Download You can calculate the kinetic energy of a moving particle, and the

Introduction
• In this chapter you will learn about calculations
involving work, energy and power
• You will learn how to use several formulae
• You will learn how to solve problems involving
kinetic and potential energy
• You will also learn about the work-energy
principle which is a common type of question on
exam papers!
Work, energy and power
You can calculate the work done by a
force when its point of application moves
by using the following formula
A box is pulled 7m across a horizontal floor by a
horizontal force of magnitude 15N. Calculate the
work done by the force
𝑾 = 𝑭𝒔
W = work done
F = magnitude of the force
s = the distance moved in the direction of
the force
For work done against gravity:
𝑾 = 𝒎𝒈𝒉
W = work done
m = mass of the object
g = gravitational constant
h = the height raised
15N
𝑊 = 𝐹𝑠
Sub in values from the question
𝑊 = 15 × 7
𝑊 = 105𝐽
Calculate
Work done is measured in Joules!
These two formulae are effectively the
same!
3A
Work, energy and power
You can calculate the work done by a
force when its point of application
moves by using the following formula
A packing case is pulled across a
horizontal floor by a horizontal rope.
The case moves at a constant speed and
there is a constant resistance to motion
of magnitude R Newtons. When the
case has moved a distance of 12m the
work done is 96J. Calculate the
magnitude of the resistance
 In this case you will need to use
more than one formula, as we do not
know either the force or the
resistance…
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Draw a diagram – we do not know the force or the
resistance, and the acceleration is 0 (constant speed)
0
RN
8N
FN
Find the force acting on the box by using one of the
formulae above…
𝑊 = 𝐹𝑠
Sub in values from the question
96 = 𝐹 × 12
𝐹 = 8𝑁
Calculate
Now use F = ma, resolving horizontally
𝐹 = 𝑚𝑎
8−𝑅 =0
𝑅 = 8𝑁
Acceleration is 0, remember to
include forces correctly
Calculate
3A
Work, energy and power
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Draw a diagram – Tension is the force in the cable. The
weight can be added to the diagram as well and
acceleration
30g
T
You can calculate the work done by a
force when its point of application
moves by using the following formula
A bricklayer raises a load of bricks of
total mass 30kg at a constant speed by
attaching a cable to the bricks.
0
Assuming the cable is vertical, calculate
the work done when the bricks are
raised a distance of 7m
If we are going to calculate
the work done, we need the
tension
 Use F = ma and resolve
vertically
30g
𝐹 = 𝑚𝑎
𝑇 − 30𝑔 = 0
𝑇 = 30𝑔
𝑊 = 𝐹𝑠
𝑊 = 30𝑔 × 7
𝑊 = 2058𝐽
Calculate
Rearrange
Sub in values (you could also have
used W = mgh)
Calculate the value in Joules
3A
Work, energy and power
You can calculate the work done by a
force when its point of application
moves by using the following formula
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Draw a diagram and label all the forces
R
Diagram of the
distance moved
P
12m
A package of mass 2kg is pulled at a
constant speed up a rough plane which
is inclined at an angle of 30° to the
horizontal. The coefficient of friction
between the package and the surface is
0.35. The package is pulled 12m up a
line of greatest slope of the plane.
Calculate:
a)
The work done against gravity
= 118𝐽
b) The work done against friction
FMAX
2g 30°
30°
2gCos30
30°
12Sin30
12Cos30
2gSin30
To calculate the work done against gravity, we need to
know the change in vertical height of the package
 You can draw a diagram to show this, with the diagonal
being 12m, and the inclination still being 30°
𝑊 = 𝑚𝑔ℎ
𝑊 = 2 × 9.8 × (12𝑆𝑖𝑛30)
Sub values in
Calculate
𝑊 = 118𝐽
3A
Work, energy and power
You can calculate the work done by a
force when its point of application
moves by using the following formula
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Draw a diagram and label all the forces
R
P
Diagram of the
distance moved
12m
A package of mass 2kg is pulled at a
constant speed up a rough plane which
is inclined at an angle of 30° to the
horizontal. The coefficient of friction
between the package and the surface is
0.35. The package is pulled 12m up a
line of greatest slope of the plane.
