matriculation chapter 5 stpm

5/24/2018 Matriculation Chapter 5 STPM

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CHAPTER 5 WORK, ENERGY AND POWER

1

CHAPTER 5:

Work, Energy and Power(3 Hours)


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At the end of this chapter, students should be able to:

(a)Define and use work done by a force.

(b) Determine work done from the force-

displacement graph.

Learning Outcome:

5.1 Work (1 hour)

sFW


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5.1 Work, W

Work done by a constant force

is defined as the product of the component of the

force parallel to the displacement times thedisplacement of a body.

OR

is defined as the scalar (dot) product between

force and displacement of a body.


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sFW

FssFW coscos

forceofmagnitude:F

bodytheofntdisplaceme:ssF

andbetweenanglethe:

Where,

Mathematically :


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It is a scalarquantity.

Dimension :

The S.I. unit of work is kg m2 s2 orjoule (J).

Thejoule (1 J) is defined as the work done by a force of 1 N

which results in a displacement of 1 m in the direction of

the force.

sFW

22

TML

W

22 smkg1mN1J1


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Work done by a variable force

Figure 5.1 shows a force, Fwhose magnitude

changes with the displacement, s.

For a small displacement, s1

the force remains

almost constant at F1

and work done therefore

becomes W1=F

1 s

1.


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To find the total work done by a variable force, Wwhen thedisplacement changes froms=s1 tos=s2, we can divide thedisplacement into N small successive displacements :

s1, s2, s3, , sN

Thus

FN

F4

s4 sNs1 s2

F/N

s0

F1

s1

W1

NN2211 sFsFsFW ...

Figure 5.1


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When N ,s 0, therefore

2

1

s

s FdsW

F/N

s/ms1 s20

Work = Area


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Applications of works equation

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

Case 2 :

Work done by a vertical force,Fon an object (Figure 4.3).

0F

sFigure 5.2

FsW cos

FsW

and

90FsW cosJ0W

andF

s

Figure 5.3


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Case 3 :

Work done by a horizontal forces,F1 andF2 on an object

(Figure 5.4).

Case 4 :

Work done by a force,Fand frictional force,fon an object(Figure 5.5).

0cos sFW 110cos sFW 22

sFWW nettnett

1F

2F

s

Figure 5.4 sFsFWWW 2121

21nett FFF sFFW 21and

cos mafFFnett sFW nettnett

sfFWnett cos masWnett

f

F

Figure 5.5 s

and

OR


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Caution :

Work done on an object is zero whenF= 0 ors= 0 and = 90.


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Sign for work.

If 0 0 (positive) work done on the system ( bythe external force) where energyis transferred to the system.

If 90


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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 :

a. Use works equation of constant force,

Example 5.1 :

m1.50s

N5.00FN1.60f

cosFsWF 0and

Example 5.1 :


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Solution :

b.

c.

fsWf

cos

fF WWW

OR

sFW nett sfFW

180and

180cos1.501.60fW

2.407.50W

1.501.605.00W


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A box of mass 20 kg moves up a rough plane which is inclined tothe horizontal at 25.0. It is pulled by a horizontal forceFofmagnitude 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 forceF,

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.

(Giveng= 9.81 m s2)

Example 5.2 :


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Solution :

a. Consider the work done along inclined plane, thusi.

m3.800.300;;N250;kg20 sFm k

sFW xF cos 0and

0cos3.8025cos250FW

25

N

F

s

gmW

yF

25cosmg

xF

25sinmg

a

25x

y


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Solution :

a. ii.

iii.

iv.

smgWg cos25sin 180and 180cos3.8025sin9.8120gW

NsWN cos 90and

sfW kf cos 180and

180cossNW kf

smgFW kf

25cos25sin 3.8025cos9.812025sin2500.300 fW


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Solution :

a. v.

b. GivenBy using equation of work for nett force,

Hence by using the equation of linear motion,

fNgF WWWWW 3230315861W

masW 3.8020223 a

asuv 22 2

0u

3.802.9320 2v


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A horizontal forceFis applied to a 2.0 kg radio-controlled car as itmoves along a straight track. The force varies with the

displacement of the car as shown in Figure 5.6. Calculate the work

done by the forceFwhen the car moves from 0 to 7 m.

Solution :

Example 5.3 :

5

47

053 6

(N)F

5 (m)s

Figure 5.6

graphunder thearea sFW

4672

15356

2

1W


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Exercise 5.1 :

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 belowthe 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.

(Giveng= 9.81 m s2)

ANS. : 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 itundergoes a displacement . Calculate

a. the work done on the trolley by the forceF,

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

trolley.ANS. : 150 J; 108

Nj

40i

30 F

mj3.0i9.0 s


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Exercise 5.1 :

3.

