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Developed by: G. Izquierdo Rodríguez & G Izquierdo Gómez Page 0

PHYSICAL SCIENCES GRADE 12 PHYSICS

NORTHERN CAPE DEPARTMENT OF EDUCATION

TERM II

WORK ENERGY AND POWER By

G. Izquierdo Rodríguez

&

G. Izquierdo Gómez

2020

Developed by: G. Izquierdo Rodríguez & G Izquierdo Gómez Page 1 Copyright reserved

Table of Contents

Lesson 1 - : Definition of work. Net work. Alternate method for determining the net work.

2

Lesson 2 - : Work-energy theorem 11

Lesson 3 -: Revision exercises. 17

Lesson 4 - : Conservative and non-conservative forces. Law of Conservation of

Mechanical Energy.Conservation of energy with non conservative forces present. 24

Lesson 5- : Revision exercises. 33

Lesson 6 -: Definition of power. Calculating power. Minimum and maximun power. 43

Lesson 7 - : Revision exercises on work-energy and power. 51

Lesson 8 -: Revision exercise 59

Lesson 9 - : Self-assessment on law of conservation of mechanical energy. 71


PHYSICS (MECHANICS)

WORK, ENERGY & POWER

LESSON 1: Definition of work. Net work. Alternate method for determining the network.

Objective:

Learners must be able to:

Define the work done on an object by a constant force 𝐹 as 𝐹∆𝑥 cos 𝜃, where F is

the magnitude of the force, ∆𝑥 the magnitude of the displacement and 𝜃 the angle between the force and the displacement.

Draw a force diagram and a free-body diagram.

Calculate the net work done on an object.

Distinguish between positive net work done and negative net work done on the system.

Introduction:

Almost all the terms we have used so far –velocity, acceleration, force and so on – convey nearly the same meaning in Physics as they do in everyday life. Now, however, we encounter a term whose meaning in Physics is distinctly deferent from its everyday meaning. That term is work. You Tube video: “What is Work: Physics for Kids. https://www.tes.co.uk/teaching-resource/what-is-work-physics-for-kids-6212881

Development:

Let us examine the situation in Figure below, where an object undergoes a displacement Δx along a straight line while acted on by a constant force F that makes an angle θ with direction of the displacement.

From the example we can conclude that work is done by a constant force acting on an object when the point of application of that force moves through some distance and the force has a component along the line of motion.

We know that W= FxΔx

But Fx = Fcos θ

Then

https://www.tes.co.uk/teaching-resource/what-is-work-physics-for-kids-6212881

https://www.tes.co.uk/teaching-resource/what-is-work-physics-for-kids-6212881


Fg

N

W = F Δx cos θ

Work done by a constant force is defined as a scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement.

W = FΔx Cosθ

It is important to note that there are TWO restrictions to use this equation.

First: The force must be constant.

Second: The object must be a particle-like. This means that the object must be rigid, all

parts must move together in the same direction.

In order for work to be done, an object must have its position changed by an amount 𝚫�⃗�

while a force, �⃗⃗� , is acting on it such that there is some non-zero component of the force in the direction of the displacement.

From this equation we can get the following conclusions:

Force and the displacement are in the same direction (figure below)

W = F Δx Cos θ if θ= 00

Then Cos 00 = 1

and so W=FΔx

The force is perpendicular to direction of the displacement (figure below).

When θ = 900

Cos 900 = 0

Then W = F Δx cos 900= F Δx (0)= 0


Force and the displacement are in opposite direction (figure below)

When θ = 1800

Cos 1800 = -1

Then W = ff Δx Cos1800= ff Δx (-1)

W = - ff Δx

SI Units

o Force F [N] o Displacement Δx [m] o Work W [J]

1J = 1 kg·m·s-2 =1N·m

EXAMPLE 1

A man cleaning a frictionless floor pulls a vacuum cleaner with a force of magnitude F = 50.0 N at an angle of 30.0° with the horizontal (Fig. below). Calculate the work done by the force on the vacuum cleaner as the vacuum cleaner is displaced 3.00 m to the right.

�⃗⃗� 𝒇 �⃗⃗� 𝒇

∆�⃗⃗�


Solution:

Step 1: Identify all the forces acting on the object (Applied force, normal force and force of gravity). Step 2:We must draw a free body diagram.

From the free body diagram we can see that the normal force and the gravitational force do not do work because they are acting perpendicularly (900) to the direction of the displacement.

The only force doing work is the applied force F.

Step 3: Using the general equation

W = F Δx Cos θ W= (50)(3,0) Cos 300

W= 129,9 J

In the case that there are more than one force acting the work done by all the forces can be calculated by the algebraic addition of all the works done by each force.

WR = W1 + W2 + ... + Wn

Also the work done by the resultant (net) force can be determined by multiplying the resultant force (vector addition of all the forces) by the displacement of the body (point of application of the force) by the cosine of the angle between the force and the direction of the displacement.

WR = FR Δx cos θ OR Wnet = Fnet Δx cos


f

F1 = 5 N

F2 = 4 N

FN = 2 N

Fr = 3 N

450

EXAMPLE 2 The sketch below shows a wooden block on a horizontal surface and 5 forces acting on it. Calculate the work done by each force and the work done by the net force when the block has covered 3 m.

Solution:

W1 = F1Δx Cos 00

W1 = (5)(3)(1) W1 = 15 J

W2 = F2Δx Cos 450 W2 = (4)(3 Cos 450 W2 = 8,49 J

WN = FNΔx Cos 900

WN = (2)(3)(0) WN = 0

Wf= FfΔx Cos180⁰

Wf= (3)(3)(-1) Wf = - 9 J

Wg = FgΔx Cos 900

Wg= (2)(3)(0) Wg= 0

Wnet= W1 + W2 + WN +Wr + Wg

Wnet = 15 J + 8,49 J + 0 – 9 + 0 Wnet= 14,49 J

EXAMPLE 3: A block of mass 2.00 kg is being pulled up a frictionless inclined plane by a force, F = 20.0 N, applied through a rope parallel to the plane. The plane makes an angle of 30.00 with the horizontal. The block is pulled a distance of 1.80 m along the plane. Calculate the work done by the net force on the block.


Solution: We must draw the free body diagram and a coordinate system if there are forces acting not in the direction of the axis like gravitational force then we have to determine its components.

Option 1:

In the Y direction the net force is zero because there is not acceleration

Wnet = FnetΔx Cos θ

Wnet= (F- Fgx) Δx Cos θ

Wnet= (F- Fg Sin 300) Δx Cosθ

Wnet= (F- mg Sin 300) Δx Cos 00

Wnet= (20 –[(2)(9,8)(0,5)] x 1,8x 1

Fnet = 10,2 x 1,8 = 18,36 J

Option 2

Wnet= WF + WFg + WN

Wnet =F Δx Cos θ + FFgΔx Cos θ + FNΔx Cos θ

Wnet = F Δx Cos 00 + mgΔx cos 1200 + FNΔx Cos 900

Wnet =20 x 1,8 x Cos 00 + 2 x 9,8 x1,8 x (-0,5) + 0= 36 – 17,64

Wnet = 18,36 J

θ θ

Y

X

The gravitational force is not in the direction of

any axis then we have to determine its

components on the X-axis and Y-axis.


Work is energy transferred to or from by an object by means of a force acting on the object. Energy transferred to the object is positive work and energy transferred from the object is negative work.

The work done by a force can then be positive or negative. This sign tells us about the direction of the energy transfer. Work is a scalar so the sign should not be misinterpreted to mean that work is a vector. Work is defined as energy transfer, energy is a scalar quantity and the sign indicates whether energy was increased or decreased.

Positive net work done on a system will increase the energy of the system and negative work done on the system will decrease the energy of the system.

EXAMPLE 4 The following graph shows a constant force that acts on an object that covers a distance of 4 m under the action of this force. Calculate the work done by the force as the particle moves from x =0 to x = 4.0 m.