Calculate:
a)
The work done against gravity
= 118𝐽
b) The work done against friction
 Find the force acting against FMAX
FMAX
30°
2g 30°
2gCos30
30°
12Sin30
12Cos30
2gSin30
We can calculate the work done against friction by using the
formula W = Fs
F = the force in the opposite direction to friction (as the
work is done AGAINST friction)
s = the distance travelled up the plane
 We therefore need to find FMAX first, and can then use it
to find the pulling force P, which is acting against friction…
3A
Work, energy and power
You can calculate the work done by a
force when its point of application
moves by using the following formula
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Draw a diagram and label all the forces
2gCos30
R
P
Diagram of the
distance moved
12m
A package of mass 2kg is pulled at a
constant speed up a rough plane which
is inclined at an angle of 30° to the
horizontal. The coefficient of friction
between the package and the surface is
0.35. The package is pulled 12m up a
line of greatest slope of the plane.
Calculate:
a)
The work done against gravity
= 118𝐽
b) The work done against friction
FMAX
0.7gCos30
2g 30°
30°
2gCos30
30°
12Sin30
12Cos30
2gSin30
The normal reaction will just be 2gCos30 as there is no
acceleration perpendicular to the plane
𝐹𝑀𝐴𝑋 = 𝜇𝑅
𝐹𝑀𝐴𝑋 = 0.35 × (2𝑔𝐶𝑜𝑠30)
𝐹𝑀𝐴𝑋 = 0.7𝑔𝐶𝑜𝑠30
Sub in values
Simplify (to ensure it
stays exact)
 Find the force acting against FMAX
3A
Work, energy and power
You can calculate the work done by a
force when its point of application
moves by using the following formula
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Draw a diagram and label all the forces
2gCos30
0.7gCos30P + 2gSin30 Diagram of the
distance moved
12m
A package of mass 2kg is pulled at a
constant speed up a rough plane which
is inclined at an angle of 30° to the
horizontal. The coefficient of friction
between the package and the surface is
0.35. The package is pulled 12m up a
line of greatest slope of the plane.
Calculate:
a)
The work done against gravity
= 118𝐽
b) The work done against friction
= 70.9𝐽
0.7gCos30
2g 30°
30°
2gCos30
30°
12Sin30
12Cos30
2gSin30
Now resolve parallel to the plane to find force P
𝐹 = 𝑚𝑎
𝑃 − 0.7𝑔𝐶𝑜𝑠30 − 2𝑔𝑆𝑖𝑛30 = 0
𝑃 = 0.7𝑔𝐶𝑜𝑠30 + 2𝑔𝑆𝑖𝑛30
𝑊 = 𝐹𝑠
𝑊 = (0.7𝑔𝐶𝑜𝑠30 + 2𝑔𝑆𝑖𝑛30) × 12
 In these types of questions, the
𝑊 = 188.9𝐽
work done against friction and the work
done against gravity give the total work 𝑊(𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛) = 70.9𝐽
done…
Sub in values, acceleration is 0.
remember to include the
gravitational part (for now…)
Work out P and leave as
an exact answer
Sub in F(P) and s
Calculate – this gives us the
TOTAL work done on the particle
Subtract the work done against
gravity (118J) to leave the work
done against friction
3A
𝑾 = 𝑭𝒔
Work, energy and power
27N
You can calculate the work done by a
force when its point of application
moves by using the following formula
A sledge is pulled 15m across a smooth
sheet of ice by a force of magnitude
27N. The force is inclined at 25° to the
horizontal. By modelling the sledge as a
particle, calculate the work done by the
force.
 As the force is at an angle to the
motion, you must split it into its
component parts
 The force will act vertically and
horizontally
𝑾 = 𝒎𝒈𝒉
25°
27Sin25
27Cos25
𝑊 = 𝐹𝑠
Sub in values
𝑊 = (27𝐶𝑜𝑠25) × 15
Calculate
𝑊 = 367𝐽
The total work done is 367J
 However, as there is no distance
travelled vertically (s = 0), there is
no work done in this direction
 Therefore, you only need the work
done horizontally…
3A
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
Work, energy and power
You can calculate the kinetic
energy of a moving particle, and
the potential energy of a particle
above ground level
Kinetic Energy
𝐾𝐸 =
1
𝑚𝑣 2
2
m is the mass of the particle
v is its velocity
Potential Energy
𝑃𝐸 = 𝑚𝑔ℎ
m is the mass of the particle
g is the gravitational constant
h is the height of the particle above
the ground (or a given fixed point)
Kinetic energy is the energy a body possesses due to
its motion
Faster movement = more Kinetic Energy
Heavier object = more Kinetic Energy
Potential energy is energy which is effectively stored
in an object and which could become active
A ball held in the air has potential energy, which will
become kinetic energy if the ball is dropped
Heavier object = more potential energy
Object held higher up = more potential energy
(This chapter focuses on gravitational potential
energy)
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can calculate the kinetic
energy of a moving particle, and
the potential energy of a particle
above ground level
The work done by a force which
accelerates a particle is connected
to the kinetic energy of the particle
Work done = Change in kinetic
energy
𝐹 = 𝑚𝑎
𝑣 2 − 𝑢2
𝐹=𝑚
2𝑠
Multiply top by m
𝑚(𝑣 2 − 𝑢2 )
𝐹=
2𝑠
𝐹𝑠 =
To show this, we will rewrite one of
the SUVAT equations to give it in
terms of a.
