Download 5. Work, Energy and Power

JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
5. Work, Energy and Power
Content:
•
•
•
•
Work
Energy Conversion and Conservation
Potential Energy and Kinetic Energy
Power
Candidates should be able to:
(a)
show an understanding of the concept of work in terms of the product of a force and
displacement in the direction of the force.
(b)
calculate the work done in a number of situations, including the work done by a gas
which is expanding against a constant external pressure: W = pΔV.
(c)
give examples of energy in different forms, its conversion and conservation, and apply
the principle of energy conservation to simple examples.
(d)
derive, from the equations of motion, the formula Ek = ½ m v2.
(e)
recall and apply the formula Ek = ½ m v2.
(f)
distinguish between gravitational potential energy, electric potential energy and elastic
potential energy.
(g)
show an understanding of, and use the relationship between, force and potential
energy in a uniform field to solve problems.
(h)
derive, from the defining equation W = Fs, the formula Ep = mgh for potential energy
changes near the Earth’s surface.
(i)
recall and use the formula Ep = mgh for potential energy changes near the Earth’s
surface.
(j)
show an appreciation for the implications of energy losses in practical devices and use
the concept of efficiency to solve problems.
(k)
define power as work done per unit time and derive power as the product of force and
velocity.
Physics Department
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
(a)
Work, Energy and Power
show an understanding of the concept of work in terms of the product of a force
and displacement in the direction of the force.
In Physics, work is done when a force moves its point of application so that some resolved
part of the displacement lies along the direction of the force.
F
θ
s
Work done by a force is the product of the force and the displacement in the direction of
the force.
W = Fs cosθ
where:
• W is the work done on the object by the constant force F, (unit: J)
• F is the constant force acting on the object by an external agent, (unit: N)
• s is the displacement of the object, (unit: m)
• θ is the angle between F and s
Work done is a scalar quantity.
Note:
In problem solving, you can either resolve force F in the direction of displacement, s or vice
versa, depending on which is more convenient in the context of the question.
7N
Example
7N
7N
25°
25°
6m
6m
6m
W = 7 (6 cos 25°)
= 38 J
Physics Department
25°
W = (7 cos 25°) 6
= 38 J
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
One joule (1 J) is the work done by a constant force of one newton (1 N) on an object when
the object moves one metre (1 m) in the direction of the force.
1 J = (1 N) (1 m) = 1 N m
The unit newton metre (N m) is used for moment of a force, while joule (J) is usually
used for work done.
Even though work done is a scalar, it can be positive or negative.
Scenario
Work done by force
Example
A man pushing a wall
(man→ agent)
Force does not move the
object
(wall→ object)
zero
force
KE remains
unchanged
wall does not move
Weight of a trolley moving along
a horizontal surface
Force is perpendicular to the
object’s movement
zero
(Earth→ agent) (trolley→ object)
KE remains
displacement
unchanged
force (weight)
A man pulling a luggage
Force has a resolved part in
the same direction as the
object’s displacement
(man→ agent) (luggage→ object)
positive
force
KE increases
displacement
Friction opposing the trolley’s
motion
Force has a resolved part in
the opposite direction to the
object’s displacement
negative
(ground→ agent) (trolley→ object)
KE decreases
force
displacement
In the above examples, positive work done on an object increases the kinetic energy of the
trolley, while negative work done decreases its kinetic energy. Zero work done means the
kinetic energy remains constant.
Physics Department
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
Worked Example 1
A man pushes a box of mass 5 kg with a force of 2 N along a horizontal frictionless surface.
The box is moved through a distance 3 m.
Calculate the work done on the box by
(a) the pushing force;
(b) the normal contact force;
(c) the weight of the box.
Solution:
[Draw a diagram to illustrate the question]
[Identify the body i.e. box]
The agent applying the force is pushing force
W = Fs cosθ = (2) (3) = 6 J
N
3m
F
mg
(a) The agent applying the force is normal contact force
W=0J
(b) The agent applying the force is weight
W=0J
Note that for both (b) and (c), applied forces are perpendicular to the object’s
displacement.
Worked Example 2
A block is pulled along a rough horizontal surface by a rope pointing at 60° above the
horizontal. The tension in the rope is 10 N and the frictional force is 5 N. The block moves a
distance of 2 m along the surface.
Calculate the work done on the box by
(a) the tension in the rope;
(b) the frictional force.