SUMMARY:

Work done by a constant force is defined as a scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement. W = FΔx cosθ

In order for work to be done, an object must have its position changed by an

amount Δx⃗ while a force, 𝐹 , is acting on it such that there is some non-zero component of the force in the direction of the displacement.

Restrictions to use the definition of work.

First: The force must be constant. Second: The object must be a particle-like. This means that the object must be

rigid, all parts must move together in the same direction.

The force acting on a particle is

constant for the first 4.0 m of motion.

The net work done by this force is the

area under the graph (line/curve).

This area is equal to the area of the

rectangular section from A to B.

The area of the rectangle is

(4.0)(5.0) = 20 J


REVISION

QUESTION 1

A man uses a rope to pull a box on a frictionless surface through a distance of 8 m. if the angle that the rope makes with the horizontal is 300, and the force exerted on the rope is 60 N.

1.1. Draw the free body diagram of all the forces acting on the box. 1.2. Calculate the work done on the box by the force apply by the man. 1.3. Does the gravitational force do work in this case? Explain your answer.

QUESTION 2

A block of mass 6.00 kg is being pulled up an inclined plane by a force, F = 40.0 N, applied through a rope parallel to the plane. The plane makes an angle of 30.00 with the horizontal. The frictional force between the block and the plane is 5 N. The block is pulled a distance of 5,0 m along the plane.

2.1. Calculate the work done by the force, F, on the block. 2.2. Calculate the work done by the force of gravity on the block. 2.3. Calculate the work done by the force of kinetic friction on the block 2.4. Calculate the work done by the net force on the block.


SOLUTIONS

QUESTION 1

1.1.

1.2 W = F Δx Cos θ W = (60)(8) x cos 300

W = 415,68 J 1.3 In this case the gravitational force does not do work because θ = 900; Cos 90° = 0, then W= 0 QUESTION 2

2.1 WF = F Δx Cosθ = 40 x 5 x Cos 00 = 200 J

2.2 Wg = FgxΔx Cosθ =(mgsin300) Δx Cos 1800= 6 x 9,8 x 0,5 x 5 x(-1)= - 147 J

OR

Wg = FGΔx Cosθ = mg Δx Cos 1200= 6 x 9,8 x 5 x (-0,5) = - 147 J

2.3 WFr = FfrΔx Cosθ = 5 x 5 Cos 1800= 25 x (-1) = - 25 J

2.4 Wnet= WF + WG+ WFr+WN

Wnet= 200 -147 – 25 = 28 J

θ θ

Y

X



LESSON 2: Work-energy theorem

Objective:


State the work-energy theorem Apply the work-energy theorem to objects on horizontal, vertical and inclined planes

(for both frictionless and rough surfaces).

Introduction:

Summary:

Work done by a constant force is defined as a scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement. W = FΔxcosθ


amount Δx⃗ while a force, �⃗⃗� , is acting on it such that there is some non-zero component of the force in the direction of the displacement.

Restrictions to use the definition of work. First: The force must be constant. Second: The object must be a particle-like. This means that the object must be rigid, all parts must move together in the same direction.

- From grade 11 we know Newton’s second law of motion.

If a resultant (net) force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the

body. The mathematical representation is: amF net

It can be difficult to use Newton’s second law to solve motion problems involving complex forces. An alternative approach is to relate the speed of a moving particle to its displacement under the influence of some net force. If the work done by the net force on a particle can be calculated for a given displacement, then the change in the particle’s speed can be easily evaluated.


Development:

Figure on the left shows a particle of mass m moving to the right under the

action of a constant net force 𝐹 𝑛𝑒𝑡. Because the force is constant, we know from Newton’s second law that the particle moves with a constant

acceleration 𝑎 . If the particle is displaced a distance Δ𝑥 , the net work done by the net force is:

o

netnet xFW 0cos

Then

xmaxFW netnet

From grade 10 we know that the following relationships are valid when a particle

(objects) undergoes constant acceleration.

t

vva

if

And t

vvx

fi

2

Where 𝒗i is the speed at t = 0 and vf is the speed at time Δt. Substituting these

expressions in the equation into the equation of work done by the net force (Wnet= maΔx)

gives:

tvv

t

vvmW

fiif

net

2

Therefore

22

22

if

net

mvmvW

The quantity 2

2

imv represents the kinetic energy associated with the motion of the

particle. The net work done on a particle by a constant net force (Fnet) acting on it equals the change in kinetic energy of the particle.

This is known as the work-energy theorem (work-kinetic energy theorem).

The net/total work done on an object is equal to the change in the object's kinetic energy. OR The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy.

In symbols: 𝑊𝑛𝑒𝑡 = ∆𝐸𝐾 = 𝐸𝐾𝑓 − 𝐸𝐾𝑖


However, the work energy theorem cannot be applied in any situation, it has limitations:

It is only applied for particles. It can be applied to real objects when they can be considered particles.

It is important to note that when we use this theorem, we must include all of the forces that do work on the particle in the calculation of the net work done. From this theorem, we see that the speed of a particle increases if the net work done on it is positive because the final kinetic energy is greater than the initial kinetic energy. The particle’s speed decreases if the net work done is negative because the final kinetic energy is less than the initial kinetic energy.

EXAMPLE 1

A 6.0 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.

Solution

We have made a drawing of this situation

Knet EW

KiKfnet EEW

220cos

22

0 if

net

mvmvxF

2

06

2

61312

22

fv

2

636

2

fv

2672 fv


6

722 fv

46,312 fv m∙s-1

Work-energy theorem with the impulse- momentum theorem.

𝑾𝒏𝒆𝒕 = ∆𝑬𝑲 (Work-energy theorem)

𝑰𝒎𝒑𝒖𝒍𝒔𝒆 = ∆𝑷 (Impulse momentum theorem)

I both cases, a force acting on a particle changes the state of the system. If the force acts

over a time interval from ti to tf, it creates an impulse that changes the particle’s

momentum. If the force acts over the spatial interval from ∆𝑥𝑖 to ∆𝑥𝑓, it does work that

changes the particle’s kinetic energy

A force acting on a particle both creates an impulse and does work, changing both the

momentum and the kinetic energy.

We can in fact, express the kinetic energy in terms of momentum as

𝐸𝐾 =1

2𝑚𝑣2 =

(𝑚𝑣)2

2𝑚=

𝑝2

2𝑚

SUMMARY.

Work done by a constant force is defined as a scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement. W = FΔxcosθ


amount 𝚫�⃗� while a force, �⃗⃗� , is acting on it such that there is some non-zero component of the force in the direction of the displacement.

The work-energy theorem (work-kinetic energy theorem) in words: The net/total work done on an object is equal to the change in the object's kinetic energy. OR The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy.

Work-energy equation:

𝑊𝑛𝑒𝑡 = ∆𝐸𝐾

𝑊𝑛𝑒𝑡 = 𝐸𝐾𝑓 − 𝐸𝐾𝑖

22

22

if

net

mvmvW

𝐹∆𝑥 cos 𝜃 =1

2𝑚𝑣𝑓

2 −1

2𝑚𝑣𝑖

2

- The teacher must orientate the homework.


REVISON

Activity 1 Multiple-choice questions

Four possible options are provided as answers. Choose the correct option by writing only the letter next to the question number.

1.1 An object moving at a constant velocity v has a kinetic energy EK. The velocity is changed to 2v. Which ONE of the following is the correct kinetic energy at this velocity?

A ¼ EK

B ½ EK

C 2 Ek

D 4 EK

1.2 The engine of a car does work (W) when the car is accelerating from 0 to v. The work done by the engine when it increases the velocity from v to 3 v, is:

A 9 W B 8 W C 3 W D W Activity 2

A 3 kg block slides at a constant velocity of 7 m∙s-1 along a horizontal surface. It then strikes a rough surface, causing it to experience a constant frictional force of 30 N. The block slides 2 m under the influence of the frictional force before it moves up a frictionless ramp as shown in the diagram below. Use the work –energy theorem to calculate the speed of the block at the bottom of the ramp.

Note: In many Physics books the work-energy theorem is called: Work-kinetic energy theorem.