Figure 5.7 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 areF1 = 3.00 N,F2 = 4.00 N andF3 = 10.0 N. Determine the totalwork done on the cargo by the three forces during the first

4.00 m of displacement.

ANS. : 15.3 J

3F

1F

2F

y

x

35

50

Figure 5.7


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(a) Define and use kinetic energy,

(b) Define and use potential energy:

i. gravitational potential energy,

ii. elastic potential energy for spring,

(c) State and use the principle of conservation of energy.

(d) Explain the work-energy theorem and use the related

equation.

Learning Outcome:

5.2 Energy And Conservation Of Energy


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Energy is defined as the systems ability to do work.

The S.I. unit for energy is same to the unit of work (joule, J).

The dimension of energy

is a scalar quantity.

Table 5.1 summarises some common types of energy.

22 TMLWorknergyE

Forms of

EnergyDescription

ChemicalEnergy released when chemical bonds between atoms

and molecules are broken.

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

HeatEnergy 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.


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Forms of

EnergyDescription

Table 5.1

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 Einsteins mass-energy

equation,E = mc

2

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.

Mechanical

a. Kineticb. 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.


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Conservation of energy

5.2.1 Kinetic energy, K is defined as the energy of a body due to its motion.

Equation :

Work-kinetic energy theorem

Consider a block with mass, m moving along the horizontalsurface (frictionless) under the action of a constant nett force,

Fnettundergoes a displacement,s in Figure 4.8.

where

s

nettF

Figure 5.8

maFFnett (1)


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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 thechange in the bodys kinetic energy.

asuv 222

s

uva

2

22 (2)

2

22

s

uvmFnett

ifnett KKmumvsF 22

2

1

2

1


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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 itis travelled upwards through 4.0 m.

(Giveng= 9.81 m s2)

Solution :

The nett force acting on the object is given by

By applying the work-kinetic energy theorem, thus

Example 5.4 :

0m;4.0;N50;kg3.0 usFm

F

s

gm

F

gm

9.813.050 mgFFnett

ifnett

KKW

02

1 2 mvsFnett

23.0

2

14.020.6 v


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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 theblock starts from rest, calculate its final speed. You can ignore the

friction. (Giveng= 9.81 m s2)

Solution :

Example 5.5 :

s

36.9

0m;0.750;kg2.00 usm

N

gm

36.9

36.9sinmg

36.9cosmg

a

x

y


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Solution :

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

0m;0.750;kg2.00 usm

36.9sin9.812.00nettF

ifnett KKW

0

2

1 2 mvsFnett

22.002

10.75011.8 v


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

by a forceF. Figure 5.9 shows howFvaries with distance

travelled,s. The speed of the object ats = 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 5.6 :

10

5

064 10

(N)F

7 (m)s

Figure 5.9


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Solution :

a.


1sm10kg;2.0 umm10tom0fromgraphunder thearea sFW

57106102

110462

1 W

if KKW 22

2

1

2

1mumvW

22

102.02

1

2.02

1

32.5 v


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Solution :

b.


m6tom0fromgraphunder thearea sFW

10462

1 W

if KKW 2

2

1muKW f

2

102.02

1

50 fK


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Exercise 5.2 :

Use gravitational acceleration,g= 9.81 m s2

1. A bullet of mass 15 g moves horizontally at velocity of250 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. : 0.278

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 s1


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5.2.2 Potential Energy

is defined as the energy stored in a body or system because

of its position, shape and state.

Gravitational potential energy, U

is defined as the energy stored in a body or system because

of its position.

Equation :

The gravitational potential energy depends only on the height

of the object above the surface of the Earth.

where

bodyaofmass:mgravitytodueonaccelerati:g


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Work-gravitational potential energy theorem

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

height, h2 as shown in the Figure 5.10.

states the change in gravitational potential energy as

the negative of the work done by the gravitational force.

1h

gm

gm

2h

s

Figure 5.10

21g hhmgmgsW

The work done by the gravitational force

(weight) is

fig UUmghmghW 21UUUW ifg

Therefore in general,


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Negative sign in the equation indicates that

When the body moves down, h decreases, the

gravitational force does positive work because U 0.

For calculation, use

where

forcenalgravitatioabydonework:W

energypotentialnalgravitatioinitial:iU


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In a smooth pulley system, a forceFis required to bring anobject 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 5.11. Determine

a. the force,F

b. the work done by the force,F.

