Download M2.4 Work and Energy

AS-Level Maths:
Mechanics 2
for Edexcel
M2.4 Work and
Energy
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Energy
Contents
Energy
Work
Conservation of energy
Work-energy and dissipative forces
Power
Examination-style questions
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Mechanical energy
Energy is a measure of a particles capacity to do work.
Energy is measured in joules or kilojoules, represented
by J or kJ.
Energy exists in many forms. These include electrical
energy, chemical energy, heat energy and light energy.
Mechanical energy is energy that a particle has due to
either its motion or its position.
Mechanical energy is the energy that will be
considered in M2. It comes in two forms – kinetic
and potential energy.
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Kinetic and potential energy
Kinetic Energy
Kinetic energy is the energy a particle possesses due to
its motion.
Kinetic Energy is defined as
½mv2
where m is mass in kg and v is speed in ms-1.
Gravitational Potential Energy
Gravitational Potential Energy is the energy a particle
possesses due to its position. It is stored energy.
Change in Gravitational Potential Energy is defined as
mgh,
where m is mass in kg, g is acceleration due to gravity and h is
the vertical distance travelled in metres.
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Kinetic energy
Find the kinetic energy of the following:
a) A particle of mass 0.5 kg travelling at 8 ms-1
b) A car of mass 900 kg travelling at 15 ms-1
c) A bullet of mass 20 g travelling at 500 ms-1
a) K.E., in joules,
b) K.E., in joules,
c) K.E., in joules,
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= ½ × m × v2
= ½ × 0.5 × 82
= 16
= ½ × m × v2
= ½ × 900 × 152
= 101 250
= ½ × m × v2
= ½ × 0.02 × 5002
= 2500
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Gravitational potential energy
Find the gravitational potential energy of the following, stating
whether it is a gain or loss in gravitational potential energy:
a) A particle of mass 2 kg raised a vertical distance of 8 m
b) A lift of mass 500 kg descending 50 m vertically
c) A man of mass 80 kg climbing a vertical distance of 75 m.
a) G.P.E., in joules, = m × g × h
= 2 × 9.8 × 8
= 156.8
This is a gain in G.P.E.
b) G.P.E., in joules, = m × g × h
= 500 × 9.8 × 50
= 245 000
This is a loss of G.P.E.
c) G.P.E., in joules, = m × g × h
= 80 × 9.8 × 75
= 58 800
This is a gain in G.P.E.
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Energy question 1
Question 1: Find the gain in kinetic energy of a car of mass
800 kg as it accelerates from 10 ms-1 to 25 ms-1.
Initial K.E.
= ½ × m × v2
= ½ × 800 × 102
= 40 000
Final K.E.
= ½ × m × v2
= ½ × 800 × 252
= 250 000
Therefore the gain in kinetic energy of the car is 210 000 J.
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Energy question 2
Question 2: A child of mass 30 kg slides down a slide of
length 4 m inclined at an angle of 30° to the horizontal. By
modelling the child as a particle and the slide as a smooth
inclined plane, find the loss of gravitational potential energy
of the child.
G.P.E.
4m
30g N
30°
=m×g×h
= 30 × 9.8 × 4sin30°
= 588
Therefore the loss in G.P.E. is 588 J.
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Work
Contents
Energy
Work
Conservation of energy
Work-energy and dissipative forces
Power
Examination-style questions
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Work
Work can be defined as force × distance
i.e., the work done by a constant force is the force (in
Newtons) multiplied by the distance moved (in metres) in the
direction of the force.
If the direction of the force is different to the direction of
motion then
Work = F × dcos = Fdcos
F

Direction of motion
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Work question 1
Question 1: Find the work done by a force of 10 N acting on
a particle that moves 3 m in the direction of the force.
Work = force × distance
= 10 × 3
= 30
Therefore the work done by the force is 30 J.
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Work question 2
Question 2: A particle of mass 2 kg is pulled 3 m along a
smooth horizontal surface by a force of 8 N acting at an angle
of 60° to the horizontal. Find the work done by the force.
8N
2 kg
60o
Direction of motion
Work done
= Fdcos
= 8 × 3 × cos60°
= 12
Therefore work done by the force is 12 J.
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Work question 3
Question 3: A stone of mass 50 g falls from the top of
a cliff of height 75 m. Find the work done by gravity as
the stone falls.
