at rest on a floor. After striking the block, the bullet is embedded in the block. The block then moves through 15 m and stops. Calculate the coefficient of kinetic friction between the block and the floor.

ANS. : 0.278ANS. : 0.278

2. A parcel is launched at an initial speed of 3.0 m s1 up a rough plane inclined at an angle of 35 above the horizontal. The coefficient of kinetic friction between the parcel and the plane is 0.30. Determine

a. the maximum distance travelled by the parcel up the plane,

b. the speed of the parcel when it slides back to the starting

point.

ANS. : 0.560 m; 1.90 m sANS. : 0.560 m; 1.90 m s11

PHYSICS CHAPTER 4

32

4.2.24.2.2 Potential EnergyPotential Energy is defined as the energy stored in a body or system because the energy stored in a body or system because

of its position, shape and stateof its position, shape and state.

Gravitational potential energy, Gravitational potential energy, UU is defined as the energy stored in a body or system because the energy stored in a body or system because

of its positionof its position. Equation :

The gravitational potential energy depends depends only on the heightheight of the object above the surface of the Earthabove the surface of the Earth.

mghU energy potential nalgravitatio : U

position initial thefrombody a ofheight : h

wherebody a of mass : m

gravity todueon accelerati : g

PHYSICS CHAPTER 4

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Work-gravitational potential energy theoremWork-gravitational potential energy theorem

Consider a book with mass, m is dropped from height, h1 to

height, h2 as shown in the figure 4.10.

states “ the change in gravitational potential energy as the change in gravitational potential energy as the negative of the work done by the gravitational forcethe negative of the work done by the gravitational force”.

1h

gm

gm

2h

s


21g hhmgmgsW

The work done by the gravitational force (weight) is

fig UUmghmghW 21

UUUW ifg

UW

Therefore in general,

PHYSICS CHAPTER 4

34

Negative sign in the equation indicates that When the body moves downmoves down, h decreases, the

gravitational force does positive workpositive work because U <0. When the body moves upmoves up, h increases, the work donework done

by gravitational force is negativenegative because U >0. For calculation, use

if UUUW

energy potential nalgravitatio final : fUwhere

force nalgravitatio aby done work : Wenergy potential nalgravitatio initial : iU

PHYSICS CHAPTER 4

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In a smooth pulley system, a force F is required to bring an object of mass 5.00 kg to the height of 20.0 m at a constant speed of 3.00 m s1 as shown in figure 4.11. Determine

a. the force, F

b. the work done by the force, F.

(Given g = 9.81 m s-2)

Example 7 :


F

m 20.0

PHYSICS CHAPTER 4

36


a. Since the object moves at the constant speed, thus

b. From the equation of work,

1s m 3.00constant m; 20.0 kg; 5.00 vhsm

0nettFmgF

N 49.1F

F

s

gm

F

gm

Constant speed

9.815.00F

θFsW cos 0θ

J 982W 20.049.1W

and

ORθFsW cos

mghUW

0θand

J 982W

PHYSICS CHAPTER 4

37

Elastic potential energy, Elastic potential energy, UUss

is defined as the energy stored in in elastic materials as the the energy stored in in elastic materials as the result of their stretching or compressingresult of their stretching or compressing.

SpringsSprings are a special instance of devicedevice which can store store elastic potential energyelastic potential energy due to its compression or compression or stretchingstretching.

Hooke’s LawHooke’s Law states “the restoring force, the restoring force, FFss of spring is of spring is

directly proportional to the amount of stretch or directly proportional to the amount of stretch or

compression (compression (extension or elongationextension or elongation), ), xx if the limit of if the limit of proportionality is not exceededproportionality is not exceeded”

OR xFs

kxFs

spring of force restoring the: sF

)(n compressioor stretch ofamount the: if -xxxconstant forceor constant spring the:k

where

PHYSICS CHAPTER 4

38

Negative signNegative sign in the equation indicates that the direction of direction of FFss

is always oppositeopposite to the directiondirection of the amount of stretch or

compression (extension), xx. Case 1:Case 1:

The spring is hung vertically and its is stretched by a suspended

object with mass, m as shown in figure 4.12.

