:
Work, Energy and Power
(3 Hours)
1
Learning Outcome:
5.1 Work (1 hour)
At the end of this chapter, students should be able to:
 Define and apply work done by a constant force.
 
W  F s

Determine work done from a forcedisplacement graph.
2
5. 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.
Equation
:
 
W  F s
W  F cos θ s  Fs cos θ
where
s : displaceme nt of the body
F : magnitude of force


θ : the angle between F and s
 It is a scalar quantity.
 The S.I. unit of work is kg m2 s2 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.
1 J  1 N m  1 kg m 2 s 2
4
Applications of work’s equation
Case 1 :
 Work done by a horizontal force, F on an object (Figure 5.2).

F
Figure 5.2
W  Fs cos θ
 W  Fs
s
and
θ  0
Case 2 :
Work
done by a vertical force, F on an object (Figure 5.3).

F
Figure 5.3
W  Fs cos θ
 W 0J
s
and
θ  90
5

Case 3 :
Work done by a horizontal forces, F1 and F2 on an object

(Figure 5.4).
W  F s cos 0
F1

F2
1
 W2  F2 s cos 0
s
Figure 5.4
W  W
 W2  F1s  F2 s 
1
W F  F s
1
1
and
2
W  W
nett
Fnett  F1  F2
 Fnett s
6
Case 4 :
Work done by a force, F and frictional force, f on an object

F
(Figure 5.5).

f
Figure 5.5
Wnett  Fnett s
and
Wnett  F cos  f s


s
Fnett  F cos θ  f  ma
OR
Wnett  mas
7
 Caution :
Work done on an object is zero when F = 0 or s = 0 and
 = 90.
8
Sign for work.
W  Fs cos
If 0< <90 (acute angle) then cos > 0 (positive
value)therefore
W > 0 (positive)  work done on the system ( by
the external force) where
energy is transferred to the system.
If 90< <180 (obtuse angle) then cos <0 (negative
value) therefore
W < 0 (negative)  work done by the system
where energy is transferred
from the system.
9
Example 5.1 :
You push your 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 :
f  1.60 N
a. Use work’s equation of constant force,
F  5.00 N
s  1.50 m
10
Solution :
b.
c.
OR
11
Example 5.2 :
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 s2)
12
Solution : m  20 kg; F

N
y
mg sin 25
x fk
25


 250 N; μk  0.300; s  3.80 m

a
Fx
Fy
25 
25  
F
mg cos 25


W  mg

s

a. Consider the work done along inclined plane, thus
i.
13
Solution :
a. ii.
iii.
iv.
14
Solution :
a. v.
b. Given u  0
By using equation of work for nett force,
Hence by using the equation of linear motion,
15
Example 5.3 :
F (N)
5
0
4
3
5
6
7
s(m)
Figure 5.6
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 :
16
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 s2)
ANS. : 31.9 J; (b) & (c) U think; 31.9 J
17
Exercise 5.2 :
2. A trolley is rolling across a parking lot of a supermarket.

You apply a constant force F  30ˆi  40ˆj N to the trolley

s
  9.0ˆi  3.0ˆj m .
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
18
Exercise 5.3 :
y

F3
3.