0.05g N
Work done
= force × distance
= (0.05 × 9.8) × 75
= 36.75
Therefore the work done by gravity is 36.75 J.
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Work question 4
Question 4: A particle of mass 5 kg is pulled 8 m up a
rough plane inclined at an angle of 60° to the horizontal.
The coefficient of friction between the plane and the
particle is 0.3.
a) Find the work done against friction.
b) Find the work done against gravity.
R
F
5g N
60°
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Work question 4
a) To find the work done against the frictional force it is first
necessary to find the frictional force.
Resolving perpendicular to the plane:
R = 5gcos60° = 2.5g
F = µR
= 0.3 × 2.5g = 7.35
The frictional force is 7.35 N.
The distance travelled in the direction opposite to the
frictional force is 8 m.
 Work done against friction = 7.35 × 8 = 58.8 J.
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Work question 4
b) To find the work done against gravity it is necessary to
calculate the vertical distance travelled.
8m
h
60°
sin60° = h ÷ 8
h = 8sin60°
 Work done against gravity = 5g × 8sin60°
= 339 J (3 s.f.)
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Conservation of energy
Contents
Energy
Work
Conservation of energy
Work-energy and dissipative forces
Power
Examination-style questions
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Conservation of energy
The total mechanical energy of a particle remains constant
if the only force acting on the particle is gravity.
The total mechanical energy of a particle is the sum of any
kinetic energy and any potential energy.
Hence, for a particle on which the only force acting is gravity,
Initial K.E. + Initial P.E. = Final K.E. + Final P.E.
Therefore
Loss in K.E. = Gain in P.E.
and
Gain in K.E. = Loss in P.E.
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Conservation question 1
Question 1: A particle of mass 5 kg slides down a smooth
slope inclined at an angle of 30° to the horizontal. Given that
the particle starts from rest, find the distance travelled along
the slope when the particle has reached a speed of 4.9 ms-1.
Since the only force acting on the particle is gravity, energy
is conserved.
5g
30
°
Initial K.E. = 0 (the particle is at rest)
Final K.E. = ½ × 5 × 4.92 = 60.025
Gain in K.E. = 60.025 J
Loss in G.P.E. = 60.025 J
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Conservation question 1
Change in G.P.E. = 5 × 9.8 × h = 60.025
60.025
h
 1.225 m
49
d
1.225 m
30
°
1.225
sin30° =
d
1.225
d 
 2.45
sin30°
Therefore the distance moved along the slope is 2.45 m.
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Conservation question 2
Question 2: A particle of mass 2 kg is projected up a rough
plane inclined at an angle of 45° to the horizontal. The
coefficient of friction between the particle and the plane is
0.15. If the particle travels 5 m up the plane before coming to
rest, find the speed with which the particle was projected.
Since the only force acting on the particle is gravity, energy
is conserved.
To calculate the gain in G.P.E. we need to calculate the
vertical height gained.
h
 h = 5sin45°
sin45° =
5
5m
Therefore the vertical distance
h
travelled is 3.54 m (3 s.f.).
45
°
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Conservation question 2
Gain in G.P.E. = mgh
= 2 × 9.8 × 5sin45
2g
= 69.3 (3 s.f.)
45
°
As energy is conserved, gain in G.P.E. = loss in K.E.
Initial K.E. = ½ × 2 × v2
Final K.E. = 0 (particle comes to rest)
Loss in K.E.
½×2×
v2
= 69.3
Gain in G.P.E.
v = 8.32 (3 s.f.)
Therefore the particle is projected up the plane with
a speed of 8.32 ms-1.
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Work-energy and dissipative forces
Contents
Energy
Work
Conservation of energy
Work-energy and dissipative forces
Power
Examination-style questions
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Work-energy principle
The relationship between work and energy is a simple one:
Work Done = Change in Kinetic Energy
This can be seen from the following application of
Newton’s Second Law.
A particle of mass m kg is moved a distance of
d m horizontally by a constant horizontal force F N.
Let the initial speed be u ms-1 and the final speed be
v ms-1.
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Work-energy principle
By Newton’s Second Law:
F = ma
 a = F/m
Since the force is assumed constant, the acceleration
can also be assumed constant.
