chapter 5 matriculation stpm

CHAPTER 5 WORK, ENERGY AND POWER

1

CHAPTER 5:

Work, Energy and Power

(3 Hours)


2

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


3

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 the

displacement of a body.

OR

is defined as the scalar (dot) product between

force and displacement of a body.


4

sFW

FssFW coscos

force of magnitude:F

body theofnt displaceme : s

sF

and between angle the:

Where,

Mathematically :


5

It is a scalar quantity.

Dimension :

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

The joule (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

22TML W

22 s m kg 1m N 1J 1


6

Work done by a variable force

Figure 5.1 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 .


7

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 , , sNThus

FN

F4

s4 sNs1 s2

F/N

s0

F1

s1

W1

NN2211 sFsFsFW ...

Figure 5.1


8

When N , s 0, therefore

2

1

s

sFdsW

graphnt displaceme-force under the area theW

F/N

s/ms1 s20

Work = Area


9

Applications of works equation

Case 1 :

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

Case 2 :

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

0F

s

Figure 5.2

FsW cosFsW

and

90FsW cosJ 0W

andF

s

Figure 5.3


10

Case 3 :

Work done by a horizontal forces, F1 and F2 on an object (Figure 5.4).

Case 4 :

Work done by a force, F and frictional force, f on an object (Figure 5.5).

0cos sFW 11

0cos sFW 22

sFWW nettnett

1F

2F

s

Figure 5.4 sFsFWWW 2121 21nett FFF sFFW 21 and

cos mafFFnett sFW nettnett sfFWnett cos masWnett

f F

Figure 5.5 s

and

OR


11

Caution :

Work done on an object is zero when F = 0 or s = 0 and = 90.


12

Sign for work.

If 0 0 (positive) work done on the system ( by the external force) where energy is transferred to the system.

If 90


13

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 :

m 1.50s

N 5.00F

N 1.60f

cosFsWF 0and

Example 5.1 :


14

Solution :

b.

c.

fsW f cos

fF WWW

OR

sFW nett sfFW

180and

180cos1.501.60fW

2.407.50W

1.501.605.00W


15

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 5.2 :


16

Solution :

a. Consider the work done along inclined plane, thus

i.

m 3.800.300; ;N 250 ;kg 20 sFm k

sFW xF cos0and

0cos3.8025cos250FW

25

25

kf

N

F

s

gmW

yF

25cosmg

xF

25sinmg

a

25x

y


17

Solution :

a. ii.

iii.

iv.

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

NsWN cos90and

sfW kf cos180and

180cossNW kf smgFW kf 25cos25sin

3.8025cos9.812025sin2500.300 fW


18

Solution :

a. v.

b. Given

By 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

3.802.93202v


19

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 5.6. Calculate the work

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

Solution :

Example 5.3 :

5

47

053 6

(N)F

5 (m)s

Figure 5.6

graph under the area sFW

4672

15356

2

1W


20

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 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 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. : 150 J; 108

N j40i30 F m j3.0i9.0 s


21

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 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 J

3F

1F

2F

y

x

35

50

Figure 5.7


22


(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:

2

2

1mvK

mghU

2

2

1kxU

5.2 Energy And Conservation Of Energy


23

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.


24

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 = 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.


25

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 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 : vbody a of mass : m

where

s

nettF

m

Figure 5.8

maFF nett (1)


26

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 change in the bodys kinetic energy.

as uv 222

s

uva

2

22 (2)

2

22

s

uvmFnett

ifnett KKmumvsF 22

2

1

2

1

KWnett


27

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)

Solution :

The nett force acting on the object is given by

By applying the work-kinetic energy theorem, thus

Example 5.4 :

0 m; 4.0 ;N 50; kg 3.0 usFm

F

s

gm

F

gm

9.813.050 mgFFnett

ifnett KKW

02

1 2 mvsFnett

23.02

14.020.6 v


28

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)

Solution :

Example 5.5 :

s

36.9

0 m; 0.750 ; kg 2.00 usm

N

gm36.9

36.9sinmg36.9cosmg

a

x

y


29

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

0 m; 0.750 ; kg 2.00 usm

36.9sin9.812.00nettF

ifnett KKW

02

1 2 mvsFnett

22.002

10.75011.8 v


30

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

by a force F. Figure 5.9 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 5.6 :

10

5

064 10

(N)F

7 (m)s

Figure 5.9


31

Solution :

a.


1s m 10 kg; 2.0 um

m 10 tom 0 fromgraph under the area sFW

57106102

11046

2

1W

if KKW

22

2

1

2

1mumvW

22 102.02

12.0

2

132.5 v


32

Solution :

b.


m 6 tom 0 fromgraph under the area sFW

10462

1W

if KKW

2

2

1muKW f

2102.02

150 fK


33

Exercise 5.2 :

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. : 0.278

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


34

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.

mghU

energy potential nalgravitatio : U

position initial thefrombody a ofheight : h

where

body a of mass : mgravity todueon accelerati : g


35

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 21 UUUW ifg

UW

Therefore in general,


36

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

if UUUW

energy potential nalgravitatio final : fUwhere

force nalgravitatio aby done work : W

energy potential nalgravitatio initial : iU


37

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 5.11. Determine

a. the force, F

b. the work done by the force, F.