3 kg

7 m.s-1

2m


Solution of the activities:

Activity 1

1.1. D 1.2. B When velocity increases from 0 to v Wnet = ΔEK

Wnet1 = EK- EKo

2

0

2

22

mmvWnet

Wnet = 2

2mv

Activity 2

Applying the work-energy theorem

knet EW

kifkr EExF 0180cos.

22)1(.

22

if

r

mvmvxF

2

)49(3

2

.3.60

2

v

5,732

.3.60

2

v

2

.3.5,7360

2v

23.25,13 vx

2327 v 29 v139 smv

When velocity increases from v to 3v Wnet = ΔEK

Wnet = EK- EKo

Wnet = (2

)3( 2vm-

2

()2m)

28

21

29

22

9 22222 mvmvmvmvmvWnet

Wnet2 = 8 Wnet1

2

)7(3

2

.3.230

22

v

x



LESSON 3: Revision exercises.

Objective:


Calculate the net work done on an object.

Apply the work-energy theorem to objects on horizontal, vertical and inclined planes (for both frictionless and rough surfaces)

Introduction:

- SUMMARY

Physical quantity Work and kinetic energy Theorem

Work Net work

Kinetic energy and potential energy

“The work done by the resultant force acting on a particle is equal to the change in the kinetic energy of the particle.”

𝑊𝑛𝑒𝑡 = ∆𝐸𝑘 𝑊𝑛𝑒𝑡 = 𝐸𝐾𝑓 − 𝐸𝐾𝑖

𝐹∆𝑥 cos 𝜃 =1

2𝑚𝑣𝑓

2

−1

2𝑚𝑣𝑖

2

𝑾 = 𝑭∆𝒙𝐜𝐨𝐬 𝜽

SI unit:

Joule [J]

𝑊𝑛𝑒𝑡 = 𝐹𝑛𝑒𝑡∆𝒙 cos 𝜽

OR

𝑊𝑛𝑒𝑡 = 𝑊1 + 𝑊2

+ ⋯

+ 𝑊𝑛

𝐸𝐾

=1

2𝑚𝑣2

𝐸𝑝 = 𝑚𝑔ℎ

SI unit:

Joule [J]


Development

REVIONS EXERCISES

QUESTION 1

A block of 85 kg stars moving from rest under the action of a force of magnitude 240 N that forms an angle of 200 with the horizontal. The coefficient of kinetic friction is 0,2 and the block moves a distance of 8 m.

1.1 Calculate the work done by the pulling force. 1.2 Calculate the work done by the kinetic frictional force. 1.3 Calculate the net work done on the block by all the forces. 1.4 State the work-energy theorem in words. 1.5 Use the WORK ENERGY THEOREM to calculate the speed of the block when it has travelled 8 m. 1.6 State Newton’s second law of motion in words. 1.5 Applying Newton’s second law of motion to calculate the magnitude of the acceleration of the block.

QUESTION 2

A crate of mass 30 kg is placed at point A on a rough inclined plane that makes an angle 250 with the horizontal. The crate slides down from rest and travels a distance 'd' before it reaches the bottom (point B), with a speed of 2 m·s-1. The crate experiences a constant frictional force of 100 N while sliding down from point A.

2.1 Draw a free-body diagram to show ALL the forces acting on the block while it moves from the position A to B. 2.2 State the WORK-ENERGY THEOREM in words. 2.3 Use the WORK-ENERGY THEOREM to find the distance ‘d’ the crate travelled from point A to point B.


QUESTION 3

A 6 kg block slides with a constant velocity of 4 m·s-1 on a horizontal surface. It hits a rough surface, causing a constant force of friction of magnitude 15 N. The block slides 2 m under the influence of this force of friction, until it moves up in Incline plane which makes an angle of 30º with the horizontal, as indicated on the diagram below.

The block moves a distance, Δx on the inclined plane before it comes to rest.

3.1 Draw a free body diagram to indicate all the forces acting on the block while moving on the frictionless inclined plane.

3.2 Show that the speed of the block at the start of the incline is 2,45 m·s-1.

3.3 Determine the distance (Δx) the block will move up the slope before coming to

stop.


SOLUTIONS:

QUESTION 1

Free body diagram:

1.1

WF= FΔxcos θ

WF= 240 x 8 xcos200

WF= + 1,8 x 103 J

1.2

0 YF

N+ Fy – Fg= 0

N = Fg - Fy

N= mg – F sin 20⁰

N= (85 x 9,8) – (240 x 0,342)

N= 750,92 N

Ff = μN

Ff = (0,2)(750,92)

Ff = 150,184 N

Wf = FfΔxcos 1800

Wf = 150,184 x 8 x (-1)

Wf = -1,20 x 103 J

1.3

Wnet= WF+ Wf + WN+ Wg

Wnet= 1,8 x 103 - 1,20 x 103

Wnet= 6 x 102 J

1.4 The net/total work done on an object is equal to the change in the object's kinetic energy. OR


The work done on an object by a resultant/net force is equal to the change in the object's kinetic energy. OR The work done by the resultant force acting on a particle is equal to the change in the kinetic energy of the particle.

1.5

22

22

if

net

mvmvW

2

)0(85

2

85106

22

2 fv

8,3fv m∙s-1

1.6 The resultant/net force acting on an object is equal to the rate of change of momentum of the object in the direction of the resultant/net force.

1.7

amF

In x direction

amfF fx

amfF f cos

a85184,15020cos240 0

a85184,15053,225

a = 0,89 m∙s-2

QUESTION 2

2.1.

2.2. The net work done on an object is equal to the change in the object's kinetic energy.


2.3.

Option 1

Wnet = ΔEK

WG + WN + WFr = 22

2

1

2

1if mvmv

mgΔxcosθ + 0 + FFrΔxcosθ= )(2

1 22

if vvm

(30)(9,8)(Δx) cos(650)+ (100)(Δx) cos(1800) = )02()30(2

1 22

124,25Δx - 100 Δx = 60

Δx = 2,47 m

Option 2

Wnet = ΔEK

WG‖ + WN + WFr= )2

1

2

1 22

if mvmv

mgsinαΔx cosθ+0+ FFrΔxcosθ= )(2

1 22

if vvm

(30)(9,8)sin(250)(Δx) cos00 + (100)(Δx) cos1800= )02()30(2

1 22

Δx= 2,47 m

QUESTION 3

3.1.

3

.

2

Fnet = ma OR wnet = ΔEk

-15 = (6)a FΔxcosθ = ½mvf2 - ½mvi

2

a = -2,5 m.s-2 (15)(2)(cos 180º) = ½(6)vf2 -

½(6)(4)2

vf2 = vi

2 + 2aΔx -30 = 3vf2 - 48

= (4)2 + 2(-2,5)(2) vf2 = 18

vf = 2,45 m·s-1 vf = 2,45 m·s-1

Fg

N


3

.

3

Wnet = ΔEk

F//Δxcosθ = ½mvf2 - ½mvi

2

(6)(9,8)(sin30º) (Δx)cos180º = 0 - ½(6)(2,45)2

-29,4 s = 0 – 18,0075

Δx = 0,6125 m

OR

Wg = ΔEk

FgΔxcosθ = Ekf - Eki

mgΔxcos120º = ½mvf2 - ½mvi

2

gΔx(-0.5) = 0 - ½vi2

Δx = vi2/ 2g(0,5)

= (2,45)2

2(9,8)(0,5)

= 0,61 m



LESSON 4: Conservative and non-conservative forces. Law of Conservation of Mechanical Energy. Conservation of energy with non-conservative forces present.

Objective:


Define a conservative force.

Define a non-conservative force.

State the principle of conservation of mechanical energy.

Solve conservation of energy problems using the equation: Wnc

= ΔEk + ΔE

p

Use the relationship above to show that in the absence of non-conservative forces, mechanical energy is conserved.

Introduction:

In Grade 10, you saw that mechanical energy was conserved in the absence of friction (non-conservative) forces. It is important to know whether a force is a conservative force or a non-conservative force in the system, because this is related to whether the force can change an object’s total mechanical energy when it does work on an object.