The spring is in equilibrium, thus

Initial position

Final position

sF

gmW

x


mgWFs

PHYSICS CHAPTER 4

39Figure 4.13Figure 4.13

(Equilibrium position)

Case 2:Case 2:

The spring is attached to an object and it is stretched and

compressed by a force, F as shown in figure 4.13.

sF

F

0x

0x

sF

F

x

x

negative is sFpositive is x

positive is sFnegative is x

0 sF0x

The spring is in equilibrium, hence

FFs

PHYSICS CHAPTER 4

40

Caution: For calculationcalculation, use : Dimension of spring constant, k :

The unit of k is kg skg s22 or N mN m11

From the Hooke’s law (without “without “” sign” sign), a restoring force, Fs against extension of the spring, x graph is shown in figure 4.14.

FkxFs

2s MTx

Fk

force applied : Fwhere

FsF

0 x1x

graph under the area xFW s

1FxW2

1 11 xkxW

2

1

s21 UkxW

2

1


PHYSICS CHAPTER 4

41

The equation of elastic potential energy, Us for compressing or

stretching a spring is

The work-elastic potential energy theoremwork-elastic potential energy theorem,

Notes : Work-energy theoremWork-energy theorem states the work done by the nett work done by the nett

forceforce on a body equals the change in the body’s total on a body equals the change in the body’s total energyenergy”

OR

xF2

1kx

2

1U s

2s

ifnett EEEW

sUW 2i

2fsisf kx

2

1kx

2

1UUW OR

PHYSICS CHAPTER 4

42

A force of magnitude 800 N caused an extension of 20 cm on a spring. Determine the elastic potential energy of the spring when

a. the extension of the spring is 30 cm.

b. a mass of 60 kg is suspended vertically from the spring.

(Given g = 9.81 m s-2)


From the Hooke’s law,

a. Given x=0.300 m,

Example 8 :

m 0.200 N; 800 xF

kxFFs 0.20800 k

13 m N104 k

2

2

1kxU s

23 0.3001042

1sU J 180sU

PHYSICS CHAPTER 4

43


b. Given m=60 kg. When the spring in equilibrium, thus

Therefore

0nettFmgFs mgkx

9.8160104 3 x

m 0.147x2

2

1kxU s

23 0.1471042

1sU

J 43.2sU

sF

gmW

x

PHYSICS CHAPTER 4

44

4.2.34.2.3 Principle of conservation of energyPrinciple of conservation of energy states “in an isolated (closed) system, the total energy of in an isolated (closed) system, the total energy of

that system is constantthat system is constant”. According to the principle of conservation of energy, we get

The initial of total energy = the final of total energyThe initial of total energy = the final of total energy

Conservation of mechanical energy Conservation of mechanical energy In an isolated system, the mechanical energy of a system is the

sum of its potential energy, U and the kinetic energy, K of the objects are constant.

OR

fi EE

constant UKE

OR

ffii UKUK

PHYSICS CHAPTER 4

45

Before After

cm 30

x


A 1.5 kg sphere is dropped from a height of 30 cm onto a spring of spring constant,

k = 2000 N m1 . After the block hits the spring, the spring experiences maximum

compression, x as shown in figure 4.15.

a. Describe the energy conversion

occurred after the sphere is

dropped onto the spring until the

spring experiences maximum

compression, x.

b. Calculate the speed of the sphere just

before strikes the spring.

c. Determine the maximum compression, x.

(Given g = 9.81 m s-2)

Example 9 :

PHYSICS CHAPTER 4

46

The spring is not stretched hence Us = 0. The sphere is

at height h1 above ground

with speed, v just before strikes the spring. Therefore

The sphere is at height h2

above the ground after compressing the spring by x. The speed of the sphere at this moment is zero. Hence

The spring is not stretched hence Us = 0. The sphere is

at height h0 above ground

therefore U = mgh0 and it is

stationary hence K = 0.