𝐹𝑠 =
𝑣 2 = 𝑢2 + 2𝑎𝑠
𝑣 2 − 𝑢2 = 2𝑎𝑠
𝑣 2 − 𝑢2
=𝑎
2𝑠
Subtract u2
Divide by 2s
Replace a with the
expression we worked
out
𝑊=
𝑚(𝑣 2
−
2
𝑢2 )
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
Final KE - Initial KE
Multiply all by s
Rewrite right
side
Fs = work done
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can calculate the kinetic
energy of a moving particle, and
the potential energy of a particle
above ground level
A particle of mass 0.3kg is moving
at a speed of 9ms-1. Calculate its
kinetic energy.
𝐾𝐸 =
𝐾𝐸 =
1
𝑚𝑣 2
2
1
(0.3)(9)2
2
Sub in values
Calculate
𝐾𝐸 = 12.15𝐽
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can calculate the kinetic
energy of a moving particle, and
the potential energy of a particle
above ground level
A box of mass 1.5kg is pulled across
a smooth horizontal surface by a
horizontal force. The initial speed
of the box is ums-1 and its final
speed is 3ms-1. The work done by
the force is 1.8J. Calculate the
value of u.
We know W, v and m, and need u
𝑊=
1.8 =
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
1
(1.5) 3
2
2
1
− (1.5)𝑢2
2
Sub in
values
Work out
parts
1.8 = 6.75 − 0.75𝑢2
Rearrange
2
0.75𝑢 = 4.95
Divide by 0.75
𝑢2 = 6.6
𝑢 = 2.57𝑚𝑠 −1
Square root
 Use the formula for the change
in kinetic energy!
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can calculate the kinetic energy of
a moving particle, and the potential
energy of a particle above ground level
A bus of mass 2000kg starts from rest at
some traffic lights. After travelling 400m
the bus’s speed is 12ms-1. A constant
resistance of 500N acts on the bus.
Calculate the driving force, P, which can
be assumed to be constant.
We know the following pieces of
information:
u = 0ms-1
v = 12ms-1
s = 400m
m = 2000kg
We also know the overall force will be the
driving force subtract the resistances
F = P - 500
500N
P
𝑊=
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
𝐹𝑠 =
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
1
𝑃 − 500 × 400 = (2000) 12
2
Replace W
with Fs
Sub in values
1
2
− (2000)(0)2
2
400 𝑃 − 500 = 144000
Calculate
parts
Divide by 400
𝑃 − 500 = 360
Add 500
𝑃 = 860𝑁
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can calculate the kinetic energy of
a moving particle, and the potential
energy of a particle above ground level
A load of bricks is lowered vertically to
the ground through a distance of 15m.
Find the loss in potential energy.
 In this case, you can use ‘h’ as the
change in height, rather than the height
of the particle
𝑃𝐸 = 𝑚𝑔ℎ
𝑃𝐸 = (30)(𝑔)(−15)
Sub in values. The height
has fallen by 15m…
Calculate
𝑃𝐸 = −4410𝐽
So the loss of potential energy is 4410J
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can calculate the kinetic energy of
a moving particle, and the potential
energy of a particle above ground level
A parcel of mass 3kg is pulled 7m up a
plane inclined at an angle θ to the
horizontal, where tanθ = 3/4. Assuming
that the parcel moves up a line of greatest
slope of the plane, calculate the potential
energy gained by the parcel.
You have seen situations like this before,
with the angle given as Tanθ. Start by
finding Sinθ and Cosθ.
