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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CHAPTER 5 WORK, ENERGY AND POWER
19
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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CHAPTER 5 WORK, ENERGY AND POWER
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At the end of this chapter, students should be able to:
(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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CHAPTER 5 WORK, ENERGY AND POWER
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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.
By using the work-kinetic energy theorem, thus
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.
By using the work-kinetic energy theorem, thus
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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CHAPTER 5 WORK, ENERGY AND POWER
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.
(Giveng= 9.81 m s-2)
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
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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.
(Giveng= 9.81 m s-2)
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
situation (2) and (3),
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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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.
(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
situation (2) and (3),
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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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At the end of this chapter, students should be able to:
(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
N4.0constant;sm5.0kg;2.0 1 fvm
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
N4.0constant;sm5.0kg;2.0 1 fvm
180andfvt
Wf cos
t
W
t
WP
fg
supplied
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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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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
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