Solution:
[Draw a diagram to illustrate the question]
[Identify the body i.e. box]
Taking rightwards as positive,
10 N
(a) W = Fs cos θ = (10 cos 60°) (2)
= 10 J
60°
(b) W = (−5)(2) = −10 J
Note: By conservation of energy,
from (a) and (b), there is no change in the KE of the
block i.e. the block moves at constant velocity.
Physics Department
5N
2m
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(b)
Work, Energy and Power
calculate the work done in a number of situations, including the work done by a
gas which is expanding against a constant external pressure: W = pΔV.
Work done by a force in displacing a body
Suppose a force F is acting on an object along the x-direction and the object moves a
distance ( x 2 − x1 ) along the same direction. A graph of F against x is plotted.
F
If F is constant,
If F varies,
work done is given by W = F (x 2 − x1 )
Work done is given by W =
x2
∫ F ⋅ dx
x1
F
F
x1
•
x
x2
x1
x2
x
x1
x2
x
Work done is equal to the area under the F – x graph.
Work done by a spring
Consider a spring which is stretched horizontally by force F as shown,
∆x
F
x=0
x=x
The work done W by F, can be represented as the area under the F - x graph and is stored
as elastic potential energy or strain energy in the deformed spring.
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Work, Energy and Power
In general, for any spring or wire,
If the spring obeys Hooke’s Law, then
Work done by force F, W = ½ Fx = ½ kx2
F
∫
Work done by force F, W =
x2
x1
F = kx
F dx
F
F
x
0
x
x
x1
Area under graph =
Area under graph = ½ kx2
•
x2
∫
x2
x1
F dx
Work done is equal to the area under the F – x graph.
Work done by a gas during expansion
In engines, gases expand and contract at different parts of the cycle of movement. When a
gas expands, it does work on its surroundings; when it is compressed, work is done on it by
its surroundings. Inside an engine, it will normally be a piston that is used to compress a gas.
The piston is part of the surroundings of the gas.
Fgas
F
gas
external pressure, p
∆x (distance moved by piston)
•
•
•
•
•
A is the cross-sectional area of piston
F is the force exerted normally by the piston on the gas = pA
Fgas is the force exerted by gas pressure on piston
p is the constant external pressure
∆x is the distance moved by piston
For a gas with pressure = external pressure p, if the piston is allowed to move outwards
slowly by a displacement ∆x, the gas will expand against the constant external pressure p,
and Fgas = F.
The work done W by the gas against this constant pressure p is given by
W = F ∆x
= (pA) ∆x= p (A∆x) = p ∆V
where ∆V is the change in volume.
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Work, Energy and Power
Hence, when a gas expands with a change in volume ∆V against constant pressure p, work
done by gas is given by:
W = p ΔV
where
• W is the work done by gas (unit: J)
• p is the external pressure (unit: Pa)
• ΔV is the change in volume of the gas (unit: m3) Note: ΔV = Vfinal – Vinitial
If p is constant,
work done is given by W = p(V2 − V1 )
work done is given by W =
∫
V2
V1
p.dV
p
p
V1
•
If p varies with V,
V2
V
V1
V2
V
Work done is equal to the area under the p – x graph.
When gas expands
When gas is compressed
(i.e. Vfinal > Vinitial )
(i.e. Vfinal < Vinitial )
Outwards (+ΔV)
Inwards (– ΔV)
in same direction
in opposite direction
Work done by gas on surroundings
positive
negative
Work done by surroundings on gas
negative
positive
Gas does work on
Surroundings do work on
surroundings
gas
Piston moves
Force exerted by gas on
surroundings and displacement
moved
We say that the
Physics Department
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Work, Energy and Power
Worked Example 3
A gas in a cylinder is kept at a constant pressure of 1.1 × 105 Pa. The gas is heated and
expands by 25 cm3.
Calculate the work done by the gas.
Solution:
1 cm = 10−2 m
1 cm3 = (10−2m) 3 = 10−6 m3
W = p ΔV = (1.1 × 105) (25 × 10−6) = 2.75 J
(c)
give examples of energy in different forms, its conversion and conservation, and
apply the principle of energy conservation to simple examples.
The stored ability to do work is called energy.
Work is done when energy is transferred; the energy stored in a compressed spring,
electrical energy stored in a battery or chemical energy in a lump of coal may be used to do
work at any time.