𝑇𝑎𝑛𝜃 =
3
4
𝑆𝑖𝑛𝜃 =
3
5
𝐶𝑜𝑠𝜃 =
4
5
Draw a diagram
7m
θ
7Sinθ
7Cosθ
𝑃𝐸 = 𝑚𝑔ℎ
𝑃𝐸 = (3)(9.8)(7𝑆𝑖𝑛𝜃)
𝑃𝐸 = (3)(9.8) 7 ×
3
5
 The change in
potential energy will be
affected by the change
in the vertical height of
the parcel
Sub in values, using
the change in height
Also use the value of
Sinθ
Calculate
𝑃𝐸 = 123.48𝐽
3B
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
“When no external forces (other than
gravity) act on a particle, the sum of its
potential and kinetic energies remain
constant.”
(This is called the principle of the
conservation of mechanical energy)
“The change in total energy of a particle
is equal to the work done on the
particle.”
(This is called the ‘work-energy’
principle)
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force (usually friction) is acting on
the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
Draw a diagram
 The normal reaction is
doing no work as there is no
movement perpendicular to
the plane
 The plane is smooth so the
particle does not have to do
any work against friction
 We can therefore use the
upper of the formulae
shown…
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
B
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
30°
Initial speed = 0
Final speed = 6
Find the increase in Kinetic energy
𝑊=
A smooth plane is inclined at 30° to the
horizontal. A particle of mass 0.5kg slides down
the slope. The particle starts from rest at
point A and at point B has a speed of 6ms-1.
Find the distance AB.
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
𝑊=
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
1
(0.5) 6
2
2
1
− (0.5)(0)2
2
Sub in values
Calculate
𝑊 = 9𝐽
𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 = 9𝐽
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
Draw a diagram
 The normal reaction is
doing no work as there is no
movement perpendicular to
x
xSin30 the plane
 The plane is smooth so the
30°
particle does not have to do
xCos30
any work against friction
 We can therefore use the
upper of the formulae
shown…
Final speed = 6
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A smooth plane is inclined at 30° to the
horizontal. A particle of mass 0.5kg slides down
the slope. The particle starts from rest at
point A and at point B has a speed of 6ms-1.
Find the distance AB.
𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 = 9𝐽
B
30°
Initial speed = 0
Find the decrease in Potential energy (find the change in vertical
height first)
 Call the diagonal distance (the one we need to find) ‘x’
𝑃𝐸 = 𝑚𝑔ℎ
Sub in values
𝑃𝐸 = (0.5)(9.8)(𝑥𝑆𝑖𝑛30)
𝑃𝐸 = 2.45𝑥
Calculate in
terms of x
𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑃𝐸 = 2.45𝑥
3C
𝑾 = 𝑭𝒔
𝑾=
𝑾 = 𝒎𝒈𝒉
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
Draw a diagram
 The normal reaction is
doing no work as there is no
movement perpendicular to
x
xSin30 the plane
 The plane is smooth so the
30°
particle does not have to do
xCos30
any work against friction
 We can therefore use the
upper of the formulae
shown…
Final speed = 6
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A smooth plane is inclined at 30° to the
horizontal. A particle of mass 0.5kg slides down
the slope. The particle starts from rest at
point A and at point B has a speed of 6ms-1.
Find the distance AB.
𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸 = 9𝐽
𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑃𝐸 = 2.45𝑥
B
30°
Initial speed = 0
𝐷𝑒𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑃𝐸 = 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝐾𝐸
2.45𝑥 = 9
𝑥 = 3.67𝑚
Sub in the values we
calculated
Divide by 2.45
This could be calculated using F = ma and
the SUVAT equations from M1, however in
M2 you will usually be asked specifically to
use these principles…
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
45°
Initial speed = 8
Final speed = 0
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A particle of mass 2kg is projected with speed
8ms-1 up a rough plane inclined at 45° to the
horizontal. The coefficient of friction between
the particle and the plane is 0.4. Calculate the
distance the particle travels up the plane
before it comes to instantaneous rest.
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
Draw a diagram
 The normal reaction is doing
no work as there is no
movement perpendicular to
the plane
 As the plane is rough, the
particle will have to do
some work against friction.
You must take this into
account.
 You will need to use the
second of the formulae to
the left
Find the kinetic energy lost
𝑊=
𝑊=
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
1
(2) 0
2
2
1
− (2)(8)2
2
Sub in values
Calculate
𝑊 = −64𝐽
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽
3C
𝑾 = 𝑭𝒔
𝑾=
𝑾 = 𝒎𝒈𝒉
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
45°
xCos45
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
xSin45
x
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
45°
Initial speed = 8
Final speed = 0
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A particle of mass 2kg is projected with speed
8ms-1 up a rough plane inclined at 45° to the
horizontal. The coefficient of friction between
the particle and the plane is 0.4. Calculate the
distance the particle travels up the plane
before it comes to instantaneous rest.
