chapter 5 matriculation stpm
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Work, energy and powerTRANSCRIPT
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CHAPTER 5 WORK, ENERGY AND POWER
1
CHAPTER 5:
Work, Energy and Power
(3 Hours)
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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.
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CHAPTER 5 WORK, ENERGY AND POWER
4
sFW
FssFW coscos
force of magnitude:F
body theofnt displaceme : s
sF
and between angle the:
Where,
Mathematically :
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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 .
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
8
When N , s 0, therefore
2
1
s
sFdsW
graphnt displaceme-force under the area theW
F/N
s/ms1 s20
Work = Area
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
11
Caution :
Work done on an object is zero when F = 0 or s = 0 and = 90.
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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 :
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CHAPTER 5 WORK, ENERGY AND POWER
14
Solution :
b.
c.
fsW f cos
fF WWW
OR
sFW nett sfFW
180and
180cos1.501.60fW
2.407.50W
1.501.605.00W
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CHAPTER 5 WORK, ENERGY AND POWER
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 :
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
22
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:
2
2
1mvK
mghU
2
2
1kxU
5.2 Energy And Conservation Of Energy
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CHAPTER 5 WORK, ENERGY AND POWER
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.
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CHAPTER 5 WORK, ENERGY AND POWER
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.
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CHAPTER 5 WORK, ENERGY AND POWER
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)
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
31
Solution :
a.
By using the work-kinetic energy theorem, thus
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
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CHAPTER 5 WORK, ENERGY AND POWER
32
Solution :
b.
By using the work-kinetic energy theorem, thus
m 6 tom 0 fromgraph under the area sFW
10462
1W
if KKW
2
2
1muKW f
2102.02
150 fK
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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,
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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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, 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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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 :
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CHAPTER 5 WORK, ENERGY AND POWER
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.
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
50
Solution :
c. Applying the principle of conservation of energy involving the
situation (2) and (3),
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
situation (2) and (3),
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
56
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 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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
59
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)
dt
dWP
vFP
t
WPav
100%input
output
P
P
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 WORK, ENERGY AND POWER
65
Solution :
a. the rate of work done against the gravitational force is given by
t
smg
t
Wg cos30sin
N 4.0 constant;s m 5.0 kg; 2.0 1 fvm
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
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CHAPTER 5 WORK, ENERGY AND POWER
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
N 4.0 constant;s m 5.0 kg; 2.0 1 fvm
180andfvt
W fcos
t
W
t
WP
fg
supplied
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CHAPTER 5 WORK, ENERGY AND POWER
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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CHAPTER 5 WORK, ENERGY AND POWER
68
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 :
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CHAPTER 5 WORK, ENERGY AND POWER
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
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CHAPTER 5 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 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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CHAPTER 5 WORK, ENERGY AND POWER
71
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