According to the principle of conservation of energy,
Energy cannot be created nor destroyed but is transformed from one form to another.
Note that the total amount of energy in an isolated system is constant. An isolated system
is one that does not interact with its surroundings. This means that if an energy calculation is
being carried out then all the energy within the system must be taken into account. No
energy must be allowed to escape.
Energy and work are both scalars, and have the unit joule.
Types of energy and their sources
Energy
Mechanical
Source
Total mechanical energy = Sum of kinetic energies and potential
energies
Kinetic Energy
Energy possessed by all objects in motion E.g. moving car
Gravitational
Potential Energy
Elastic Potential
Energy
Energy possessed by an object by virtue of its position in a gravitational
field. E.g. object on top of a building
Energy possessed by an object by virtue of its state of deformation. Eg.
Compressed or stretched springs, a bent diving board, stretched elastic
band of a catapult.
Electrical
Due to charge or current. E.g. current in closed circuit
Chemical
Energy possessed by a fuel by virtue of its chemical composition
E.g. oils, wood, food, chemicals in electric cells.
Physics Department
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Nuclear
Radiant
Internal energy
Work, Energy and Power
Energy in nucleus of atoms due to nuclear composition.
E.g. energy released in atomic bombs, produced by nuclear reactors.
Energy in the form of electromagnetic waves.
E.g. visible light, radiowaves, ultraviolet, x-rays.
Energy possessed by atoms/molecules of matter in the form of kinetic
energy due to motion of the particles in the matter and potential energy
which depends on the separation between atoms / molecules.
Renewable and Non-renewable energy
Renewable sources of energy are those that can be replaced or replenished each day by
the Earth's natural processes. Eg. wind, geothermal, solar and tidal energy.
Non-renewable sources of energy are those that are finite or exhaustible because it takes
several million years to replace them, e.g. fossil fuels like coal, oil and natural gas, energy
sources that are tapped from minerals e.g. nuclear energy from the fission of uranium nuclei.
Worked Example 4
Discuss the energy changes which take place in the following systems:
a) diver jumping off a diving board
•
•
•
The diver uses his gravitational potential energy to do work in bending the diving
board.
The work done is stored as elastic potential energy, which is then converted into
kinetic energy of the diver as he is pushed upwards and off the diving board.
At the same time, some of the elastic potential energy is lost as thermal and sound
energy due to dissipative forces in the diving board.
b) burning of fossil fuel
•
When fuels such as oil, coal and wood are burnt, the chemical energy stored in these
materials is converted into thermal and light energy.
c) hammering a nail into a wooden block
•
•
•
•
A person uses the chemical energy in his muscles to work against the gravitational pull
in order to lift the hammer.
The work done is converted into the gravitational potential energy of the hammer in its
raised position.
As the hammer falls, its gravitational potential energy is converted into kinetic energy.
When the hammer hits the nail, its kinetic energy is used to do work in driving the nail
into the wooden block, producing sound energy in the air and thermal energy in the
block, nail and hammer.
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Work, Energy and Power
d) bouncing ball
•
•
•
•
•
As the ball falls, its gravitational potential energy is converted into its kinetic energy.
When the ball hits the ground, the ball is deformed during the collision. Its kinetic energy
is converted into elastic potential energy. Some kinetic energy may be lost as thermal
or sound energy.
The elastic potential energy is converted back into kinetic energy as the ball regains
its original shape.
The kinetic energy is converted into gravitational potential energy as the ball bounces
upwards, until it reaches its highest position.
During the flight, presence of air resistance will cause kinetic energy to be dissipated as
thermal energy, thus reducing the total energy in the ball and its subsequent height after
each bounce.
e) falling plasticine
•
(d)
(e)
During impact, all kinetic energy is converted into thermal and sound energies as the
plasticine is permanently deformed.
derive, from the equations of motion, the formula Ek = ½ m v2.
recall and apply the formula Ek = ½ m v2.
The kinetic energy (Ek OR K.E.) is a positive scalar quantity that represents the energy
associated with the body due to its motion.
It is equal to the work done by the resultant force in accelerating the body moving
horizontally from rest to an instantaneous speed v.
Ek =
1
mv 2
2
where
• Ek is the kinetic energy of the body moving at speed v (unit: joule, J)
• m is the mass of the body (unit: kg)
• v is the speed of the body (unit: m s−1)
Derive, from the equations of motion, the formula Ek = ½ m v2.