When the only forces doing work are conservative forces (for example, gravitational and spring forces), energy changes forms - from kinetic to potential (or vice versa); yet the total amount of mechanical energy (EK + EP) is conserved. For example, as an object falls in a gravitational field from a high elevation to a lower elevation, some of the object’s potential energy is changed into kinetic energy. However, the sum of the kinetic and potential energies remain constant.

Development:

A force is anything that can cause a change to objects.

Some of these changes are:

change the shape of an object, accelerate or stop an object, and change the direction of a moving object.

Force is the push or pull required to change the state of motion of an object, as defined by Newton’s second law it is a vector quantity.

There are two categories of forces we will consider, conservative and non-conservative.

YouTube video (conservative force): http://youtube.com/watch?v=iba4gUeQN0w

Conservative force is a force for which the work done in moving an object between two points is independent of the path taken.

http://youtube.com/watch?v=iba4gUeQN0w


Examples are:

Gravitational force The elastic force in a spring The electrostatic forces (coulomb forces)

A conservative force results in stored or potential energy and we can define a potential energy (Ep ) for any conservative force.

The gravitational force is a conservative force and we studied gravitational potential energy in Grade10.

The total work done by a conservative force results in a change in potential energy ∆𝐸𝑝. If the conservative force does positive work then the change in potential energyis negative. Therefore:

𝑊𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = −∆𝐸𝑝

YouTube video (non-conservative force): http://youtube.com/watch?v=TXV4HynolaY

All forces, which their work depends only on the final and initial position but do not depend on the path are called conservatives forces. For a conservative force the work done in a closed trajectory is zero.

A non-conservative force is a force for which the work done in moving an object between two points depends on the path taken.

Examples are:

frictional force air resistance tension in a chord

Friction is a good example of a non-conservative force because if removes energy from the system so the amount of mechanical energy is not conserved. Non-conservative forces can also do positive work thereby increasing the total mechanical energy of the system.

The energy transferred to overcome friction depends on the distance covered and is converted to thermal energy which can’t be recovered by the system.

It is very important to know that non-conservative forces do not imply that total energy is not conserved. Total energy is always conserved. Non-conservative forces mean that mechanical energy isn’t conserved in a particular system which implies that the energy has been transferred in a process that isn’t reversible.

You tube video (conservation of mechanical energy): http://youtube.com/watch?v=Eeqg8xukCEo

http://youtube.com/watch?v=TXV4HynolaY

http://youtube.com/watch?v=Eeqg8xukCEo


The principle of conservation of mechanical energy state that: The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant.

Remember that a system is isolated when the resultant/net external force acting on the system is zero.

Lest say that a suitcase on the cupboard falls. Consider the mechanical energy of the suitcase at the top and at the bottom. We can say:

𝐸𝑀1 = 𝐸𝑀2 𝐸𝐾1 + 𝐸𝑃1 = 𝐸𝐾2 + 𝐸𝑃2

1

2𝑚𝑣1

2 + 𝑚𝑔ℎ1 =1

2𝑚𝑣2

2 + 𝑚𝑔ℎ2

- The teacher must explain how to solve conservation of energy problems using the equation: W

nc= ΔE

k + ΔE

p

We know that the net work done, will be the sum of the work done by all of the individual forces (conservative forces and non-conservative forces):

𝐹𝑛𝑒𝑡 = 𝐹𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 + 𝐹𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 (the work done by these forces yields)

𝑊𝑛𝑒𝑡 = 𝑊𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 + 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒

(using the work-energy theorem) 𝑊𝑛𝑒𝑡 = ∆𝐸𝐾

∆𝐸𝐾 = 𝑊𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 + 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒

Remember that 𝑊𝑐𝑜𝑛𝑐𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = −∆𝐸𝑝

So:

∆𝐸𝐾 = −∆𝐸𝑝 + 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒

If we rearrange the equation

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = ∆𝐸𝐾 + ∆𝐸𝑝

The total mechanical energy (𝐸𝑀1 = 𝐸𝐾1 + 𝐸𝑃1) at the top

The total mechanical energy (𝐸𝑀2 = 𝐸𝐾2 + 𝐸𝑃2) at the

bottom

The total mechanical energy will remain constant

throughout the motion


When the non-conservative forces oppose the motion, the work done by the non-conservative forces is negative, causing a decrease in the mechanical energy of the system. When the non-conservative forces do positive work, energy is added to the system. If the sum of the non-conservative forces is zero then mechanical energy is conserved.

0 = ∆𝐸𝐾 + ∆𝐸𝑝

∆𝐸𝑀 = 0

The total mechanical energy of an object remains constant as the object moves, provided that that the next work done external non-conservative forces is zero.

In conclusion: In an isolated system the total mechanical energy of the system is conserved.

In this case we understand that a closed system is a system in which only internal forces (conservative forces) act on the system. It means that a total mechanical energy of a system is conserved whether: • Only conservative forces acting on the system (non

conservatives forces are not acting on the system). Example: The motion of a projectile (free fall) • Non conservative forces are acting but in equilibrium

(balanced). Example: When a block is pushed at a constant velocity on a

horizontal rough surface. • The resultant (net force) of the non-conservative forces

acting on the system does not do work. Examples: Circular motion of an electron inside a magnetic field - the magnetic force does not do work. Another case is when a body rotates without sliding (pure rotation) on an inclined plane with friction, the frictional force does not do work.

EXAMPLE 1 (Example 9 page 242, Siyabula Book Grade 12):

Consider the situation shown where a football player slides to a stop on levelground. Using energy considerations, calculate the distance the 65,0 kg football player slides, given that his initial speed is 6,00 m∙s-1 and the magnitude of the constant frictional force is 450 N.

SOLUTION:

Step 1: Analyse the problem and determine what is given

Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by


friction is negative, because Ff is in the opposite direction of the motion (that is, θ= 1800, and so cosθ = -1). Thus Wnon-conservative = -FfΔx.

There is no change in potential energy.

Step 2: Next we calculate the distance using the conservation of energy:

We begin with conservation of energy:


The equation expands to:


= 𝐸𝐾𝑓 − 𝐸𝐾𝑖 + (0)

−𝐹𝑓∆𝑥 = 𝐸𝐾𝑓 − 𝐸𝐾𝑖

−𝐹𝑓∆𝑥 = 0 − 𝐸𝐾𝑖

𝐹𝑓∆𝑥 =1

2𝑚𝑣𝑖

2

∆𝑥 =𝑚𝑣𝑖

2

2𝐹𝑓

∆𝑥 =(65,0)(6,0)2

(2)(450)

∆𝑥 = 2,60 𝑚

Step 3: Quote the final answer

The footballer comes to a stop after sliding for 2,60 m.

Summary: YouTube video: http://youtube.com/watch?v=0g31hczz7sA&t=84s

There are two categories of forces we will consider, conservative and non-conservative.

o Conservative force is a force for which the work done in moving an object between two points is independent of the path taken.

If the conservative force does positive work then the change in potential energy is

negative.(𝑊𝑐𝑜𝑛𝑐𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = −∆𝐸𝑝). A non-conservative force is a force for which the work done in moving an object

between two points depends on the path taken. The principle of conservation of mechanical energy state that: The total mechanical

energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant.


http://youtube.com/watch?v=0g31hczz7sA&t=84s


REVISON:

QUESTION 1 (Exercise 5 – 3.1 page 244, Siyavulla book, Grade 12) Adapted:

A 60,0 kg skier with an initial speed of 12,0 m∙s-1 coasts up a 2,50 m-high rise as shown in the figure below. Find her final speed at the top, given that the coefficientof friction between her skis and the snow is 0,0800. (The distance traveled up the incline is 4,36 m)

QUESTION 2 (Exercise 5 – 3.1 page 244, Siyavulla book, Grade 12):

A 60,0 kg skier with an initial speed of 12,0 m∙s-1 coasts up a 2,50 m-high rise as shown in the figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0,0800. (Hint: Find the distance travelled up the incline assuming a straight-line path as shown in the figure.)