F1
35 
x

F2
50 
Figure 5.7
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
19
Work done by a variable force
F/N
FN
Figure 5.7
F4
F1
 W1
0
s1
s1
s4
s
sN 2 s
W  F1s1  F2 s2  ...  FN s N
20
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 .
To find the total work done by a variable force, W when
the displacement changes from s=s1 to s=s2, we can divide
the displacement into N small successive displacements :
s1 , s2 , s3 , …, sN
Thus
When N  , s  0, therefore
s2
W   Fds
s1
21
s2
W   Fds
s1
F/N
Work = Area under the graph
0 s1
s2 s/m
Example 5.4 :
A force , F acting on a particle varies with the displacement
x as shown in figure below.
Calculate the work done by the force as the particle moves
from x=0 to x=6 m.
F (N)
Solution :
5
The Work done = area under graph
4
6
x (m)
23
Learning Outcome:
5.2 Energy and Conservation of energy (1 hour)
At the end of this chapter, students should be able to:
Define and use kinetic energy,
Define and use potential energy:
i. gravitational potential energy,
ii. elastic potential energy for spring,
1 2
K  mv
2
U  mgh
1
U  kx2
2
State and use the principle of conservation of energy.
State the work-energy theorem and use the related equation.
5.2 Energy and Conservation of energy
Energy
is defined as the system’s ability to do work.
The S.I. unit for energy is same to the unit of work
(joule, J).
is a scalar quantity.
Table 5.1 summarises some common types of
energy.
Forms of
Energy
Description
Chemical
Heat
Electrical
Internal
26
Forms of
Energy
Nuclear
Mass
Description
Energy released by the splitting of heavy nuclei.
Energy released when there is a loss of small amount
of mass in a nuclear process. The amount of energy
can be calculated from Einstein’s mass-energy
equation, E = mc2
Radiant Heat Energy associated with infra-red radiation.
Sound
Mechanical
a. Kinetic
b. Gravitational
potential
c. Elastic
potential
Energy transmitted through the propagation of a series
of compression and rarefaction in solid, liquid or gas.
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.
Table 5.1
27
Kinetic energy, K
 is defined as the energy of a body due to its motion.

Equation :
where
1 2
K  mv
2
K : kinetic energy of a body
m : mass of a body
v : speed of a body
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 5.8.

Fnett
m

s
Figure 5.8
F  F
nett
 ma
By using an equation of linear motion:
v2  u 2
a
2s
(1)
v 2  u 2  2as
(2)
By substituting equation (2) into (1), we arrive
Fnett
 v2  u 2 

 m
 2s 
1 2 1 2
Fnett s  mv  mu  K f  K i
2
2
Therefore
Wnett  K
 states “the work done by the nett force on a
body equals the change in the body’s kinetic
energy”.
30
Example 5.5 :
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 s2)
m  3.0 kg ; F  50 N; s  4.0 m; u  0
Solution :

F
 
s F
The nett force acting on the object is given by

mg

mg
By applying the work-kinetic energy theorem, thus
Example 5.6 :
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 s
m  2.00 kg ; s  0.750 m; u  0

N

a
y
mg sin 36.9 
mg cos 36.9 
36.9  
mg

s
x
36.9 
Solution : m  2.00 kg ; s  0.750 m; u  0
Since the motion of the block along the incline surface thus nett
force is given by
By using the work-kinetic energy theorem, thus
33
Example 5.7 :
F (N)
10
0
4
6
5
7
10
s(m)
Figure 5.9
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 s1.
Determine
a. the speed of the object at s = 10 m,
b. the kinetic energy of the object at s = 6.0 m.
34
Solution : m  2.0 kg; u  10 m s 1
a.
W  area under the F  s graph from 0 m to 10 m
By using the work-kinetic energy theorem, thus
35
Solution :
b. W  area under
the F  s graph from 0 m to 6 m
By using the work-kinetic energy theorem, thus
36
Exercise 5.4 :
Use gravitational acceleration, g = 9.81 m s2
1. A bullet of mass 15 g moves horizontally at velocity of
250 m s1.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.
Exercise 5.5 :
ANS. : 0.278
2. A parcel is launched at an initial speed of 3.0 m s1 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 s1
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 :
U  mgh
where
U : gravitatio nal potential energy
m : mass of a body
g : acceleration due to gravity
h : height of a body from the initial position

The gravitational potential energy depends only on the height
of the object above the surface of the Earth.
38

Work-gravitational potential energy theorem
◦ Consider a book with mass, m is dropped from height,
h1 to height, h2 as shown in the Figure 4.10.
The work done by the gravitational force (weight) is
s

mg
h2

mg
h1
Figure 4.10
Wg  mgs  mg h1  h2 
Wg  mgh1  mgh2  U i  U f
Wg  U f  U i   U
Therefore in general,
W  U
◦states
“ the change in gravitational potential energy as the
negative of the work done by the gravitational force”.
39
◦Negative
sign in the equation indicates that
the body moves down, h decreases, the gravitational
force does positive work because U <0.
When
 When the body moves up, h increases, the work done by
gravitational force is negative because U >0.
◦For
W  U  U f  U i
calculation, use
where
U f : final gravitatio nal potential energy
U i : initial gravitatio nal potential energy
W : work done by a gravitatio nal force
40
Example 5.8 :