Consider a situation where a constant resultant force F
acts on a stationary object of mass m. The object is
displaced by a displacement s in the direction of the
force. The final velocity is v.
Physics Department
m
F
v
u=0
s
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Work, Energy and Power
By Newton’s 2nd law, F = ma Æ a =
F
m
v 2 = u 2 + 2as Æ v 2 = 0 + 2 ⎛⎜ F ⎞⎟ s
From
⎝m⎠
ÆF=
mv
2s
2
mv 2
(s)
Using work done on object = Fs =
2s
=
1
2
m v2
By conservation of energy, since Fs is the work done on the object, which is equal to the
increase in kinetic energy (Ek – 0), Ek = 21 m v2
Note:
• Ek is a scalar quantity and has the same units as work i.e. joules, J.
• By conservation of energy,
o Work done by external forces (e.g. frictional force, push/pull force)
= change in total energy (e.g. kinetic, gravitational potential, elastic
potential, thermal energy)
Worked Example 5
A box of mass 10 kg is accelerated from rest to a speed of 5.0 m s−1 along a smooth
horizontal road by a horizontal pulling force of 200 N.
Calculate the distance travelled by the box in reaching its final speed.
Solution:
Taking rightwards as positive,
By conservation of energy,
WD by pulling force = change in total energy
W = mv − mu
1
2
2
1
2
( 200 ) s = (10 )( 5.0 )
1
2
2
2
Alternative method
Using Equations of Motion:
v 2 = u 2 + 2as
⎛ 200 ⎞
52 = 0 + 2 ⎜
⎟s
⎝ 10 ⎠
s = 0.625 m
−0
s = 0.625 m
Worked Example 6
A bullet of mass 2.0 g moving with a velocity of 500 m s−1 hits a wooden pole. It comes to rest
after penetrating the pole to a depth of 100 mm.
Calculate the average retarding force during the passage of the bullet within the pole.
Solution:
By conservation of energy,
WD by retarding force = change in total energy = final energy − initial energy
WD by retarding force = 12 mv 2 − 12 mu 2
F (100 × 10 −3 ) = 0 − 21 ( 2 × 10 −3 ) ( 500 )
2
F = −2500 N
(negative sign indicates retarding force)
Physics Department
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(f)
Work, Energy and Power
distinguish between gravitational potential energy, electric potential energy and
elastic potential energy.
Potential energy
The energy possessed by a system by virtue of the relative
positions of its component parts.
Gravitational potential
energy
Arises from interaction between masses, where gravitational forces
involved are always attractive in nature.
Electric potential
energy
Arises from interaction between charges, where electric forces
involved can be either attractive or repulsive.
Elastic potential energy
Arises from deformation of a material.
Worked Example 7
A spring, fixed at one end, has a mass attached to the other end. The mass bounces up and
down. It is shown in the diagram at three positions X, Y and Z.
Which line gives the kinetic, gravitational potential and elastic potential energies of the
spring-mass system?
Kinetic Energy
A
B
C
D
zero at the top
maximum in centre
zero at bottom
maximum in centre
Gravitational
Potential Energy
maximum at the top
zero at bottom
zero at bottom
maximum at the top
Elastic
Potential Energy
maximum at the top
maximum in centre
zero at the top
maximum at bottom
Solutions:
The kinetic energy is zero at top (X) and bottom (Z) and is maximum at the centre (Y)
The gravitational potential energy is maximum at the top (X) and zero at the bottom (Z) (take
reference point to be at the bottom)
The elastic potential energy is maximum at the bottom (Z) with the largest extension.
Ans: (D)
Physics Department
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
(g)
Work, Energy and Power
show an understanding of, and use the relationship between, force and potential
energy in a uniform field to solve problems.
In a uniform field e.g. near the surface of the Earth at g = 9.81 m s-2, a body experiences the
same force F (due to the field) at all points.