(2)

v

1h

(3)

x

2h

cm 30h

0h

(1)

01 mghE 212 mv

2

1mghE 2

23 kx2

1mghE


a.

PHYSICS CHAPTER 4

47


b. Applying the principle of conservation of energy involving the

situation (1) and (2),

210 mvhhmg

2

1

21 EE2

10 mvmghmgh2

1

0.309.812 v

1m N 2000 m; 0.30 kg; 1.5 khm

and 10 hhh

1s m 2.43 v

ghv 2

PHYSICS CHAPTER 4

48


c. Applying the principle of conservation of energy involving the


2221 kxmvhhmg

2

1

2

1

32 EE

22

21 kxmghmvmgh

2

1

2

1

1m N 2000 m; 0.30 kg; 1.5 khm

and 21 hhx

m 107.43 2x

04.4314.71000 2 xx

22 20002

12.431.5

2

19.811.5 xx

PHYSICS CHAPTER 4

49

A bullet of mass, m1=5.00 g is fired into a wooden block of mass,

m2=1.00 kg suspended from some light wires as shown in figure

4.16. The block, initially at rest. The bullet embeds in the block, and

together swing through a height, h=5.50 cm. Calculate

a. the initial speed of the bullet.

b. the amount of energy lost to the surrounding.

(Given g = 9.81 m s2)

Example 10 :


1m2m

21 mm

h1u

PHYSICS CHAPTER 4

50

(1)

1m2m

1u02u

(3)

h

21 mm

012v

(2)

21 mm 12u

m 105.50 kg; 1.00kg; 105.00 23 hmm 21

32 EE

ghmmumm 211221 2

2

1

2105.509.8122 ghu12

UK

1s m 1.04 12u


a.

Applying the principle of conservation of energy involving the


PHYSICS CHAPTER 4

51


Applying the principle of conservation of linear momentum

involving the situation (1) and (2),

b. The energy lost to the surrounding, Q is given by

m 105.50 kg; 1.00kg; 105.00 23 hmm 21

21 pp

122111 ummum 1.041.00105.00105.00 33

1u1s m 209 1u

21 EEQ

2

2

1

2

11221

211 ummumQ

2323 1.041.00105.002

1209105.00

2

1 Q

J 109Q

PHYSICS CHAPTER 4

52

Objects P and Q of masses 2.0 kg and 4.0 kg respectively are connected by a light string and suspended as shown in figure 4.17. Object Q is released from rest. Calculate the speed of Q at the instant just before it strikes the floor.

(Given g = 9.81 m s2)

Example 11 :


P

Q

m 2

Smooth pulley

PHYSICS CHAPTER 4

53


Applying the principle of conservation of mechanical energy,

0 m; 2 kg; 4.0kg; 2.0 QP uhmm

fi EE2

Q2

PPQ 2

1

2

1vmvmghmghm

QPPQ KKUU

1s m 3.62 v

InitialInitial

P

Q

m 2

Smooth pulley

P

Qm 2

Smooth pulley

v

v

FinalFinal

22 4.02

12.0

2

129.812.029.814.0 vv

PHYSICS CHAPTER 4

54

Exercise 4.2.2 :Use gravitational acceleration, g = 9.81 m s2

1. If it takes 4.00 J of work to stretch a spring 10.0 cm from its initial length, determine the extra work required to stretch it an additional 10.0 cm.

ANS. : 12.0 JANS. : 12.0 J

2. A book of mass 0.250 kg is placed on top of a light vertical spring of force constant 5000 N m1 that is compressed by 10.0 cm. If the spring is released, calculate the height of the book rise from its initial position.

ANS. : 10.2 m ANS. : 10.2 m

3. A 60 kg bungee jumper jumps from a bridge. She is tied to a bungee cord that is 12 m long when unstretched and falls a total distance of 31 m. Calculate

a. the spring constant of the bungee cord.

b. the maximum acceleration experienced by the jumper.

ANS. : 100 N mANS. : 100 N m11; 22 m s; 22 m s22

PHYSICS CHAPTER 4

55

Exercise 4.2.2 :4.