F
20.0 m
Figure 5.11
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 s1 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)
41
Solution : m  5.00 kg; s  h  20.0 m; v  constant  3.00 m s 1

a. Since the object moves at the constant
speed, thus
F
Constant
speed

F

mg

s

mg
b. From the equation of work,
OR
42

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.
 Hooke’s 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”

where
Fs   x
Fs : the restoring force of spring
Fs  kx
k : the spring constant or force constant
x : the amount of stretch or compressio n ( x f -xi )

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
Figure 5.12
Initial position

Fs
x
Final position
thus
Fs  W  mg


W  mg

Case 2:
The spring is attached to an object and it is stretched and
compressed by a force, F as shown in Figure 5.13.

Fs is negative F
s
x is positive

F
The spring is in
equilibrium, hence
x
x0
Fs  0
x0


Fs  F
(Equilibrium position)

F
x 0
Fs Fs is positive
x is negative
x
Figure 5.13
45

Caution:
For calculation, use : Fs  kx  F where F : applied force
◦ The unit of k is kg s2 or N m1

From the Hooke’s law (without “” sign), a restoring force, Fs
against extension of the spring, x graph is shown in Figure 5.14.
Fs
F
W  area under the Fs  x graph
1
W  Fx1
2
0
Figure 5.14
x1
x
1
W  kx1 x1
2
1 2
W  kx1  U s
2

The equation of elastic potential energy, Us for compressing
or stretching a spring is
1 2 1
U s  kx  Fs x
2
2

The work-elastic potential energy theorem,
W  U s
OR
1 2 1 2
W  U sf  U si  kx f  kxi
2
2
 Notes :
◦ Work-energy theorem states the work done by the nett force
on a body equals the change in the body’s total energy”
Wnett  E 
E  E
f
i
47
Example 5.9 :
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 Hooke’s law
a. Given x=0.300 m
,
48
Solution :
b. Given m=60 kg. When the spring in
equilibrium, thus

Fs
x
Therefore


W  mg
49
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
E  E
i
f
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 the objects are constant.
E  K  U  constant
Ki  U i  K f  U f
Example 5.10 :
A 1.5 kg sphere is dropped from a height of
30 cm onto a spring of spring constant,
k = 2000 N m1 . 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.
30 cm
x
Before
After
Figure 5.15
(Given g = 9.81 m s-2)
51
Solution :
a.
h  30 cm
v
h0
x
h1
h2
(1)
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.
E
1
 mgh0
(3)
(2)
The spring is not stretched The sphere is at height h2
hence Us = 0. The sphere is above the ground after
compressing the spring by x.
at height h1 above ground
The speed of the sphere at
with speed, v just before
this moment is zero. Hence
strikes the spring. Therefore

1
E2  mgh1  mv 2
2

1 2
E3  mgh2  kx
2 52

Solution : m  1.5 kg; h  0.30 m; k  2000 N m 1
b. Applying the principle of conservation of energy involving the
situation (1) and (2),
53
Solution :
m  1.5 kg; h  0.30 m; k  2000 N m 1
c. Applying the principle of conservation of energy involving the
situation (2) and (3),
54
Example 5.11 :
m1
u1
m1  m2
m2
h
Figure 5.16
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 s2)
55
Solution :
m1  5.00 10 3 kg; m2  1.00 kg; h  5.50 10 2 m
a
.
v12  0
u2  0
m1
u1
m2
m1  m2
(1)
(2)
m1  m2
u12
h
(3)
Applying the principle of conservation of energy involving the
situation (2) and (3),
Solution : m1  5.00  10 3 kg; m2  1.00 kg; h  5.50  10 2
Applying the principle of conservation of linear momentum
involving the situation (1) and (2),


p1 

m

p2
m1u1  m1  m2 u12
5.00 10 3 u1  5.00 10 3  1.00 1.04
u1  209 m s 1

 

b. The energy lost to the surrounding, Q is given by
Q
E  E
1
2
1
1
2
2
Q  m1 u1  m1 m 2 u12 
2
2
1
1
2
2
3
Q  5.00  10 209   5.00 10 3  1.00 1.04 
2
2
Q  109 J