F
Uniform
field
Δx
Surface of the Earth
If this force F moves the body along a distance Δx in its direction, then the work done W by
this force F is W = F Δx. (+W since F and Δx are in the same direction)
By conservation of energy, this work done must be compensated by a decrease in potential
energy, −ΔU. (Ufinal < Uinitial as the gravitational potential energy decreases as height
decreases)
W = F Δx = − ΔU
F =−
ΔU
Δx
In general, in any field of force (uniform or non-uniform), the relationship between the force
exerted by the field on a body and the potential energy possessed by the body is given by:
F = −
where:
•
•
dU
dx
F is the force acting on the body (a point mass or charge) placed at that particular
point due to the field. (unit: N)
dU
is the change in the potential energy of the body with a variation of the
dx
distance from the source of the field (unit: J m-1)
If the potential energy U of a body varies with distance from a reference point in a field as
shown in the graph below, then the force F exerted by the field on the body at a point Xp in
the field is found from the negative of the gradient of tangent to the curve at that point.
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U
Work, Energy and Power
Xp
x
F=−
dU
dx
Worked Example 8
A mass experiences a gravitational force of 24 N in a uniform gravitational field. Calculate the
change in its gravitational potential energy if it moves a distance of 5.0 m
(a)
(b)
(c)
along the same direction as the gravitational force;
along the direction opposite to the gravitational force;
in a direction perpendicular to the gravitational force.
For each case, indicate whether the change is an increase or decrease.
Solution:
Taking upwards as positive,
(a)
(b)
Δx
ΔU = −F Δx = − (−24) (−5.0)
= −120 J (decrease)
F
Δx
ΔU = −F Δx = − (−24) (5.0)
= +120 J (increase)
F
(c)
Δx
ΔU = −F Δx = − (−24) (0)
= 0 J (no change)
F
Physics Department
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Work, Energy and Power
Worked Example 9
The potential energy of a body when it is at a point P a distance x from a reference point O is
given by V = kx2, where k is a constant. What is the force acting on the body when it is at P?
A
2 kx in the direction OP
B
kx in the direction OP
C
zero
D
kx in the direction PO
E
2 kx in the direction PO
Solution:
dV
Recall F = −
dx
P
V = kx2
x
dV
= 2kx
dx
dV
F= = - 2kx
dx
O
F
F has a magnitude of 2 kx and acts in the direction opposite to the direction of x. Hence
direction of F is from P to O. Ans: (E)
(h)
(i)
derive, from the defining equation W = Fs, the formula Ep = mgh for the potential
energy changes near the Earth’s surface.
recall and use the formula Ep = mgh for potential energy changes near the
Earth’s surface.
Derive, from the defining equation W = Fs, the formula Ep = mgh
Consider a situation where a force F acts on an object of
mass m to move the object vertically upwards.
The object is displaced at a constant speed (no change
in Ek) by a displacement h in the direction of the force.
Since the object moves at constant speed, the upward
force, F, must be equal to the weight of the object, mg.
F
mg
= Fh
= (mg)h
Since the object’s velocity is constant, its kinetic energy
is also constant.
Hence work done on object
h
By conservation of energy, since Fh is the work done on
the object and is equal to the increase in gravitational
potential energy, Ep = mgh
Physics Department
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Work, Energy and Power
Worked Example 10
A car travels upwards along a straight road inclined at an angle of 5.0o to the horizontal, as
shown in the figure below.
The length of the road is 450 m and the car has a mass 800 kg.
What is the gain in gravitational potential energy of the car when it reaches the top of the
slope?
Solution:
Vertical rise in height, h = (450) (sin 5.0o)
Gain in G.P.E = mgh = (800) (9.81) (450) (sin 5.0o) = 3.1 x 105 J
Worked Example 11
A block weighing 800 N moves down a frictionless slope inclined at 20o to the horizontal of
height 10.0 m as shown below. Determine the speed of the block at the bottom of the slope,
assuming that the block moves off from the top at an initial speed of 5.0 m s-1.
10.0 m
θ
Solution:
By conservation of energy,
total initial energy of the block (at the top) = total final energy of the block (at the bottom)
KEi + GPEi = KEf + GPEf
1
1
mv i2 + mghi = mv f2 + mghf
2
2
1
1
mghi − mghf = mv f2 − mv i2
2
2
(GPE loss = KE gain by block)
1
1
ghi − ghf = v f2 − v i2
2
2
1
1 2
(9.81)(10.0) − (9.81)(0) = (v f2 ) − ( 5 )
2
2
-1
v f = 14.9 m s
Physics Department
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Work, Energy and Power
Worked Example 12
A 50 kg block is released from rest at a height of 5.00 m above the ground. It then travels a
distance of 10.0 m along a curved slope to the ground as shown in the figure below. The final
speed of the block at the end of the slope is only 4.90 m s-1 because a constant resistive
force acts on it during the descent.
