## 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

F_{MAX}

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

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 F_{MAX}

### 3A

### =

### =

30°

2g 2gCos30 2gSin30

30°

R

F_{MAX}

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

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 F_{MAX}

### 3A

### =

### =

30°

2g 2gCos30 2gSin30

30°

R

F_{MAX}

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

A package of mass 2kg is pulled at a constant speed up a rough plane which

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

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### − ;

4### 2

### =

### =

### 3

4

_{− ;}

4
### 2

### =

### (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;

4### 0.75;

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
F_{MAX}

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 }

_{. 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

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.

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