SELF-ASSESSMENT 1 MARKS: 10

QUESTION 1

A crate of mass 2 kg is initially at rest at the bottom of a runway. It is pull up the runway

(inclined uniformly at 25o with the horizontal) by a constant force of 100 N acting along

the inclined. The crate experiences a constant frictional force of magnitude 50 N.

1.1 Define the work done on an object by a constant force in words. (2)

1.2 State the work-energy theorem in words: (2)

1.3 Calculate the velocity of the crate when it has travelled a distance of 8 m up the

inclined. (6)

[10]


SOLUTIONS

QUESTION 1

Data:

𝑚 = 60 𝑘𝑔

𝑣𝑖 = 12,0 𝑚 ∙ 𝑠−1 ℎ𝑖 = 0 𝑚 ℎ𝑓 = 2,50 𝑚

𝑣𝑓 =?

𝜇 = 0,0800 ∆𝑥 = 4,36 𝑚

Solution:


𝐹𝑓∆𝑥 cos 𝜃 = (𝐸𝐾𝑓 − 𝐸𝐾𝑓) + (𝐸𝑃𝑓 − 𝐸𝑃𝑖)

𝐹𝑓∆𝑥 cos 𝜃 = (1𝑚𝑣𝑓

2

2−

1𝑚𝑣𝑖2

2) + (𝑚𝑔ℎ𝑓 − 𝑚𝑔ℎ𝑖)

We need the frictional force:

𝐹𝑓 = 𝜇𝑁 = 𝜇𝐹𝑔𝑦 = 𝜇𝑚𝑔 cos 𝜃 = (0,0800)(60)(9,8) cos 35° = 38,53 𝑁

Let’s substitute in the equation:

(38,53)(4,36)(−1) = (60𝑣𝑓

2

2−

1(60)(12,0)2

2) + ((60)(9,8)(2,5) − (60)(9,8)(0))

−167,99 =60𝑣𝑓

2

2− 4320 + 1470

−167,99 + 4320 − 1470 =60𝑣𝑓

2

2

2(2682,01)

60= 𝑣𝑓

2

𝑣𝑓 = √89,40

𝑣𝑓 = 9,46 𝑚 ∙ 𝑠−1


QUESTION 2:

Data:

𝑚 = 60 𝑘𝑔

𝑣𝑖 = 12,0 𝑚 ∙ 𝑠−1 ℎ𝑖 = 0 𝑚 ℎ𝑓 = 2,50 𝑚

𝑣𝑓 =?

𝜇 = 0,0800

Solution:




2

2−

1𝑚𝑣𝑖2


We need the frictional force and the displacement:


The length of the slope will be: ∆𝑥 =ℎ

sin𝜃=

2,50

sin35°= 4,36 𝑚


(38,53)(4,36)(−1) = (60𝑣𝑓

2

2−

1(60)(12,0)2

2) + ((60)(9,8)(2,5) − (60)(9,8)(0))

−167,99 =60𝑣𝑓

2

2− 4320 + 1470

−167,99 + 4320 − 1470 =60𝑣𝑓

2

2

2(2682,01)

60= 𝑣𝑓

2

𝑣𝑓 = √89,40

𝑣𝑓 = 9,46 𝑚 ∙ 𝑠−1


MEMO SELF-ASSESSMENT 1 MARKS: 10

QUESTION 1

1.1 The work done on an object by a constant force is equal to the product of the

magnitude of the force by the magnitude of the displacement by the angle between the

force and the displacement. (2)

1.2 The net/total work done on an object is equal to the change in the object's kinetic

energy OR

The work done on an object by a resultant/net force is equal to the change in the object's

kinetic energy. (2)

1.3

W=F∆xcosθ

Wnet=100 ×8× cos 0o

+FN ×8× cos 90o

+50×8× cos 180o

+2×9,8 ×8× cos 245o

Wnet=333,73 J

𝑊𝑛𝑒𝑡 = ∆𝐸𝐾

𝑊𝑛𝑒𝑡 = 𝐸𝐾𝑓 − 𝐸𝐾𝑖

𝑊𝑛𝑒𝑡 =1

2𝑚𝑣𝑓

2 −1

2𝑚𝑣𝑖

2

333,73 𝐽 =1

2(2)𝑣2 −

1

2(2) × 02

𝑣𝑓 = 18,27 𝑚 ∙ 𝑠−1 (6)

[10]

Any one



LESSON 5: Revision exercises.

Objective:



= ΔEk + ΔE

p

Use the relationship above to show that in the absence of non-conservative forces, mechanical energy is conserved

Introduction:

- There are two categories of forces we will consider, conservative and non-conservative. 1. Define conservative force.

Conservative force is a force for which the work done in moving an object between two points is independent of the path taken. 2. Define non-conservative.

A non-conservative force is a force for which the work done in moving an object between two points depends on the path taken. 3. State the law of conservation of mechanical energy in words:

The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. If the conservative force does positive work then the change in potential energy is negative. (𝑊𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = −∆𝐸𝑝).


Development:

PROBLEM-SOLVING STRATEGY:

Model. Identify which objects are part of the system and what are in the environment. Some problems may need to be subdivided into two or more parts. Visualize. Draw a free-body-diagram showing all the forces acting on the object (objects). And identify the type of force (conservative or non-conservative) Solve. Collect the data and write it in symbolic form. If the system is isolated and non-dissipative then the mechanical energy is conserved.

𝐸𝐾𝑖 + 𝐸𝑃𝑖 = 𝐸𝐾𝑓 + 𝐸𝑃𝑓 = 𝐸𝑀 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

If there are non-conservative forces acting (external or dissipative) then use equation



Kinematics equations and some other laws must be needed for some problems.

Assess. Check that your result has the correct units, is reasonable and answers the question.

EXAMPLE 1 [ (Exercise 5 – 3.1 page 244, Siyavula book, Grade 12)]:

A 60,0 kg skier with an initial speed of 12,0 m∙s-1 coasts up a 2,50 m -high rise as shown in the figure. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0,0800. (Hint: Find the distance travelled up the incline assuming a straight-line path as shown in the figure.)

Solution: Model: There is a sketch of the situation then we do not need to do it: Object: Skier Environment: Incline Visualize. Free body diagram

Solve: There is friction which is a non- conservative force (dissipative) then mechanical energy is not conserved. In this case Normal force is conservative (does not do work) Gravitational force is always a conservative force. Data:

𝑚 = 60 𝑘𝑔

𝑣𝑖 = 12,0 𝑚 ∙ 𝑠−1 ℎ𝑖 = 0 𝑚 ℎ𝑓 = 2,50 𝑚

𝐹 𝑔 (𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒)

𝑓 𝑓

(𝑛𝑜 − 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒)

�⃗⃗� (𝐼𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑑𝑜 𝑤𝑜𝑟𝑘 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒)


𝑣𝑓 =?

𝜇 = 0,0800




2

2−

1𝑚𝑣𝑖2


We need the frictional force and the displacement:


The length of the slope will be: ∆𝑥 =ℎ

sin𝜃=

2,50

sin35°= 4,36 𝑚


(38,53)(4,36)(−1) = (60𝑣𝑓

2

2−

1(60)(12,0)2

2) + ((60)(9,8)(2,5) − (60)(9,8)(0))

−167,99 =60𝑣𝑓

2

2− 4320 + 1470

−167,99 + 4320 − 1470 =60𝑣𝑓

2

2

2(2682,01)

60= 𝑣𝑓

2

𝑣𝑓 = √89,40

𝑣𝑓 = 9,46 𝑚 ∙ 𝑠−1

Assess - The answer has sense and the units are correct.

REVISON EXERCISES

QUESTION 1 (Activity 6 page 75, Mind the Gap, Grade 12, 2015)

A box of mass 100 kg slides down a slope. Its velocity increases from 0 m·s−1 at point A to 4 m·s−1 at point B as in the diagram. Calculate the work done by the non-conservative force while it slides from A to B.