A 2.00 kg block is pushed against a light spring of the force

constant, k = 400 N m-1, compressing it x =0.220 m. When the block is released, it moves along a frictionless horizontal

surface and then up a frictionless incline plane with slope =37.0 as shown in figure 4.18. Calculate

a. the speed of the block as it slides along the horizontal

surface after leaves the spring.

b. the distance travelled by the block up the incline plane before

it slides back down.

ANS. : 3.11 m sANS. : 3.11 m s11; 0.81 m; 0.81 m


PHYSICS CHAPTER 4

56

Exercise 4.2.2 :5.

A ball of mass 0.50 kg is at point A with initial speed, u =4 m s1 at a height of 10 m as shown in figure 4.19 (Ignore the frictional force). Determine

a. the total energy at point A,

b. the speed of the ball at point B where the height is 3 m,

c. the speed of the ball at point D,

d. the maximum height of point C so that the ball can pass over

it.

ANS. : 53.1 J; 12.4 m sANS. : 53.1 J; 12.4 m s11; 14.6 m s; 14.6 m s11; 10.8 m; 10.8 m

u

m 10

A

B

C

DFigure 4.19Figure 4.19

PHYSICS CHAPTER 4

57

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Define and useDefine and use power: power:

Average power, Average power,

Instantaneous Power, Instantaneous Power,

Derive and applyDerive and apply the formulae the formulae Define and useDefine and use mechanical efficiency, mechanical efficiency,

and the consequences of heat dissipationand the consequences of heat dissipation..

Learning Outcome:

4.3 Power and mechanical efficiency (1 hour)

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

ww

w.k

mp

h.m

atri

k.ed

u.m

y/p

hys

ics

dt

dWP

vFP

t

WPav

100%input

output

P

Pη

PHYSICS CHAPTER 4

58

4.3 Power and mechanical efficiency4.3.1 Power, P is defined as the rate at which work is donethe rate at which work is done.

OR the rate at which energy is transferredthe rate at which energy is transferred. If an amount of work, W is done in an amount of time t by a

force, the average poweraverage power, PPavav due to force during that time

interval is

The instantaneous powerinstantaneous power, PP is defined as the instantaneous the instantaneous rate of doing workrate of doing work, which can be write as

t

E

t

WPav

dt

dW

t

WP

0t

limit

PHYSICS CHAPTER 4

59

is a scalar quantity. The dimension of the power is

The S.I. unit of the power is kg mkg m22 s s33 or J sJ s11 or watt (W)watt (W). Unit conversionUnit conversion of watt (WW), horsepower (hp) and foot pounds

per second (ft. lb sft. lb s11)

Consider an object that is moving at a constant velocity v along a frictionless horizontal surface and is acted by a constant force,

F directed at angle above the horizontal as shown in figure

4.20. The object undergoes a displacement of ds.

3222

TMLT

TML

t

WP

1s lb ft. 550 W746hp 1


F

sd

PHYSICS CHAPTER 4

60

Therefore the instantaneous power, P is given by

OR

dt

dWP

vFP

dsθFdW cos

θFvP cos

and

dt

dsθFP

cos

dt

dsv and

where

vFθ

and between angle the:

force of magnitude:F velocityof magnitude : v

PHYSICS CHAPTER 4

61

An elevator has a mass of 1.5 Mg and is carrying 15 passengers through a height of 20 m from the ground. If the time taken to lift the elevator to that height is 55 s. Calculate the average power

required by the motor if no energy is lost. (Use g = 9.81 m s2 and the average mass per passenger is 55 kg)


M = mass of the elevator + mass of the 15 passengers

M = 1500 + (5515) = 2325 kg

According to the definition of average power,

Example 12 :

t

MghPav

t

EPav

W8294avP

55

209.812325avP

s 55 m; 20 Δth

PHYSICS CHAPTER 4

62

An object of mass 2.0 kg moves at a constant speed of 5.0 m s1 up a plane inclined at 30 to the horizontal. The constant frictional force acting on the object is 4.0 N. Determine

a. the rate of work done against the gravitational force,

b. the rate of work done against the frictional force,

c. the power supplied to the object. (Given g = 9.81 m s2 )