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
Draw a diagram
 The normal reaction is doing
no work as there is no
movement perpendicular to
the plane
 As the plane is rough, the
particle will have to do
some work against friction.
You must take this into
account.
 You will need to use the
second of the formulae to
the left
Find the potential energy gained
 As in the last example, call the distance moved up the plane ‘x’,
and work out the vertical change, based on this…
𝑃𝐸 = 𝑚𝑔ℎ
Sub in values
𝑃𝐸 = (2)(9.8)(𝑥𝑆𝑖𝑛45)
𝑃𝐸 = 9.8 2𝑥
Calculate in
terms of x
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
x
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
45°
Initial speed = 8
Final speed = 0
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A particle of mass 2kg is projected with speed
8ms-1 up a rough plane inclined at 45° to the
horizontal. The coefficient of friction between
the particle and the plane is 0.4. Calculate the
distance the particle travels up the plane
before it comes to instantaneous rest.
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
Draw a diagram
 The normal reaction is doing
no work as there is no
movement perpendicular to
the plane
 As the plane is rough, the
particle will have to do
some work against friction.
You must take this into
account.
 You will need to use the
second of the formulae to
the left
 This time, we cannot just set these values equal to each other, as
some energy will be lost to friction
 Find an expression for the loss of energy by using the highlighted
formula
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐾𝐸 𝑙𝑜𝑠𝑡 − 𝑃𝐸 𝑔𝑎𝑖𝑛𝑒𝑑
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥
Sub in values to find
an expression for the
loss of energy
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥
3C
𝑾 = 𝑭𝒔
𝑾=
𝑾 = 𝒎𝒈𝒉
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
2gCos45
R
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
FMAX
7.84Cos45
If gravity is the only force acting on a particle:
45°
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
Initial speed = 8
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A particle of mass 2kg is projected with speed
8ms-1 up a rough plane inclined at 45° to the
horizontal. The coefficient of friction between
the particle and the plane is 0.4. Calculate the
distance the particle travels up the plane
before it comes to instantaneous rest.
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥
Draw a diagram
 The normal reaction is doing
x
no work as there is no
movement perpendicular to
the plane
2gCos45

As the plane is rough, the
45°
particle will have to do
2g
some work against friction.
2gSin45 You must take this into
account.

You will need to use the
Final speed = 0
second of the formulae to
the left
The energy lost will all have been used against friction
 We need to find an expression for the work done by friction,
and set it equal to the loss of energy
 We will first need to find the normal reaction, then find the
maximum frictional force
𝐹𝑀𝐴𝑋 = 𝜇𝑅
𝐹𝑀𝐴𝑋 = (0.4)(2𝑔𝐶𝑜𝑠45)
𝐹𝑀𝐴𝑋 = 7.84𝐶𝑜𝑠45
Sub in values
Rewrite
3C
𝑾 = 𝑭𝒔
𝑾=
𝑾 = 𝒎𝒈𝒉
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
2gCos45
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
7.84Cos45
If gravity is the only force acting on a particle:
45°
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
Initial speed = 8
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
Draw a diagram
 The normal reaction is doing
x
no work as there is no
movement perpendicular to
the plane
2gCos45

As the plane is rough, the
45°
particle will have to do
2g
some work against friction.
2gSin45 You must take this into
account.

You will need to use the
Final speed = 0
second of the formulae to
the left
Now we can calculate the work done against friction, by using
one of the formulae from earlier in the chapter
 The frictional force acts over a distance ‘x’
A particle of mass 2kg is projected with speed
8ms-1 up a rough plane inclined at 45° to the
horizontal. The coefficient of friction between
the particle and the plane is 0.4. Calculate the
distance the particle travels up the plane
before it comes to instantaneous rest.
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥
This is the energy lost
𝑊 = 𝐹𝑠
𝑊 = (7.84𝐶𝑜𝑠45)(𝑥)
𝑊 = 7.84𝑥𝐶𝑜𝑠45
Sub in F and s
Rewrite in terms of x
This is the work done
against friction
These expressions will be equal as all the energy
lost has been working against friction!
3C
𝑾 = 𝑭𝒔
𝑾=
𝑾 = 𝒎𝒈𝒉
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
2gCos45
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
7.84Cos45
If gravity is the only force acting on a particle:
45°
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
Draw a diagram
 The normal reaction is doing
x
no work as there is no
movement perpendicular to
the plane
2gCos45

As the plane is rough, the
45°
particle will have to do
2g
some work against friction.