[4]


QUESTION 2 A 2 kg block initially at 4 m height is released and slides downhill from rest on a frictionless ramp and then moves along a horizontal surface. It then moves on a 10 m length rough horizontal surface with coefficient of friction µ=0,2 until reaches a rough ramp with the same coefficient of friction and slides on it 5 m until it stops. 2.1. State the law (principle) of conservation of mechanical energy in words. 2.2. Use the law of conservation of mechanical energy to calculate the speed of the block at the bottom of the ramp. 2.3. State the work energy theorem in words. 2.4. Use the WORK ENERGY THEOREM to calculate the speed of the block when passing position B. 2.5. Use the LAW OF CONSERVATION OF ENERGY to calculate the height reached by the block until it comes to rest. QUESTION 3 (Activity 7 page 76, Mind the Gaps, Grade 12, 2015)

A toy truck, mass 1,4 kg, moving down an inclined track, has a speed of 0,6 m·s–1 at point P, which is at a height of 1,5 m above the ground level QR. The curved section of the track, PQ, is 1,8 m long. When the truck reaches point Q it has a speed of 3 m·s–1. There is friction between the track and the truck.

3.1 State the principle of conservation of mechanical energy in words. (2) 3.2 Is mechanical energy conserved? Explain. (2) 3.3 Assume that the average frictional force between the track and the truck is constant along PQ and calculate the average frictional force experienced by the truck as it moves along PQ. (6) [10]

10 m

B

12o

h

5 m


QUESTION 4:

A boy of mass 55 kg is travelling at 6m·s-1on a skateboard of mass 1 kg, when he reaches a ramp of length 8m, he keeps moving and reaches the top of the ramp with velocity of 1 m.s-1. The constant friction that he experiences is 6 N.

4.1 Explain whether his mechanical energy is conserved or not. 4.2. Calculate the kinetic energy of the boy when reaches the ramp. 4.3. Calculate the height of the ramp.

QUESTION 5:

John applies a force F to help his friend in a wheelchair to move up a ramp of length 10 m and a vertical height of 1,5 m, as shown in the diagram below. The combined mass of his friend and the wheelchair is 120 kg. The frictional force between the wheels of the wheelchair and the surface of the ramp is 50 N. The rotational effects of the wheels of the wheelchair may be ignored.

The wheelchair moves up the ramp at constant velocity of 1,5 m·s-1.

5.1 What is the magnitude of the net force acting on the wheelchair as it moves up the ramp? Give a reason for your answer. (2) 5.2 What is the magnitude of the net work done on the wheelchair on reaching the top of the ramp? (1) 5.3 Calculate the following: 5.3.1 Work done on the wheelchair by force F. (5) 5.3.2 The magnitude of force F exerted on the wheelchair by John (4) [12]

8 m

h = 1,5 m

10 m

F


SOLUTIONS

QUESTION 1 1:

Data:

𝑚 = 100 𝑘𝑔

𝑣𝑖 = 0 𝑚 ∙ 𝑠−1 𝑣𝑓 = 4 𝑚 ∙ 𝑠−1

ℎ𝑖 = 1 𝑚 ℎ𝑓 = 0 𝑚

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 =?

Solution:


𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = (𝐸𝐾𝑓 − 𝐸𝐾𝑓) + (𝐸𝑃𝑓 − 𝐸𝑃𝑖)

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = (𝑚𝑣𝑓

2

2−

𝑚𝑣𝑖2


𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = ((100)(4)2

2−

(100)(0)2

2)+ ((100)(9,8)(0) − (100)(9,8)(1))

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = 800 − 980 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = −180 𝐽 [4]

QUESTION 2:

2.1 The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. 2.2 ΔEm = 0 Emo = Em

EKo + EPo = Ek+ EP

0 + mgh = EKo + 0

mgh = 2

2

fmv

v2= gh2

v = gh2

v = 48,92 xx

v = 8,85 m∙s-1


2.3 The net / total work done on an object is equal to the change in the object’s kinetic

energy.

2.4 knet EW

kifkNFgfr EEWWW

kifkgr EENFxF 000 90cos90cos180cos

kifkr EEmgmgxF 000 90cos90cos180cos

22)1(

22

if mvmvxN

22.

22

if mvmvxmg

22.

22

if vvxg

).2( 22 xgvv if

)108,92,02)85,8( 22 fv

2,3932,782 fv

12,39fv

1.25,6 smv f

2.5.


𝑓𝑓∆𝑥𝑐𝑜𝑠 180 = (𝐸𝐾𝑓 − 𝐸𝐾𝑓) + (𝐸𝑃𝑓 − 𝐸𝑃𝑖)

𝜇𝑁∆𝑥𝑐𝑜𝑠 180 = (𝑚𝑣𝑓

2

2−

𝑚𝑣𝑖2


𝜇(𝑚𝑔 cos ∝)∆𝑥𝑐𝑜𝑠 180 = (𝑚𝑣𝑓

2

2−

𝑚𝑣𝑖2


0,2 × 2 × 9,8 × cos 12𝑜 (5)(−1) = (2×02

2−

2×(6.25)2

2) + (2 × 9,8 × ℎ𝑓 − 2 × 9,8 × 0)

−19.17 = 0 − 39,0625 + 19,6ℎ

19.89 = 19,6ℎ h = 1, 01482 m

𝐹 𝑔 (𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒)

𝑓 𝑓

(𝑛𝑜𝑛 − 𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒)

�⃗⃗� (𝐼𝑡 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑑𝑜 𝑤𝑜𝑟𝑘 𝑖𝑛 𝑡ℎ𝑖𝑠 𝑐𝑎𝑠𝑒)


ℎ = ∆𝑥 sin 𝜃 0,99 = 5 sin 𝜃 sin 𝜃 = 0,198

θ = 11,450

QUESTION 3:

3.1 The total mechanical energy in an isolated system remains constant or is conserved. (2)

3.2 The mechanical energy is not conserved due to the presence of non-conservative force (frictional force). (2)

3.3




2

2−

𝑚𝑣𝑖2


𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = ((1,4)(3)2

2−

(1,4)(0,6)2

2) + ((1,4)(9,8)(0) − (1,4)(9,8)(1,5))

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = 6,048 − 20,58 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = −14,53 𝐽

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = 𝐹𝑓∆𝑥 cos 𝜃

−14,53 = 𝐹𝑓(1,8)(−1)

𝐹𝑓 =−14,53

−1,8

𝐹𝑓 = 8,07 𝑁 (6) [10]

QUESTION 4:

4.1

No, because there is frictional force acting which is non-conservative. Or No because the system is not isolated/closed, frictional force is acting.

4.2

2

2mvEK

2

6)155( 2KE


2

3656KE

2

2016KE

EK= 1008 J 4.3 Wnon cons = ∆EM Wnon cons = ∆EMf - ∆EMi Wnon cons = (Epf + EKf)–(Epi + EKi)

Ki

fo Emv

mghxFf 2

180cos

2

10082

156)8.956(86

2

h

100828)8.548(48 h

8,548

10082848 h

8,548

932h

mh 70.1 QUESTION 5: 5.1 0 (N)/Zero . No acceleration/constant velocity 5.2 0 (J)/Zero Option 1:

Mnoncon EW

MiMfFaFf EEWW

)()( PiKiPfKfFaFf EEEEWW

)0()( KiPfKfFaFf EEEWW KfKi EE

mghWxFf Fa cos

5.18.9120)1(1050 FaW

JWFa 2264 (2,26 x 103 J)

Option 2: For equilibrium: F = f + wparallel to incline= f + mgsin ( - angle of incline with horizontal)


F = 50 + (120)(9,8)(10

51,)

F = 226,4 N

Wapplied = F xcos = (226,4)(10)(cos0°) OR (226,4)(10) = 2 264 J (2,26 x 103 J) Option 3:

Wnet= K

Wapplied + Wfriction+ Wgravity = K Wapplied + (50)(10)(-1) - (120)(9,8)(1,5) = 0

Wapplied = 2 264 J (2,26 x 103 J) Option 4:

W(external forces) = U + K OR

Wapplied + Wfriction = (Uf – Ui) + (Kf – Ki)/U + K Wapplied + (50)(10)(-1) = (120)(9,8)(1,5) - 0) + 0

Wapplied = 2 264 J (2,26 x 103 J) 5.3.2.

