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
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 = work done
F = magnitude of the force
s = the distance moved in the direction of the force
For work done against gravity:
W = work done m = mass of the object g = gravitational constant
h = the height raised
These two formulae are effectively the same!
3A
=
=
15N
A box is pulled 7m across a horizontal floor by a horizontal force of magnitude 15N. Calculate the
work done by the force
=
= 15 × 7
= 105
Sub in values from the question
Calculate
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…
3A
=
=
FN RN
0
Draw a diagram – we do not know the force or the resistance, and the acceleration is 0 (constant speed)
=
96 = × 12
= 8
Sub in values from the question
Calculate
Find the force acting on the box by using one of the formulae above…
8N
Now use F = ma, resolving horizontally
=
8 − = 0
= 8
Acceleration is 0, remember to include forces correctly
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 bricklayer raises a load of bricks of total mass 30kg at a constant speed by
attaching a cable to the bricks. Assuming the cable is vertical, calculate
the work done when the bricks are raised a distance of 7m
3A
=
=
Draw a diagram – Tension is the force in the cable. The weight can be added to the diagram as well and
acceleration
30g T
0
If we are going to calculate the work done, we need the
tension
Use F = ma and resolve
vertically
=
− 30 = 0
= 30
Calculate Rearrange 30g
=
= 30 × 7
= 2058
Sub in values (you could also have used W = mgh)
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 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 b) The work done against friction
3A
=
=
30°
2g 2gCos30 2gSin30
30°
R
FMAX
Draw a diagram and label all the forces
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° 12m
30° 12Sin30
12Cos30
Diagram of the distance moved
P
=
= 2 × 9.8 × (12 !"30)
= 118
Sub values in
Calculate
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 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 b) The work done against friction
Find the force acting against FMAX
3A
=
=
30°
2g 2gCos30 2gSin30
30°
R
FMAX
Draw a diagram and label all the forces
12m
30° 12Sin30
12Cos30
Diagram of the distance moved
P
= 118
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
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 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 b) The work done against friction
Find the force acting against FMAX
3A
=
=
30°
2g 2gCos30 2gSin30
30°
R
FMAX
Draw a diagram and label all the forces
12m
30° 12Sin30
12Cos30
Diagram of the distance moved
P
= 118
$%&
= '
$%&
= 0.35 × (2()30)
$%&
= 0.7()30
The normal reaction will just be 2gCos30 as there is no acceleration perpendicular to the plane
Sub in values
Simplify (to ensure it stays exact)
0.7gCos30
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 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 b) The work done against friction
In these types of questions, the work done against friction and the work done against gravity give the total work
done…
3A
=
=
30°
2g 2gCos30 2gSin30
30°
Draw a diagram and label all the forces
12m
30° 12Sin30
12Cos30
Diagram of the distance moved
P
= 118
=
Now resolve parallel to the plane to find force P
Sub in values, acceleration is 0. remember to include the gravitational part (for now…)
Work out P and leave as an exact answer
0.7gCos30
2gCos30
* − 0.7()30 − 2 !"30 = 0 * = 0.7()30 + 2 !"30
0.7gCos30 + 2gSin30
=
= (0.7()30 + 2 !"30) × 12
= 188.9
(!", ./!0,!)") = 70.9
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
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 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
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
=
=
25°
27N
27Sin25 27Cos25
=
= (27()25) × 15
= 367
Sub in values
Calculate
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
m is the mass of the particle v is its velocity
Potential Energy
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)
3B
=
=
12 =
1
2
3
4
*2 =
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
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
To show this, we will rewrite one of the SUVAT equations to give it in
terms of a.
3B
=
=
56 =
7
8
9
8
:6 =
=
3
4= ;
4+ 2
3
4− ;
4= 2
3
4− ;
42
=
=
3
4
− ;
42
=
(3
4
− ;
4)
2
=
(3
4
− ;
4)
2
=
1
2
3
4
−
1
2
;
4
Subtract u2
Divide by 2s
Replace a with the expression we worked
out
Multiply top by m
Multiply all by s
Rewrite right side
=
1
2
3
4
−
1
2
;
4
Fs = work done
Final KE - Initial KE
=
7
8
9
8
−
7
8
<
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.
3B
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
12 =
1
2
3
4
12 =
1
2
(0.3)(9)
4
12 = 12.15
Sub in values
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
Use the formula for the change
in kinetic energy!