2gSin45 You must take this into
account.

You will need to use the
Final speed = 0
second of the formulae to
the left
Initial speed = 8
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A particle of mass 2kg is projected with speed
8ms-1 up a rough plane inclined at 45° to the
horizontal. The coefficient of friction between
the particle and the plane is 0.4. Calculate the
distance the particle travels up the plane
before it comes to instantaneous rest.
𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑡 = 64𝐽
𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝑔𝑎𝑖𝑛𝑒𝑑 = 9.8 2𝑥
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 64 − 9.8 2𝑥
𝑊 = 7.84𝑥𝐶𝑜𝑠45
Set the two expressions equal to each other and solve for the
distance, ‘x’
64 − 9.8 2𝑥 = 7.84𝑥𝐶𝑜𝑠45
Add 9.8√2x
64 = 7.84𝑥𝐶𝑜𝑠45 + 9.8 2𝑥
64 = 7.84𝐶𝑜𝑠45 + 9.8 2 𝑥
64
7.84𝐶𝑜𝑠45 + 9.8 2
=𝑥
Factorise right
side
Divide by the
bracket
Calculate
3.3𝑚 = 𝑥
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A skier passes a point A on a ski-run, moving
downhill at 6ms-1. After descending 50m
vertically, the run starts to ascend. When the
skier has ascended 25m to point B her speed is
4ms-1. The skier and skis have a combined mass
of 55kg. The total distance travelled from A to
B is 1400m. The resistances to motion are
constant and have a magnitude of 12N.
B
50m
25m
Initial speed = 6
Final speed = 4
The work done by the skier will be equal to the work done against
resistances, subtract the total loss of energy
Calculate the loss of kinetic energy (it is a loss as speed has fallen)
𝑊=
𝑊=
1
1
𝑚𝑣 2 − 𝑚𝑢2
2
2
1
(55) 4
2
2
1
− (55)(6)2
2
Sub in values
Calculate
𝑊 = −550𝐽
Calculate the work done by the skier
𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A skier passes a point A on a ski-run, moving
downhill at 6ms-1. After descending 50m
vertically, the run starts to ascend. When the
skier has ascended 25m to point B her speed is
4ms-1. The skier and skis have a combined mass
of 55kg. The total distance travelled from A to
B is 1400m. The resistances to motion are
constant and have a magnitude of 12N.
B
50m
25m
Initial speed = 6
Final speed = 4
The work done by the skier will be equal to the work done against
resistances, subtract the total loss of energy
Calculate the gain of potential energy (it is actually a loss as the
height has fallen!)
𝑃𝐸 = 𝑚𝑔ℎ
Sub in values
𝑃𝐸 = (55)(9.8)(−25)
𝑃𝐸 = −13475𝐽
Calculate
Calculate the work done by the skier
𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽
𝑃𝐸 𝐺𝑎𝑖𝑛𝑒𝑑 = −13475𝐽
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A skier passes a point A on a ski-run, moving
downhill at 6ms-1. After descending 50m
vertically, the run starts to ascend. When the
skier has ascended 25m to point B her speed is
4ms-1. The skier and skis have a combined mass
of 55kg. The total distance travelled from A to
B is 1400m. The resistances to motion are
constant and have a magnitude of 12N.
Calculate the work done by the skier
𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽
𝑃𝐸 𝐺𝑎𝑖𝑛𝑒𝑑 = −13475𝐽
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 14025𝐽
B
50m
25m
Initial speed = 6
Final speed = 4
The work done by the skier will be equal to the work done against
resistances, subtract the total loss of energy
Calculate the total loss of energy
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 𝐾𝐸 𝑙𝑜𝑠𝑡 − 𝑃𝐸 𝑔𝑎𝑖𝑛𝑒𝑑
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 550 − (−13475)
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 14025𝐽
Sub in values
Calculate
It makes sense that these are added
together, as we have lost both Kinetic
and Potential energies!