Wapplied = F xcos

2 264 J = F(10)(1)

F = 226,4 N (2,26 x 102 N) OR F = f + wpar to incline= f + mgsin ( - angle of incline with horizontal)

F = 50 + (120)(9,8)(10

51,)

F = 226,4 N (2,26 x 102 N)



LESSON 6: Definition of power. Calculating power. Minimum and maximum power.

Objective:


Define power

Calculate the power involved when work is done

Perform calculations using Pave= Fvave when an object moves at a constant speed

along a rough horizontal surface or a rough inclined plane.

Calculate the power output for a pump lifting a mass (e.g. lifting water through a height at constant speed).

Introduction:

- Imagine two identical models of an automobile: one with a base-priced four-cylinder engine; and the other with the highest-priced optional engine, a mighty eight cylinder power plant. Despite the differences in engines, the two cars have the same mass.

Both cars climb a roadway up a hill, but the car with the optional engine takes much less time to reach the top. Both cars have done the same amount of work against gravity, but in different time periods. From a practical viewpoint, it is interesting to know not only the work done by the vehicles but also the rate at which it is done. In taking the ratio of the amount of work done to the time taken to do it, we have a way of quantifying this concept. This new physical quantity is called power.

Development:

Yube video (power). http://youtube.com/watch?v=WNTbbF3bivg&t=27s

If an external force is applied to an object (which we assume acts as a particle), and if the work done by this force in the time interval Δt is W, then the average power expended during this interval is defined as:

𝑃 =𝑊

∆𝑡

The work done on the object contributes to the increase in the energy of the object. Therefore, a more general definition of power is the time rate of energy transfer.

Or

Power as the rate at which work is done or energy is expended.

The SI unit of power is joules per second (J/s), also called the watt (W) (after James Watt, the inventor of the steam engine):

http://youtube.com/watch?v=WNTbbF3bivg&t=27s


1 W = 1 J/s = N·m·s-1 =1 kg·m2.s-3

When several forces act on a body and the resultant (net) force is zero the body is stationary or moving with a constant velocity.

𝑃𝑎𝑣 =𝑊

∆𝑡

and

𝑃𝑎𝑣 =𝐹∆𝑥

∆𝑡

If the body possesses rectilinear uniform motion (v = constant) then

But t

xv

Hence

𝑃𝑎𝑣𝑒 = 𝐹𝑣𝑎𝑣𝑒

EXAMPLE 1

The following figure shows an elevator that has a mass of 1 000 kg and is carrying passengers having a combined mass of 800 kg. A constant frictional force of 4 000 N retards its motion upward.

1.1 What must be the minimum power delivered by the motor to lift the elevator car at a constant speed of 3.00 m·s-1?

1.2 What power must the motor deliver at the instant its speed is v if it is designed to provide an upward acceleration of 1.00 m·s-2?


Solution:

1.1

The motor must supply the force of magnitude T that pulls the elevator upward. Reading that the speed is constant provides the hint that a = 0, and therefore we know from Newton’s second law that ƩFy = 0. We have drawn a free-body diagram and we choose the upward direction as positive.

From Newton’s first law we obtain:

0 yF

0 FgFfT

0 MgFfT

Where M is the total mass of the system (elevator plus passengers), equal to 1 800 kg. Therefore,

MgfT f

)8.91800(4000 T

NT 41016.2

FvP

00.31016.2 4 P

WP 41048.6

1.2

Now we expect to obtain a value greater than we did in 1.1, where the speed was constant, because the motor must now perform the additional task of accelerating the car. The only change in the setup of the problem is that now a > 0.

Applying Newton’s second law to the car gives:

aMFy


MaFgfT f

MaMgfT f

MaMgfT f

)( gaMfT

)8.91(108.1104 33 T

31044.23 T

NT 41034.2

Therefore, using Equation P = Fv, we obtain for the required power

𝑃 = 𝑇𝑣

𝑃 = 234 × 104𝑣) W

Where v is the instantaneous speed of the car in meters per second. The power is less

than that obtained in part 1.1 as long as the speed is less than T

P= 2.77 m·s-1, but it is

greater when the elevator’s speed exceeds this value.

EXAMPLE 2 (Example 14, page 248, Siyabula book, Grade 12):

What is the power required to pump water from a borehole that has a depth h = 15,0 m at a rate of 20,0 L∙s-1?

Solution:

Step 1: Analyse the question

We know that we will have to do work on the water to overcome gravity to raise it a certain height. If we ignore any inefficiency we can calculate the work, and power, required to raise the mass of water the appropriate height.

We know how much water is required in a single second. We can first determine the

mass of water: 20,0 ×1 𝑘𝑔

1 𝐿= 20,0 𝑘𝑔.

The water will also have non-zero kinetic energy when it gets to the surface because it needs to be flowing. The pump needs to move 20,0 kg from the depth of the borehole every second, we know the depth so we know the speed that the water needs to be

moving is 𝑣 =∆𝑥

∆𝑡=

ℎ

∆𝑡=

15,0

1= 15,0 𝑚 ∙ 𝑠−1

Step 2: Work done to raise the water:

We can use





2

2− 0) + (𝑚𝑔ℎ𝑓 − 0)

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 =(20)(15)2

2+ (20)(9,8)(15)

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = 2,25 × 103 + 2,94 × 103

𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = 5,19 × 103 𝐽

Step 3: Calculate power:

𝑃 =𝑊

𝑡=

5,19×103

1= 5,19 × 103 𝑊

Step 4: Quote the final answer:

The minimum power required from the pump is 5,19 × 103 𝑊.

Summary:

Power as the rate at which work is done or energy is expended

In symbols 𝑃 =𝑊

∆𝑡.

We can calculate the average power using Pave= Fvave.

REVISON

QUESTION 1 Multiple choice questions

Four options are provided as possible answer to the following question. There is only ONE correct answer. Choose the correct answer.

1.1. When an object is traveling at constant velocity on a horizontal surface, the equation 𝑃 = 𝐹𝑣 can be used. Here the force refers to:

A The resultant force acting on the object. B The external force applied to the object. C The frictional force acting on the object. D The gravitational force action on the object.

QUESTION 2

An elevator, m = 800 kg has a maximum load of 600 kg. The elevator goes up 30 m at a constant speed of 4 m·s-1. What is the average power output of the elevator motor if the elevator is fully loaded with its maximum weight? (Neglect friction).


QUESTION 3

A block of 200 kg is pulled along the floor at a constant speed by an electric motor. The coefficient of friction between the block and the floor is 0,2.

3.1 Calculate the frictional force experienced by the block. 3.2 Calculate the power the motor must deliver if the block is to move at a constant speed of 6 m.s-1. 3.3 How much work is done by the motor in 60 s? 3.4 What is the net work done on the block in the 60 s?


SOLUTIONS

QUESTION 1:

B

QUESTION 2:

Option 1 Option 2

Find the work done by the motor and divide it by the time it takes:

𝑃 =𝑊

∆𝑡

We need the time:

t

xv

t

304

∆𝑡 = 7,5 𝑠

Work done by the gravitational force

𝑊 = − ∆𝐸𝑝 𝑊 = − (𝐸𝑝𝑓 − 𝐸𝑝𝑖) 𝑊 = −𝑚𝑔(ℎ𝑓 − ℎ𝑖)

Level zero on the ground hi = 0

𝑊 = − (800 + 600)(9,8)(30) 𝑊 = − 4,12 × 105 𝐽

As the motion is with constant velocity

𝑇 = 𝐹𝑔 then the Work done by the

elevator is 4,12 × 105 𝐽

𝑃 =𝑊

∆𝑡=

4,12×105

7,5= 54900 𝑊 = 54,9 𝑘𝑊

Use the velocity and the force that the motor must exert:

𝑃 = 𝐹𝑣

0 gFT

Working with the projections of the vector forces on the Y-axis we have:

T- Fg = 0 T = Fg T = mg

F = T 𝑃 = 𝐹𝑣 = 𝑚𝑔𝑣 = (800 +600)(9,8)(4) 𝑃 = 54900 𝑊 = 54,9 𝑘𝑊


QUESTION 3:

3.1

N = mg Ff = μmg Ff = (0,2)(200)(9,8) = 392 N

3.2

P = Fv P = (392)(6) = 2352 W

3.3

P = t

W

2352 = s

W

60

W = 141120 J

3.4

Wnet = ∆Ek = Ekf - Eki Since velocity is constant then ∆Ek = 0 Hence W = 0 J



LESSON 7: Revision exercises on work-energy and power.