5.00 m
Determine the resistive force acting on the block.
Solution:
By conservation of energy,
WD by friction = change in total energy = final energy − initial energy
⎛1
⎞ ⎛1
⎞
W = ⎜ mv 2 + mghf ⎟ − ⎜ mu 2 + mghi ⎟
⎝2
⎠ ⎝2
⎠
⎛1
⎞
F (10) = ⎜ (50)(4.90)2 + 0 ⎟ − ( 0 + (50)(9.81)(5.00) )
2
⎝
⎠
F = −185 N
(negative sign indicates retarding force)
(j)
show an appreciation for the implications of energy losses in practical devices
and use the concept of efficiency to solve problems.
For practical devices to work, energy input is needed. Most modern practical devices run on
electrical energy (e.g. television, computer) or chemical energy (e.g. vehicle).
When a practical device works, it converts the energy input into both useful energy output
and wasted energy.
If the output is intended in our design or operation of the device, then it is useful.
If it is not intended, then it is considered as wasted energy.
Efficiency of a practical device is a measure of how much useful work that device produces
from a given amount of energy input.
Its value does depend on what energy output we consider as useful.
Efficiency is dimensionless and can be expressed as a ratio or percentage.
useful energy output
x 100%
energy input
useful power output
=
x 100%
power input
Efficiency, η =
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Work, Energy and Power
We can never make a practical device with 100% efficiency because:
a) we have limited control over physical processes (e.g. a filament bulb must heat up
before it produces light, but the heat produced becomes wasted energy);
b) Dissipative forces tend to convert useful energy into heat, which is a form of wasted
energy. Mechanical devices are especially susceptible to this.
Worked Example 13
A car has a mass of 800 kg and the efficiency of its engine is rated at 18%. Determine the
amount of fuel used to accelerate the car from rest to 60 km h−1, given that the energy
supplied by one litre of fuel is 1.3 × 108 J.
Solution:
Note: In this case, the useful energy output is defined to be the change in kinetic energy of
the car as it accelerates from rest to 60 km h-1.
2
Useful energy output = K.E. of car =
η=
useful energy output
× 100%
energy input
1
60000 ⎞
5
( 800 ) ⎛⎜
⎟ = 1.11× 10 J
2
⎝ 60 × 60 ⎠
1.11× 105
× 100%
energy input
energy input = 6.17 x 105 J
⇒
18% =
⇒
Amount of fuel =
(k)
6.17 x 105
= 4.7 x10-3 litres
1.3 × 108
define power as work done per unit time and derive power as the product of
force and velocity.
Power is defined as:
i.
the rate of work done, or
ii.
the work done per unit time, or
iii.
the rate of energy conversion.
P =
where
•
•
•
•
W E
=
t
t
P is the power (unit: W)
W is the work done (unit: J)
E is the energy converted (unit: J)
t is the time taken (unit: s)
The S.I. unit for power is the watt.
One watt is defined as the power when one joule of work is done (or one joule of energy is
converted) per second.
1 W = (1 J) / (1 s)
Physics Department
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
Derive power as the product of force and velocity.
Suppose a constant force F displaces an object, moving at a constant velocity v, by a
distance s over a time interval t, and that F, v and s all point along the same line.
v
v
F
F
s
t
Power P developed by the constant force F,
P=
W Fs
⎛s⎞
=
= F ⎜ ⎟ = Fv
t
t
⎝t ⎠
P = Fv
Note: In order for the object to move at constant velocity, by Newton’s 1st law, there must be
an equal and opposite force f (i.e. drag force) acting on it.
•
Power developed by the drag force Pf = fv, where F = f
If force F and/or velocity v is/are not constant, then
total work done
total time taken
•
the average power <P> is given by < P > =
•
the instantaneous power is given by P = Fv,
where F and v take the values at the instant considered
Worked Example 14
A car moves along a horizontal road at a constant velocity of 15 m s−1. If the total resistive
force experienced by the car is 5000 N, calculate the power output of the car’s engine.
Solution:
At constant velocity, By Newton’s 1st law,
Fnet = 0
∴ F = 5000 N
P = Fv
= (5000) (15)
= 75000 W
Physics Department
F
5000 N
Page 19 of 22
JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
Guided Worked Example
A trolley of mass 7.0 kg is initially rest at t = 0 s.