3B
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
=
1
2
3
4
−
1
2
;
4
1.8 =
1
2
(1.5) 3
4
−
1
2
(1.5);
4
1.8 = 6.75 − 0.75;
40.75;
4= 4.95
;
4= 6.6
; = 2.57
>?Sub in values
Work out parts
Rearrange
Divide by 0.75
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 van’s speed is 12ms-1. A constant
resistance of 500N acts on the van. 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
3B
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
=1 23
4−1
2;
4
P 500N
=1 23
4−1
2;
4
* − 500 × 400 =1
2(2000) 12
4−1
2(2000)(0)
4
400 * − 500 = 144000
* − 500 = 360
* = 860
Replace W with Fs
Sub in values
Calculate parts
Divide by 400
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
3B
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
*2 =
*2 = (30)()(−15)
*2 = −4410
Sub in values. The height has fallen by 15m…
Calculate
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.
3B
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
7m
"@ = 3
4 !"@ = 3
5 ()@ =
4 5
Draw a diagram
The change in
potential energy will be affected by the change in the vertical height of
the parcel 7Sinθ
7Cosθ θ
*2 =
*2 = (3)(9.8)(7 !"@)
*2 = (3)(9.8) 7 ×
3
5
*2 = 123.48
Sub in values, using the change in height
Also use the value of Sinθ
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)
3C
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
If gravity is the only force acting on a particle:
ABCDBEB FG :6 = HGCDBEB FG 56
If another force (usually friction) is acting on
the particle:
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
30°
Initial speed = 0 Final speed = 6
Find the increase in Kinetic energy
=1 23
4−1
2;
4
=1
2(0.5) 6
4−1
2(0.5)(0)
4
= 9
Sub in values
Calculate
Q"0/RR !" 12 = 9
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
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
30°
30°
x
xCos30
xSin30
Initial speed = 0 Final speed = 6
Q"0/RR !" 12 = 9
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
B
Find the decrease in Potential energy (find the change in vertical height first)
Call the diagonal distance (the one we need to find) ‘x’
*2 =
*2 = (0.5)(9.8)(S !"30)
*2 = 2.45S
Sub in values
Calculate in terms of x
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
30°
30°
x
xCos30
xSin30
Initial speed = 0 Final speed = 6
Q"0/RR !" 12 = 9
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
B
TR0/RR !" *2 = 2.45S
TR0/RR !" *2 = Q"0/RR !" 12
2.45S = 9
S = 3.67
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
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
45°
Initial speed = 8 Final speed = 0
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
=1 23
4−1
2;
4
= 1 2(2) 0
4−1
2(2)(8)
4
= −64
1!"R,!0 2"R/U V), = 64
Sub in values
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8:6 =
=
7
8
9
8
−
7
8
<
8
45°
Initial speed = 8 Final speed = 0
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…
1!"R,!0 2"R/U V), = 64
45° x xCos45 x S in 4 5 *2 =
*2 = (2)(9.8)(S !"45)
*2 = 9.8 2S
*),R",!V 2"R/U !"RW = 9.8 2S
Sub in values
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
45°
Initial speed = 8 Final speed = 0
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
1!"R,!0 2"R/U V), = 64
*),R",!V 2"R/U !"RW = 9.8 2S
),V V) ). R"R/U = 12 V), − *2 !"RW
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
),V V) ). R"R/U = 64 − 9.8 2S
Sub in values to find an expression for the
loss of energy
),V V) ). R"R/U = 64 − 9.8 2S
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8:6 =
=
7
8
9
8
−
7
8
<
8
45°
Initial speed = 8 Final speed = 0
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 45° 2g 2gCos45 2gSin45 R 2gCos45 FMAX
1!"R,!0 2"R/U V), = 64
*),R",!V 2"R/U !"RW = 9.8 2S ),V V) ). R"R/U = 64 − 9.8 2S
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
$%& = '
$%& = (0.4)(2()45)
$%& = 7.84()45
Sub in values
Rewrite x
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8:6 =
=
7
8
9
8
−
7
8
<
8
45°
Initial speed = 8 Final speed = 0
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 45° 2g 2gCos45 2gSin45 2gCos45
1!"R,!0 2"R/U V), = 64
*),R",!V 2"R/U !"RW = 9.8 2S ),V V) ). R"R/U = 64 − 9.8 2S
x
7.84Cos45
=
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’
= (7.84()45)(S)
= 7.84S()45
Sub in F and s
Rewrite in terms of x
This is the work done against friction
This is the energy lost
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:
If another force is acting on the particle:
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.