Example 13 :

N 4.0 constant;s m 5.0 kg; 2.0 1 fvm

30f

N

gmW

30cosmg

30sinmg

v

30x

y

s

PHYSICS CHAPTER 4

63


a. the rate of work done against the gravitational force is given by

t

θsmg

t

Wg cos30sin


180θand

t

smg

t

Wg 30sin

t

sv and

W49.1

t

Wg

vmgt

Wg 30sin

5.030sin9.812.0

t

Wg

OR θvFt

Wg

g cos

180cos30sin vmgt

Wg

W49.1

t

Wg

PHYSICS CHAPTER 4

64


b. The rate of work done against the frictional force is

c. The power supplied to the object, Psupplied

= the power lost against gravitational and frictional forces, Plost


180θand

W20.0

t

W f

θfvt

W f cos

180cos5.04.0

t

W f

t

W

t

WP fg

supplied

20.049.1supplied P W69.1supplied P

PHYSICS CHAPTER 4

65

4.3.24.3.2 Mechanical efficiency, Mechanical efficiency, Efficiency is a measure of the performance of a machines,

engine and etc... The efficiency of a machine is defined as the ratio of the the ratio of the

useful (output) work done to the energy inputuseful (output) work done to the energy input. is a dimensionless quantity (no unit). Equations:

100% in

out

E

Wη

OR

100% in

out

P

Pη

where system by the producedpower :outPsystem a tosuppliedpower : inP

PHYSICS CHAPTER 4

66

Notes :

In practice, PPoutout< P< Pinin hence < < 100% 100%.

The system loses energyloses energy to its surrounding becausebecause it may have encountered resistancesencountered resistances such as surface friction or air resistance.

The energyenergy which is dissipateddissipated to the surroundings, may be in the form of heat or soundheat or sound.

A 1.0 kW motor is used to lift an object of mass 10 kg vertically

upwards at a constant speed. The efficiency of the motor is 75 %.

Determine

a. the rate of heat dissipated to the surrounding.

b. the vertical distance travelled by the object in 5.0 s.

(Given g = 9.81 m s2 )

Example 14 :

PHYSICS CHAPTER 4

67


a. The output power of the motor is given by

Therefore the rate of heat dissipated to the surrounding is

b.

Since the speed is constant hence the vertical distance in 5.0 s

is

W1000 75%; kg; 10.0 inPηm

%100in

out

P

Pη

1001000

75 outP W750outP

7501000dissipatedheat of Rate outin PP W250dissipatedheat of Rate

θFvPout cos 0θwhere and mgF 0cosmgvPout

1s m 7.65 v v9.8110.0750

t

hv

5.07.65

h

m 38.3d

PHYSICS CHAPTER 4

68

Exercise 4.3 :

Use gravitational acceleration, g = 9.81 m s2

1. A person of mass 50 kg runs 200 m up a straight road inclined at an angle of 20 in 50 s. Neglect friction and air resistance. Determine

a. the work done,

b. the average power of the person.

ANS. : 3.36ANS. : 3.36101044 J; 672 W J; 672 W

2. Electrical power of 2.0 kW is delivered to a motor, which has an efficiency of 85 %. The motor is used to lift a block of mass 80 kg. Calculate

a. the power produced by the motor.

b. the constant speed at which the block being lifted vertically

upwards by the force produced by the motor.

(neglect air resistance)

ANS. : 1.7 kW; 2.17 m sANS. : 1.7 kW; 2.17 m s11

PHYSICS CHAPTER 4

69

Exercise 4.3 :3.

A car of mass 1500 kg moves at a constant speed v up a road with an inclination of 1 in 10 as shown in figure 4.21. All resistances against the motion of the car can be neglected. If the engine car supplies a power of 12.5 kW, calculate the

speed v.

ANS. : 8.50 m sANS. : 8.50 m s11


10 1

70

PHYSICS CHAPTER 4

THE END…Next Chapter…

CHAPTER 5 :Static

Note Chapter4 SF017

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