Using v2 = u2 + 2as
v2 = u2 + 2 × F/m × d
2Fd
m
½mv2 = ½mu2 + Fd
v2 = u2 +
This gives
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Remember:
s = distance
u = initial velocity
v = final velocity
Fd = ½mv2 – ½mu2
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Work-energy question 1
Question 1: A particle of mass 3 kg is pulled 6 m by a
horizontal force of 10 N across a smooth horizontal surface.
If the particle started from rest, find its speed when it has
travelled 6 m.
Work done = Fd = 10 × 6 = 60
Direction of motion
Fd = ½mv2 – ½mu2
Work done = ½mv2 – ½mu2 = 60
3 kg
10 N
½ × 3 × v2 – 0 = 60
1.5v2 = 60
v = 6.32 (3 s.f.)
Therefore the speed of the particle when it has
travelled 6 m is 6.32 ms-1.
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Work-energy question 2
Question 2: Find the horizontal force that causes a particle
of mass 2.5 kg to increase speed from 2 ms-1 to 5 ms-1 over
a distance of 10 m on a smooth horizontal surface.
Work done = change in kinetic energy
Initial K.E. = ½ × 2.5 × 22 = 5
Final K.E. = ½ × 2.5 × 52 = 31.25
Change in kinetic energy = 26.25
Work done = Fd = 10F = 26.25
F = 2.625
Therefore the required force is 2.625 N.
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Work and dissipative forces
A dissipative force is a force which causes a particle to
lose energy.
This means that the principle of conservation no longer
applies, as the total mechanical energy of the particle is
not constant.
Friction and air resistance are examples of dissipative forces.
Work done against a dissipative force is equal to the loss of
the total mechanical energy of the system.
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Work and dissipative forces
Example: A particle of mass 3 kg is moving down a rough
slope inclined at an angle of 30° to the horizontal. The
particle passes a point A at 3 ms-1 and a point B 2 m down
the slope at 4 ms-1. Calculate the work done against friction.
Gain in K.E. = ½ × 3 × 42 – ½ × 3 × 32
= 10.5 J
sin30° = h ÷ 2
 h = 2sin30° = 1
2m
h
30°
Loss in G.P.E. = 3 × g × 1 = 29.4 J
Loss of energy = 29.4 – 10.5 = 18.9
 Work done against friction = 18.9 J
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Power
Contents
Energy
Work
Conservation of energy
Work-energy and dissipative forces
Power
Examination-style questions
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Power
Power is defined as the rate of doing work, with respect
to time.
Work (in joules)
Power is calculated by:
Time (in seconds)
The S.I. unit of power is the watt, which is a rate of work of
1 joule per second.
The watt is a relatively small unit. You will more commonly
meet the larger unit, the kilowatt.
1 kilowatt (abbreviated to 1 kW) = 1000 watts
= 1000 joules per second
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Power
Consider a force F acting for a small time interval t over a
small distance d.
Work (in joules)
Work done = F × d
Power =
Time (in seconds)
So, Power
=
Fd
t
d
= F
= Fv
t
So, the power of an engine or machine moving at a speed v
subject to a driving force F is Fv.
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Power question 1
Question 1: A car exerts a driving force of 1500 N.
If the car is travelling at a constant speed of 35 ms-1,
calculate the power of the engine.
Power, in watts = Fv
= 1500 × 35 = 52 500
Therefore the power produced by the car engine is 52.5 kW.
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Power question 2
Question 2: The engine of a car is generating power of
18 kW. Find the driving force produced when the car is
travelling at a speed of:
a) 12 ms-1
b) 20 ms-1
c) 30 ms-1
Power = Force × Velocity
Power
So, Force =
Velocity
a) Force produced = 18 000 ÷ 12 = 1500 N
b) Force produced = 18 000 ÷ 20 = 900 N
c) Force produced = 18 000 ÷ 30 = 600 N
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Power question 3
Question 3: A cyclist is travelling along a horizontal road
at a constant speed of 7 ms-1.
If the cyclist experiences a total resistance to motion of
20 N, find the power produced by the cyclist.
Since the speed is constant there is no acceleration.
Therefore, driving force = resistance force.
Driving force = 20 N
Velocity = 7 ms-1
 Power = 20 × 7 = 140 W
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Examination-style questions
Contents
Energy
Work
Conservation of energy
Work-energy and dissipative forces
Power
Examination-style questions
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Exam question 1
Examination-style question 1: A car of mass 1000 kg
moves along a straight, horizontal road. The car engine is
working at a constant rate of 45 kW and the total resistance
to the motion of the car is 500 N.
a) Find the acceleration of the car when its speed is 15 ms-1.