57
Example 5.12 :
Smooth
pulley
Q
P
2m
Figure 5.17
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 s2)
58
Solution :
mP  2.0 kg; mQ  4.0 kg; h  2 m; u  0
Smooth
pulley
Q
P
2m
Initial
Smooth
pulley
v
P
2m Q
v
Final
Applying the principle of conservation of mechanical energy,
59
Exercise 5.6 :
Use gravitational acceleration, g = 9.81 m s2
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
Exercise 5.7 :
2. A book of mass 0.250 kg is placed on top of a light vertical
spring of force constant 5000 N m1 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
Exercise 5.8 :
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 m1; 22 m s2
Exercise 5.9 :
4.
Figure 5.18
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 s1; 0.81 m
61
Exercise 5.10 :
C
u
5.
A
10 m
B
D
Figure 5.19
A ball of mass 0.50 kg is at point A with initial speed, u =4 m s1
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 s1; 14.6 m s1; 10.8 m
62
Learning Outcome:
5.3 Power and mechanical efficiency (1 hour)
At the end of this chapter, students should be able to:
Define and use power:
Average power,
Instantaneous Power,
W
Pav 
t
P
Derive and apply the formulae
dW
dt
 
P  F v
Define and use mechanical efficiency,
and the consequences of heat dissipation
η
Poutput
Pinput
 100%
5.3 Power and mechanical efficiency
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 a force, the average power, Pav due to force during that
time interval is
W E
Pav 

t
t
The instantaneous power, P is defined as the instantaneous
rate of doing work, which can be write as
W dW
P  limit

t 0 t
dt
 is a scalar quantity.

The S.I. unit of the power is kg m2 s3 or J s1 or watt (W).

Unit conversion of watt (W), horsepower (hp) and foot
pounds per second (ft. lb s1)
1 hp  746 W  550 ft. lb s 1

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.
F

Figure 5.20

ds
Therefore the instantaneous power, P is given by
dW
P
dt
and
dW  F cos θ ds

F cos θ ds
P
dt
P  Fv cos θ
OR
where
ds
and v 
dt
 
P  F v
F : magnitude of force
v : magnitude of velocity


θ : the angle between F and v
66
Example 5.13 :
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
s2 and the average mass per passenger is 55 kg)
Solution :
M = mass of the elevator + mass of the 15 passengers
M = 1500 + (5515) = 2325 kg
According to the definition of average power,
Example 5.14 :
An object of mass 2.0 kg moves at a constant speed of 5.0 m s1
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 s2 )
Solution :
1
m  2.0 kg;
v

5.0
m
s
  constant; f  4.0 N

v
N

s
y
mg sin 30 
x f
30

30 
mg cos 30


W  mg

Solution :
m  2.0 kg; v  5.0 m s 1  constant; f  4.0 N
a. the rate of work done against the gravitational force is
given by
OR
Solution :
m  2.0 kg; v  5.0 m s 1  constant; f  4.0 N
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
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:
Wout
η
 100%
Ein
where
OR
Pout
η
 100%
Pin
Pout : power produced by the system
Pin : power supplied to a system

◦In
Notes :
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.
Example 5.15 :
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 s2 )
Solution :
m  10.0 kg; η  75%; Pin  1000 W
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
Exercise 5.11 :
Use gravitational acceleration, g = 9.81 m s2
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,
ANS. : 3.36104 J; 672 W
b. the average power of the person.
Exercise 5.12 :
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 s1
Exercise 5.13 :
1
10
Figure 5.21
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 s1
THE END…
Next Chapter…
CHAPTER 6 :
Circular Motion
76