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8:6 =
=
7
8
9
8
−
7
8
<
8
45°
Initial speed = 8 Final speed = 0
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 45° 2g 2gCos45 2gSin45 2gCos45
1!"R,!0 2"R/U V), = 64
*),R",!V 2"R/U !"RW = 9.8 2S ),V V) ). R"R/U = 64 − 9.8 2S
x
7.84Cos45
= 7.84S()45
64 − 9.8 2S = 7.84S()45
64 = 7.84S()45 + 9.8 2S
64 = 7.84()45 + 9.8 2 S
64
7.84()45 + 9.8 2= S
Add 9.8√2x
Factorise right side
Divide by the bracket
3.3 = S
Calculate
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:
If another force is acting on the particle:
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
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
A
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)
=1 23
4−1
2;
4
=1
2(55) 4
4−1
2(55)(6)
4
= −550
12 X), = 550
Sub in values
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:
If another force is acting on the particle:
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
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
A
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!)
12 X), = 550
*2 =
*2 = (55)(9.8)(−25)
*2 = −13475
Sub in values
Calculate
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:
If another force is acting on the particle:
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
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
A
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
12 X), = 550 *2 Y!"RW = −13475
),V V) ). R"R/U = 12 V), − *2 !"RW
),V V) ). R"R/U = 550 − (−13475)
),V V) ). R"R/U = 14025
Sub in values
Calculate
It makes sense that these are added together, as we have lost both Kinetic
and Potential energies!
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:
If another force is acting on the particle:
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
3C
ABCDBEB FG :6 = HGCDBEB FG 56
IJKEL LJ JM BGBDN = 56 LJK − :6 EFGBO JDP OJGB = 56 LJK − :6 EFGBO
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
A
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
12 X), = 550 *2 Y!"RW = −13475 ),V V) ). R"R/U = 14025
Calculate the total work done against resistances
=
= (12)(1400)
= 16800
Sub in values – the resistances of 12N act over 1400m
Calculate
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
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 = power (W)
F = the driving force of the engine (N) v = velocity (ms-1)
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
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.
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
* = 3
* = (2000)(8)
* = 16000
* = 16Z
Sub in values
Calculate
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
b) The maximum speed of the car
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
Draw a diagram and show forces
T
600N
v = 6
P = 24000W
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
* = 3
24000 = (6)
4000 =
Sub in values
Divide by 6
T is often used as the ‘tractive’ force of the
engine
4000N
=
4000 − 600 = (1250)
2.72 =
Resolve horizontally and sub in values
Calculate a
At a velocity of 6ms-1, the acceleration is 2.72ms-2
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
b) The maximum speed of the car
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
Draw a diagram and show forces
600N
When the car is at its maximum speed, the resultant force will be 0
The driving force must be 600N!
We can use this to calculate the velocity at this point…
4000N
= 2.72
>4* = 3
600N
24000 = (600)3
40 = 3
Sub in values
Calculate v
So the maximum speed of the car is 40ms-1 Important points to note:
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)
This is the maximum speed for the given power level. It is
Work, energy and power
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 non-gravitational resistances to motion 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
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
7˚
7˚ 1100g
1100gCos7 1100gSin7 R
T N
As the speed is constant, the driving force must be equal to the forces opposing motion (gravity and anything else) Find the driving force first
* = 3
24000 = (15)
= 1600
Sub in values
Calculate F
Work, energy and power
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 non-gravitational resistances to motion 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
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
7˚
7˚ 1100g
1100gCos7 1100gSin7 R
N
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
1600N
=
1600 − − 1100 !"7 = 0
1600 − 1100 !"7 =
= 286.2
Sub in values. Remember acceleration is 0
Rearrange to find R
Calculate 268.2N
Work, energy and power
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 non-gravitational resistances to motion 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
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
7˚
7˚ 1100g
1100gCos7
1100gSin7
N
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
1600N
268.2N
= 286.2
* = 3 28000 = (15)
= 1867
Sub in values
Calculate (remember to use the exact value later on)
Work, energy and power
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 non-gravitational resistances to motion 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
3D
=
=
56 =
7
8
9
8
:6 =
=
7
8
9
8
−
7
8
<
8
: = 9
7˚
7˚ 1100g
1100gCos7
1100gSin7
N
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…
268.2N
= 286.2
1867N
=
1867 − 268.2 − 1100 !"7 = 1100
0.242 =
Sub in values
Divide by 1100