The car comes to a hill which is inclined at an angle of 
to the horizontal, where sin = 0.1. The resistance to the
motion of the car is unchanged.
b) Find the maximum speed of the car up the hill when it
is working at a rate of 45 kW.
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Exam question 1
a) To calculate the acceleration
it is first necessary to
calculate the driving force.
Power = Force × Velocity
a
500 N
D
1000
Power
So, Force =
Velocity
45 000
D=
= 3000
15
Applying Newton’s Second Law, F = ma
3000 – 500 = 1000a
1000a = 2500
 a = 2.5
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Therefore the acceleration of
the car when it is travelling at
15 ms-1 is 2.5 ms-2.
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Exam question 1
b) Before calculating the
maximum speed of the car
it is first necessary to
calculate the driving force.
R
D
500 N

1000g
Applying Newton’s Second Law up the slope,
D – 500 – 1000gsin = 0
D = 500 + 1000g × 0.1 = 1480
The driving force of the car is 1480 N.
Power
Power = Force × Velocity  Velocity =
Force
45 000
v
 30.4 (3 s.f.)
1480
The maximum speed of the car is 30.4 ms-1.
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Exam question 2
Examination-style question 2: A stone of mass 0.5 kg is
sliding down a rough plane inclined at an angle of 15° to the
horizontal.
The stone passes a point A with a speed of 10 ms-1 and a
point B with a speed of 6 ms-1.
a) Find the loss of mechanical energy of the stone as it
moves from point A to point B.
b) Calculate the coefficient of friction between the stone and
the plane.
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Exam question 2
a) Loss of energy of the stone = loss of K.E. + loss of G.P.E.
Initial K.E. = ½ × 0.5 × 102
Final K.E. = ½ × 0.5 × 62
Loss of K.E. = 25 – 9 = 16 J
10
15°
sin15° 
h
h
10
 h = 10sin15°
Loss of G.P.E. = 0.5 × 9.8 × 10sin15° = 12.7 (3 s.f.)
16 + 12.7 = 28.7
Therefore the loss of energy of the stone is 28.7 J.
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Exam question 2
b) To calculate the coefficient of
friction, it is necessary to
calculate both the normal contact
force and the frictional force.
R = 0.5gcos15° = 4.73 (3 s.f.)
R
F
15°
0.5g
Loss of energy of the stone found in part a) is
equal to the work done against friction.
The work done against friction over a distance of
10 m is 28.7 J.
Work done = force × distance
force = 28.7 ÷ 10 = 2.87
F 2.87
µ  
 0.61 (2 s.f.)
R 4.73
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 the coefficient of
friction is 0.61.
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Exam question 3
Examination-style question 3: A parcel of mass 30 kg is
hauled up a slope inclined at an angle of 20° to the
horizontal. The parcel is hauled by means of a constant force
of 300 N at an angle of 30° to the line of greatest slope of the
plane.
The acceleration of the parcel up the plane is 0.1 ms-2 and
the total resistance to motion is RN.
a) Find the work done by the force in hauling the parcel,
from rest, 10 m up the slope.
b) By considering work and energy, find the value of R.
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Exam question 3
a) Work done = force × distance
Since the force acts at an angle
of 30° to the slope, the
component of force acting along
the slope must be calculated.
300 N
30°
c
c
cos30° =
300
c = 300cos30o = 260 (3 s.f)
Work done = 260 × 10 = 2600
So, work done by the force is 2600 J.
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Exam question 3
h
b) sin20° =
10
so, h = 10sin20°
10
h
20°
Gain in G.P.E. = 30 × 9.8 × 10sin20o
= 1006 (4 s.f)
To find gain in K.E. it is
necessary to find the
speed of the parcel after
it has travelled 10 m.
v2 = u2 + 2as
= 0 + 2 × 0.1 × 10 = 2
v = 1.41 ms-1 (3 s.f.)
Gain in K.E. = ½ × 30 × 2 = 30
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Exam question 3
1006 + 30 = 1036
Therefore the increase in energy of the
system is 1036 J (4 s.f).
The difference between the work done by the force acting up
the slope and the gain in energy of the system is
2600 – 1036 = 1564 J.
This is the work done by the resistance force.
Work done = force × distance
 R = 1564 ÷ 10 = 156.4 N
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