(Giveng= 9.81 m s-2)

Example 5.7 :

Figure 5.11

F

m20.0


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Solution :

a. Since the object moves at the constant

speed, thus

b. From the equation of work,

1sm3.00constantm;20.0kg;5.00 vhsm

0nettFmgF

F

s

gm

F

gm

Constant

speedFsW cos 0and

OR

FsW cosmghUW

0and


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Elastic potential energy, Us

is defined as the energy stored in in elastic materials as the

result of their stretching or compressing.

Springs are a special instance of device which can store

elastic potential energy due to its compression or

stretching.

Hookes Law states the restoring force, Fs

of spring is

directly proportional to the amount of stretch or

compression (extension or elongation), xif the limit of

proportionality is not exceeded

OR xFs

where


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Negative sign in the equation indicates that the direction of Fs

is always opposite to the direction of the amount of stretch or

compression (extension), x.

Case 1:

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

object with mass, m as shown in Figure 5.12.

The spring is in equilibrium, thus

Initial position

Final position

sF

gmW

x

Figure 5.12

mgWFs


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41Figure 5.13

(Equilibrium position)

Case 2:

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

compre5sed by a force,Fas shown in Figure 5.13.

sF

F

0x

0x

sF

F

x

x

negativeissFpositiveisx

positiveissFnegativeisx

0sF0x

The spring is in equilibrium,

hence

FFs


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Caution:

For calculation, use :

Dimension of spring constant, k:

The unit of kis kg s2 or N m1

From the Hookes law (without sign), a restoring force,Fsagainst extension of the spring,x graph is shown in Figure 5.14.

FkxFs

2s MT

x

Fk

forceapplied:Fwhere

F

sF

0 x

1x

graphunder thearea xFW s

1FxW21 11 xkxW

21

s21 UkxW

2

1

Figure 5.14


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The equation of elastic potential energy, Us for compressing orstretching a spring is

The work-elastic potential energy theorem,

Notes :

Work-energy theorem states the work done by the nett

force on a body equals the change in the bodys totalenergy

OR

OR


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A force of magnitude 800 N caused an extension of 20 cm on a

spring. Determine the elastic potential energy of the spring whena. the extension of the spring is 30 cm.

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


Solution :From the Hookes law,

a. Givenx=0.300 m,

Example 5.8 :

m0.200N;800 xF

kxFFs 0.20800 k

2

2

1kxUs

23 0.3001042

1sU

CHAPTER O G O


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Solution :

b. Given m=60 kg. When the spring in

equilibrium, thus

Therefore

0nettFmgFs mgkx

9.81601043

x

2

2

1kxUs

23 0.14710421 sU

sF

gmW

x

CHAPTER 5 WORK ENERGY AND POWER


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5.2.3 Principle of conservation of energy

states in an isolated (closed) system, the total energy of

that system is constant.

According to the principle of conservation of energy, we get

The initial of total energy = the final of total energy

Conservation of mechanical energy

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

sum of its potential energy, Uand the kinetic energy,Kof theobjects are constant.

OR

OR



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Before After

cm30

x

Figure 5.15

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 thespring, the spring experiences maximum

compression,x as shown in Figure 5.15.

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.


Example 5.9 :



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The spring is not stretched

hence Us = 0. The sphere is

at height h1 above groundwith speed, vjust beforestrikes the spring. Therefore

The sphere is at height h2above the ground after

compressing the spring byx.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 henceK= 0.

(2)

v

1h

(3)

x

2h

cm30h

0h

(1)

01 mghE

212

mv2

1mghE 223 kx

2

1mghE

Solution :

a.



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Solution :

b. Applying the principle of conservation of energy involving the

situation (1) and (2),

210 mvhhmg21

21 EE2

10 mvmghmgh

2

1

0.309.812v

1mN2000m;0.30kg;1.5 khm

and 10 hhh

ghv 2



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Solution :

c. Applying the principle of conservation of energy involving the


2221 kxmvhhmg2

1

2

1

32 EE2

22

1 kxmghmvmgh

2

1

2

1

1mN2000m;0.30kg;1.5 khm

and 21 hhx

22 20002

12.431.5

2

19.811.5 xx



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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

5.16. 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.

(Giveng= 9.81 m s2)

Example 5.10 :

Figure 5.16

1m 2m

21 mm

h1u



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(1)

1m 2m1u

02u

(3)

h

21 mm

012v

(2)

21 mm 12u

m105.50kg;1.00kg;105.00 23 hmm 21

32

EE

ghmmumm 211221 2

2

1

2105.509.8122 ghu12

UK

Solution :

a.