A cyborg pushes this trolley with a constant force of 95 N along a horizontal floor.
The frictional force acting on the trolley is 11 N.
Calculate
(i)
the acceleration of the trolley;
(ii)
the speed of the trolley at t = 4.0 s;
(iii)
the kinetic energy of the trolley at t = 4.0 s;
(iv) the distance travelled during the first 4.0 s;
(v)
the instantaneous power supplied to the trolley by the cyborg at t = 4.0 s;
(vi) the average power supplied to the trolley by the cyborg during the first 4.0 s;
(vii) the average power dissipated by friction during the first 4.0 s;
(viii) the net average power gained by the trolley during the first 4.0 s, and hence, the total
energy gained by the trolley during the first 4.0 s.
Solution:
[Draw a diagram to illustrate the question]
[Identify the direction of acceleration, which is in the same direction
as the resultant force according to Newton’s 2nd law]
(i)
Fnet = ma ⇒ 95 − 11 = 7a ⇒ a = 12 m s−2
11 N
(Note: the 7 kg is accelerating to the right at 12 m s-2)
95 N
[For constant a and linear motion, apply the Equations of Motion to find v]
(ii)
v = u + at = 0 + 12(4) = 48 m s−1
[Use kinetic energy formula]
(iii)
Ek = ½ mv2 = ½ (7)(48)2 = 8064 J ≈ 8.1 x 103 J
[Use the Equations of Motion to find s]
(iv) s = ut + ½ at2 = 0 + ½ (12)(4)2 = 96 m
[Use instantaneous power formula to find power by cyborg]
(v)
P = Fv = (95)(48) = 4560 W ≈ 4.5 x 103 W
[Use average power formula to find average power by the cyborg]
(vi)
<Psupplied> =
total WD
Fs (95)(96)
=
=
= 2280 W ≈ 2.3 x 103 W
total time taken
4
t
[Use average power formula to find average power by friction]
(vii)
<Pdissipated> =
Wfriction
fs (11)(96 )
=
=
= 264 W ≈ 2.6 x 102 W
4
t
t
(Note: work done by friction is negative as frictional force is opposite to its displacement
i.e. there is energy loss by the trolley. Correspondingly, power dissipated suggests power
loss by the trolley)
,
(viii) <Pnet> = <Psupplied> − <Pdissipated> = 2280 − 264 = 2016 W ≈ 2.0 x 103 W
Total energy gained = <Pnet>t = (2016)(4) = 8064 J = K.E. gain by trolley in part (iii)
Physics Department
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
Reference Textbooks:
1) Serway/Faughn, College Physics 6th Ed.(Int. Student Ed), Thomson Learning 2003
2) Loo Kwok Wai, Longman Advanced Level Physics, Pearson Ed. 2006
3) Robert Hutchings, Physics 2nd Ed., Nelson 2000
4) Paul G. Hewitt, Conceptual Physics 9th Ed., Addison Wesley 2002
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JJ 2014 H1/H2 PHYSICS (8866 & 9646)
Work, Energy and Power
Supplementary Readings:
Other units of energy:
1.
Using kilowatt hour (kWh) as a unit of energy
Cost of electricity consumption is commonly calculated in kilowatt-hours (kWh) of
electrical energy used. kWh is sometimes known as the domestic unit of electricity
1 kWh is the energy used by a device at a rate of 1000 watts in 1 hr.
∴ 1 kWh = (1000 W) (60 min) (60 sec) = 3.6 x 106 J = 3.6 MJ
The local utility companies (such as Singapore Power) calculate household electricity
consumption based on how many kWh of electrical energy has been used. They
charge a certain rate per unit (i.e. per kWh) of electricity consumed by taking into
account the factors affecting the cost of producing that unit of electricity.
Example (taken from the NEA’s Green Label energy labelling scheme):
(Please log on to www.nea.gov.sg for more details.)
2.
Horsepower (hp)
For motors and engines, power is often measured in horse-power (hp) where
1 hp = 746 W
Example:
The Chevrolet Corvette Z-06 accelerates from 0 to100 kmh-1 in 3.7 s. The engine of the
Chevrolet Corvette Z-06 has an output of 505 hp and is capable of producing
(505) (746) = 377 kW
i.e. the motor is capable of doing 377 kJ of work each second.
Physics Department
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