(Given g = 9.81 m s-2)

Example 5.7 :

Figure 5.11

F

m 20.0


38

Solution :

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

F

s

gm

F

gm

Constant

speedFsW cos 0and

OR

FsW cosmghUW

0and


39

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), x if the limit of proportionality is not exceeded

OR xFs

kxFs

spring of force restoring the: sF

)(n compressioor stretch ofamount the: if -xxx

constant forceor constant spring the:k

where


40

Negative sign in the equation indicates that the direction of Fsis 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


41Figure 5.13

(Equilibrium position)

Case 2:

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

compre5sed by a force, F as shown in Figure 5.13.

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


42

Caution:

For calculation, use :

Dimension of spring constant, k :

The unit of k is 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 MTx

Fk

force applied : Fwhere

F

sF

0 x1x

graph under the area xFW s

1FxW2

1 11 xkxW

2

1

s21 UkxW

2

1

Figure 5.14


43

The equation of elastic potential energy, Us for compressing or stretching 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 total energy

OR

xF2

1kx

2

1U s

2s

ifnett EEEW

sUW 2

i

2

fsisf kx2

1kx

2

1UUW OR


44

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)

Solution :

From the Hookes law,

a. Given x=0.300 m,

Example 5.8 :

m 0.200 N; 800 xF

kxFFs 0.20800 k

2

2

1kxU s

23 0.3001042

1sU


45

Solution :

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

Therefore

0nettFmgFs mgkx

9.8160104 3 x

2

2

1kxU s

23 0.1471042

1sU

sF

gmW

x


46

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, U and the kinetic energy, K of theobjects are constant.

OR

fi EE

constant UKE

OR

ffii UKUK


47

Before After

cm 30

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 the spring, 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.

(Given g = 9.81 m s-2)

Example 5.9 :


48

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 h2above 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 mv2

1mghE 223 kx

2

1mghE

Solution :

a.


49

Solution :

b. Applying the principle of conservation of energy involving the

situation (1) and (2),

210 mvhhmg2

1

21 EE2

10 mvmghmgh2

1

0.309.812 v

1m N 2000 m; 0.30 kg; 1.5 khm

and 10 hhh

ghv 2


50

Solution :

c. Applying the principle of conservation of energy involving the


2221 kxmvhhmg2

1

2

1

32 EE2

22

1 kxmghmvmgh2

1

2

1

1m N 2000 m; 0.30 kg; 1.5 khm

and 21 hhx

22 20002

12.431.5

2

19.811.5 xx


51

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.

(Given g = 9.81 m s2)

Example 5.10 :

Figure 5.16

1m 2m

21 mm

h1u


52

(1)

1m 2m1u

02u

(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

Solution :

a.

Applying the principle of conservation of energy involving the



53

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

m 105.50 kg; 1.00kg; 105.00 23 hmm 21

21 pp

122111 ummum 1.041.00105.00105.00 33 1u

21 EEQ

22

1

2

11221

211 ummumQ

2323 1.041.00105.002

1209105.00

2

1 Q


54

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

5.17. 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 5.11 :

Figure 5.17

P

Q

m 2

Smooth

pulley


55

Solution :

Applying the principle of conservation of mechanical energy,

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

fi EE2

Q2

PPQ2

1

2

1vmvmghmghm

QPPQ KKUU

Initial

P

Q

m 2

Smooth

pulley

P

Qm 2

Smooth

pulley

v

v

Final

22 4.02

12.0

2

129.812.029.814.0 vv


56

Exercise 5.3 :


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 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

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 m1; 22 m s2


57

Exercise 5.3 :

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 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


58

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

m 10

A

B

C

DFigure 5.19


59


(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)

dt

dWP

vFP

t

WPav

100%input

output

P

P


60

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 t by aforce, the average power, Pav due to force during that timeinterval is

The instantaneous power, P is defined as the instantaneousrate of doing work, which can be write as

t

E

t

WPav

dt

dW

t

WP

0tlimit


61

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 along a frictionless horizontal surface and is acted by a constant force,

F directed at angle above the horizontal as shown in Figure 5.20. The object undergoes a displacement of ds.

3222

TMLT

TML

t

WP

1s lb ft. 550 W746hp 1

Figure 5.20

F

sd


62

Therefore the instantaneous power, P is given by

OR

dt

dWP

vFP

dsFdW cos

FvP cos

and

dt

dsFP

cos

dt

dsv and

where

vF

and between angle the:

force of magnitude:F velocityof magnitude : v


63

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)

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

MghPav

t

EPav

s 55 m; 20 th


64

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 )

Solution :

Example 5.13 :

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

30

f

N

gmW

30cosmg

30sinmg

v

30x

y

s


65

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

sv and

vmgt

Wg 30sin

5.030sin9.812.0

t

Wg

OR vFt

Wg

gcos

180cos30sin vmgt

Wg


66

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

W fcos

t

W

t

WP

fg

supplied


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:

100% in

out

E

W

OR

100% in

out

P

P

where system by the producedpower :outP

system a tosuppliedpower : inP


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Notes :

In practice, Pout< Pin hence < 100%. The system loses energy to its surrounding because it may

have encountered resistances such as surface friction or

air resistance.

The energy which is dissipated to the surroundings, may

be in the form of heat 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 5.14 :


69

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

W1000 75%; kg; 10.0 inPm

%100in

out

P

P

1001000

75 outP

7501000dissipatedheat of Rate outin PP

FvPout cos0where and mgF

0cosmgvPout

t

hv


70

Exercise 5.4 :


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. : 3.36104 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 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