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
Work, energy and power
𝟏
𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑲𝑬 =
A
You can use the principle of the
conservation of mechanical energy and
the work-energy principle to solve
problems involving a moving particle
If gravity is the only force acting on a particle:
𝑫𝒆𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑷𝑬 = 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 𝒊𝒏 𝑲𝑬
If another force is acting on the particle:
𝑻𝒐𝒕𝒂𝒍 𝒍𝒐𝒔𝒔 𝒐𝒇 𝒆𝒏𝒆𝒓𝒈𝒚 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
𝑾𝒐𝒓𝒌 𝒅𝒐𝒏𝒆 = 𝑲𝑬 𝒍𝒐𝒔𝒕 − 𝑷𝑬 𝒈𝒂𝒊𝒏𝒆𝒅
A skier passes a point A on a ski-run, moving
downhill at 6ms-1. After descending 50m
vertically, the run starts to ascend. When the
skier has ascended 25m to point B her speed is
4ms-1. The skier and skis have a combined mass
of 55kg. The total distance travelled from A to
B is 1400m. The resistances to motion are
constant and have a magnitude of 12N.
Calculate the work done by the skier
𝐾𝐸 𝐿𝑜𝑠𝑡 = 550𝐽
𝑃𝐸 𝐺𝑎𝑖𝑛𝑒𝑑 = −13475𝐽
𝑇𝑜𝑡𝑎𝑙 𝑙𝑜𝑠𝑠 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 = 14025𝐽
B
50m
25m
Initial speed = 6
Final speed = 4
The work done by the skier will be equal to the work done against
resistances, subtract the total loss of energy
Calculate the total work done against resistances
𝑊 = 𝐹𝑠
𝑊 = (12)(1400)
Sub in values – the resistances of
12N act over 1400m
Calculate
𝑊 = 16800𝐽
16800J of energy has been used against the resistances.
 The loss of kinetic and potential energy of 14025J has
contributed to this
 The rest will be work done by the skier
16800 − 14025 = 2775𝐽
3C
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
You can calculate the power developed
by an engine and solve problems about
moving vehicles
Power is the rate of doing work
 It is measured in Watts (W), where 1
watt = 1 joule per second
 Often an engine’s power will be given
in kilowatts (kW) where 1kW =
1000W
The power developed by an engine is given
by the following formula:
𝑷 = 𝑭𝒗
P = power (W)
F = the driving force of the engine (N)
v = velocity (ms-1)
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
You can calculate the power developed
by an engine and solve problems about
moving vehicles
A truck is being pulled up a slope at a
constant speed of 8ms-1 by a force of
magnitude 2000N acting parallel to the
direction of motion of the truck.
Calculate the power developed in
kilowatts.
𝑃 = 𝐹𝑣
Sub in values
𝑃 = (2000)(8)
Calculate
𝑃 = 16000𝑊
Change to kilowatts
𝑃 = 16𝑘𝑊
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
You can calculate the power developed
by an engine and solve problems about
moving vehicles
Draw a diagram and show forces
600N
T4000N
A car of mass 1250kg is travelling along a
horizontal road. The car’s engine is
working at 24kW. The resistance to
motion is constant and has magnitude
600N. Calculate:
v=6
a)
24000 = 𝐹(6)
The acceleration of the car when it is
travelling at 6ms-1
𝑎 = 2.72𝑚𝑠
−2
b) The maximum speed of the car
To calculate the acceleration we can use
the formula F = ma. However, we do not
know the driving force from the engine
yet.
 We can calculate the driving force
from the information given
T is often used as the
‘tractive’ force of the
engine
P = 24000W
𝑃 = 𝐹𝑣
Sub in values
Divide by 6
4000𝑁 = 𝐹
𝐹 = 𝑚𝑎
4000 − 600 = (1250)𝑎
Resolve horizontally and
sub in values
Calculate a
2.72 = 𝑎
At a velocity of 6ms-1, the acceleration is 2.72ms-2
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
You can calculate the power developed
by an engine and solve problems about
moving vehicles
A car of mass 1250kg is travelling along a
horizontal road. The car’s engine is
working at 24kW. The resistance to
motion is constant and has magnitude
600N. Calculate:
a)
The acceleration of the car when it is
travelling at 6ms-1
𝑎 = 2.72𝑚𝑠 −2
Draw a diagram and show forces
600N
600N
4000N
𝑃 = 𝐹𝑣
Sub in values
24000 = (600)𝑣
Calculate v
40 = 𝑣
So the maximum speed of the car is 40ms-1
b) The maximum speed of the car
Important points to note:
When the car is at its maximum speed,
the resultant force will be 0
 As the velocity of the car increases, the driving force falls
(it is harder for a car to accelerate more if it is already at a
high speed)
 The driving force must be 600N!