Applying the principle of conservation of energy involving the




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Solution :

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

m105.50kg;1.00kg;105.00 23 hmm 21

21 pp

122111 ummum 1.041.00105.00105.00 33 1u

21 EEQ

221

2

11221211 ummumQ

2323 1.041.00105.002

1209105.00

2

1 Q



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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 Figure5.17. Object Q is released from rest. Calculate the speed of Q at

the instant just before it strikes the floor.

(Giveng= 9.81 m s2)

Example 5.11 :

Figure 5.17

P

Q

m2

Smooth

pulley



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Solution :

Applying the principle of conservation of mechanical energy,

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

fi EE

2Q

2PPQ

2

1

2

1vmvmghmghm

QPPQ KKUU

Initial

P

Q

m2

Smoothpulley

P

Qm2

Smoothpulley

v

v

Final

22 4.02

12.0

2

129.812.029.814.0 vv



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Exercise 5.3 :

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 itsinitial length, determine the extra work required to stretch it an

additional 10.0 cm.

ANS. : 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 10.0cm. If the spring is released, calculate the height of the book rise

from its initial position.

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 totaldistance of 31 m. Calculate

a. the spring constant of the bungee cord.

b. the maximum acceleration experienced by the jumper.

ANS. : 100 N m1; 22 m s2



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Exercise 5.3 :

4.

A 2.00 kg block is pushed against a light spring of the forceconstant, k= 400 N m-1, compressing itx =0.220 m. When theblock is released, it moves along a frictionless horizontal surface

and then up a frictionless incline plane with slope =37.0 as

shown in Figure 5.18. 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 s1

; 0.81 m

Figure 5.18



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Exercise 5.3 :

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 5.19 (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 s1; 14.6 m s1; 10.8 m

u

m10

A

B

C

D

Figure 5.19



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(a) Define and use power:

Average power,

Instantaneous Power,

(b) Derive and apply the formulae

(c) Define and use mechanical efficiency,

and the consequences of heat dissipation.

Learning Outcome:

5.3 Power and mechanical efficiency (1 hour)



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5.3 Power and mechanical efficiency

5.3.1 Power, P is defined as the rate at which work is done.

OR the rate at which energy is transferred.

If an amount of work, W is done in an amount of time tby a

force, the average power,Pavdue to force during that timeinterval is

The instantaneous power,Pis defined as the instantaneous

rate of doing work, which can be write as



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is a scalar quantity.

The dimension of the power is

The S.I. unit of the power is kg m2 s3 or J s1 or watt (W). Unit conversion of watt (W), horsepower (hp) and foot pounds

per second (ft. lb s1)

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

Fdirected at angle above the horizontal as shown in Figure5.20. The object undergoes a displacement of ds.

3222

TMLTTML

tWP

Figure 5.20

sd



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Therefore the instantaneous power,Pis given by

OR

dt

dW

P and

dt

dsFP

cos

dt

dsvand

where

vF

andbetweenanglethe:

forceofmagnitude:F velocityofmagnitude:v



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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 liftthe elevator to that height is 55 s. Calculate the average power

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

Solution :

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

M= 1500 + (5515) = 2325 kg

According to the definition of average power,

Example 5.12 :

t

Mgh

Pav

t

E

Pav

s55m;20 th



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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 frictionalforce 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. (Giveng= 9.81 m s2 )

Solution :

Example 5.13 :

N4.0constant;sm5.0kg;2.0 1 fvm

30

f

N

gmW

30cosmg

30sinmg

v

30x

y

s



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Solution :

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

t

smg

t

Wg cos30sin


180and

t

smg

t

Wg 30sin

t

svand

vmgt

Wg 30sin

5.030sin9.812.0

t

Wg

OR vFt

Wg

gcos

180cos30sin vmgt

Wg



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Solution :

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


180andfvt

Wf cos

t

W

t

WP

fg

supplied



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67

5.3.2 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 useful

(output) work done to the energy input.

is a dimensionless quantity (no unit).

Equations:

OR

where systemby theproducedpower:outP

systematosuppliedpower:in

P



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Notes :

In practice, Pout< P

inhence


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Solution :

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

W100075%;kg;10.0 inPm

%100in

out

PP

1001000

75 outP

7501000dissipatedheatofRate outin PP

FvPout cos 0where and mgF0cosmgvPout

t

hv



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WORK, ENERGY AND POWER

70

Exercise 5.4 :

Use gravitational acceleration,g= 9.81 m s2

1. A person of mass 50 kg runs 200 m up a straight road inclinedat 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. : 3.36104 J; 672 W2. 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 s1



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,

Exercise 5.4 :

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 5.21. 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 s1

Figure 5.21

10 1

matriculation chapter 5 stpm

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total work

f1and work

unit of work

positive work

n force

constant force

s1the force

vertical force