We can use this to calculate the velocity
at this point…
 This is the maximum speed for the given power level. It is
possible to increase the power in an engine (for example by
changing gear), and hence the top speed will increase
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
N
T
1600N
You can calculate the power developed
by an engine and solve problems about
moving vehicles
R
1100g 7˚
1100gCos7
A car of mass 1100kg is travelling at a
constant speed of 15ms-1 along a straight
road which is inclined at 7˚ to the
horizontal. The engine is working at a
rate of 24kW.
As the speed is constant, the driving force must be equal
to the forces opposing motion (gravity and anything else)
a)
Find the driving force first
Calculate the magnitude of the nongravitational resistances to motion
The rate of working of the engine is now
increased to 28kW. Assuming the
resistances to motion are unchanged:
𝑃 = 𝐹𝑣
24000 = 𝐹(15)
7˚
1100gSin7
Sub in values
Calculate F
𝐹 = 1600𝑁
b) Calculate the initial acceleration of the
car
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
N
1600N
You can calculate the power developed
by an engine and solve problems about
moving vehicles
A car of mass 1100kg is travelling at a
constant speed of 15ms-1 along a straight
road which is inclined at 7˚ to the
horizontal. The engine is working at a
rate of 24kW.
a)
Calculate the magnitude of the nongravitational resistances to motion
𝑅 = 286.2𝑁
The rate of working of the engine is now
increased to 28kW. Assuming the
resistances to motion are unchanged:
b) Calculate the initial acceleration of the
car
268.2N
R
7˚
1100g 7˚
1100gCos7
1100gSin7
As the speed is constant, the driving force must be equal
to the forces opposing motion (gravity and anything else)
Now you have the driving force, resolve parallel to the
plane
𝐹 = 𝑚𝑎
1600 − 𝑅 − 1100𝑔𝑆𝑖𝑛7 = 0
1600 − 1100𝑔𝑆𝑖𝑛7 = 𝑅
Sub in values. Remember
acceleration is 0
Rearrange to find R
Calculate
𝑅 = 286.2𝑁
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
N
1600N
1867N
You can calculate the power developed
by an engine and solve problems about
moving vehicles
A car of mass 1100kg is travelling at a
constant speed of 15ms-1 along a straight
road which is inclined at 7˚ to the
horizontal. The engine is working at a
rate of 24kW.
a)
Calculate the magnitude of the nongravitational resistances to motion
𝑅 = 286.2𝑁
The rate of working of the engine is now
increased to 28kW. Assuming the
resistances to motion are unchanged:
268.2N
7˚
1100g 7˚
1100gCos7
1100gSin7
As the speed is constant, the driving force must be equal
to the forces opposing motion (gravity and anything else)
For part b), we need to start again by calculating the
tractive force of the vehicle
𝑃 = 𝐹𝑣
28000 = 𝐹(15)
𝐹 = 1867𝑁
Sub in values
Calculate (remember to use
the exact value later on)
b) Calculate the initial acceleration of the
car
3D
𝑾 = 𝑭𝒔
𝑾 = 𝒎𝒈𝒉
𝟏
𝑲𝑬 = 𝒎𝒗𝟐
𝟐
𝑷𝑬 = 𝒎𝒈𝒉
𝑾=
𝟏
𝟏
𝒎𝒗𝟐 − 𝒎𝒖𝟐
𝟐
𝟐
𝑷 = 𝑭𝒗
Work, energy and power
N
1867N
You can calculate the power developed
by an engine and solve problems about
moving vehicles
A car of mass 1100kg is travelling at a
constant speed of 15ms-1 along a straight
road which is inclined at 7˚ to the
horizontal. The engine is working at a
rate of 24kW.
a)
Calculate the magnitude of the nongravitational resistances to motion
𝑅 = 286.2𝑁
The rate of working of the engine is now
increased to 28kW. Assuming the
resistances to motion are unchanged:
b) Calculate the initial acceleration of the
car
268.2N
7˚
1100g 7˚
1100gCos7
1100gSin7
As the speed is constant, the driving force must be equal
to the forces opposing motion (gravity and anything else)
Now use the F = ma formula again with the updated
information…
𝐹 = 𝑚𝑎
1867 − 268.2 − 1100𝑔𝑆𝑖𝑛7 = 1100 𝑎
Sub in values
Divide by 1100
0.242 = 𝑎
The initial acceleration will be 0.242ms-2
3D
Summary
• We have learnt how to solve problems
involving work done on or by a particle
• We have seen how to calculate Kinetic and
Potential energies, and how these link
together
• We have seen how to perform calculations
involving the power of an engine and its
driving force