Objective:


Calculate the net work done on an object

Apply the work-energy theorem to objects on horizontal, vertical and inclined planes (for both frictionless and rough surfaces).


= ΔEk + ΔE

p



Perform calculations using Pave= Fvave when an object moves at a constant speed along a rough horizontal surface or a rough inclined plane.


-


INTRODUCTION SUMMARY

Physical quantity Work-energy theorem and the principle of conservation of energy

Law of conservation of mechanical energy Work Kinetic energy Potential energy

Mechanical energy

Power Classification of

forces

Work done by a constant force is defined as the scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement.

𝑾 = 𝑭∆𝒙 𝒄𝒐𝒔 𝜽 SI unit: joule [J]

Kinetic energy is the energy an object has due to its motion.

𝑬𝑲 =𝟏

𝟐𝒎𝒗𝟐

SI unit: joule [J]

Gravitational potential energy is the energy an object has due to its position in a gravitational field relative to some reference point.

𝑬𝒑 = 𝒎𝒈𝒉

SI unit: joule [J]

Total Mechanical energy is equal to the sum of the gravitational potential energy plus the kinetic energy.

𝑬𝑴 = 𝑬𝑲 + 𝑬𝒑

SI unit: joule [J]

Power is the rate at which work is done or energy is expended.

𝑷 =𝑾

∆𝒕

If the body possesses rectilinear uniform motion (v = constant), then: 𝑷𝒂𝒗𝒆 = 𝑭𝒗𝒂𝒗𝒆 SI unit: watt

[W]

A conservative force is a force for which the work done in moving an object between two points is independent of the path taken. Examples:

gravitational force, electrostatic force and elastic force. A non-

conservative force is a force for which the work done in moving an object between two points depends on the path taken. Examples: Frictional

force, tension, etc

Work- energy theorem: The work done by the resultant (net) force acting on an object is equal to the change in the kinetic energy of the object.

𝑾𝒏𝒆𝒕 = 𝜟𝑬𝑲 𝑾𝒏𝒆𝒕 = 𝑬𝑲𝒇 − 𝑬𝑲𝒊

𝑊𝑛𝑒𝑡 = 𝑊𝑐 + 𝑊𝑛𝑐 Using the work- energy

theorem: 𝑊𝑐 + 𝑊𝑛𝑐= ∆𝐸𝐾 But: 𝑊𝑐𝑜𝑛𝑠 = −∆𝐸𝑝

−∆𝐸𝑝 + 𝑊𝑛𝑐= ∆𝐸𝐾

Hence: 𝑾𝒏𝒄 = ∆𝑬𝑲 + ∆𝑬𝒑

This equation can be used to solve conservation of energy problems and is called the principle of conservation of energy.

Total mechanical energy in an isolated system remains constant (is conserved).

𝑬𝑴(𝒊𝒏𝒊𝒕𝒊𝒂𝒍)

= 𝑬𝑴(𝒇𝒊𝒏𝒂𝒍)

In an isolated system, only conservative forces are acting. OR The total mechanical energy of an object remains constant as the object moves, provided that the net work done by external non-conservative forces is zero.

0 = ∆𝑬𝑲 + ∆𝑬𝒑

Developed by: G. Izquierdo Rodríguez & G Izquierdo Gómez Page 53

Development: REVISON EXCERCISES

QUESTION 1:

A horse pulls a cart of mass 500 kg on a frictionless horizontal surface from rest and covers a distance of 5 m in 10 s. If the force applied is in the same direction of the displacement.

Calculate:

1.1 The work done by the force applied by the horse on the cart. 1.2 The power developed by the horse. 1.3 The change in kinetic energy.

QUESTION 2:

An airplane is 5000 m high and it is flying in a horizontal direction at a speed of 900 km.h-

1. From this height a parcel of 20 kg is dropped. Ignore air resistance and calculate:

2.1 The initial mechanical energy of the parcel. 2.2 The work done by the gravitational force. 2.3 Speed of the parcel when it hits the ground.

QUESTION 3:

A water pump is required to lift water from a well that is 8 m below ground level. If 2 kg of water are pumped to the surface in one second.

3.1 Calculate the work done by the gravitational force on the 2 kg of water. 3.2 Calculate the minimum force required to pump the water. 3.3 Calculate the minimum power developed by the pump in one second.

QUESTION 4:

Water is pumped out of a borehole 25 m deep with an electric pump. 4.1 Define power in words.

4.2 What must the minimum power output for the electric pump be to lift 500 kg water every minute at a speed of 7 m·s-1?


QUESTION 5:

The maximum power output of an electric pump is 60 000 W. It is used to pump water from a borehole 20 m below the ground. If 200 kg of water is required every second, calculate the speed at which the water is discharged if the pump operates at maximum power output.

QUESTION 6:

A 3 kg block slides at a constant velocity of 7 ms-1 along a horizontal surface. It then strikes a rough surface, causing it to experience a constant frictional force of 30 N. The block slides 2 m under the influence of the frictional force before it moves up a frictionless ramp as shown in the diagram below. Show by calculation that the speed of the block at the bottom of the ramp is 3 m∙s-1.

SELF-ASSESSMENT 2 MARKS: 10

QUESTION 1

A mass piece is attached to a short length of string that is attached to the ceiling, and then allowed to swing. In the absence of air friction the pendulum reaches maximum height of 30 cm.

1.1. Define a conservative force in words. (2) 1.2 State the Law of Conservation of Mechanical Energy in words. (2) 1.3 Draw a free-body diagram at position A. (2) 1.4 Use the law of conservation of mechanical energy to determine the speed of a pendulum at a height C, 10 cm above the lowest point in its swing B. (4) [10]

3 kg

7 m.s-1

2m

B

A

C 30 cm

10 cm

D


SOLUTIONS

QUESTION 1

Data: m = 500 kg Δx = 5 m t = 10 s W -? P -? ΔEk - ?

1.1 1.2 1.3

W = F Δx cos00 F = ma W = ma Δx

a = 2

2

t

x

W = m 2

2

t

xΔx

W = 2

2)(2

t

xm

2

2

10

)5)(500)(2(W

100

)25)(1000(W

100

25000W

W = 250 J

P = t

W

10

250P

P = 25 W

Wnet = ΔΕK = 250 J

QUESTION 2:

2.1 2.2 2.3

PiKiMi EEE

iM mghmv

E i 2

2

50008.9202

)250(20 2

ME

9800002

6250020

ME

980000625000ME

JEM 1605000

W = - (EP2 – EP1) We can take the level zero the surface of the Earth (zero gravitational potential energy) Ep=0 W = - (0 – EP1) W = EPi

W = mgh W = (20 x 9,8 x 5000)

W = 980000J

Wnet = ΔEK Wnet = Ekf - EKi

22

22

if

net

mvmvW

6250002

20980000

2

fv

v= 400,62 m.s-1


QUESTION 3:

3.1

3.2

3.3

QUESTION 4: 4.1 Power is the rate at which work is done or energy is expended.

4.2 PKnc EEW

)(2

1 22

ifnc vvmW )( if hhmg

)07)(500(2

1 2 ncW )025)(8,9(500

Wnc = 12250 + 122500= 134750 J

t

WP

60

134750P

WP 83,2245

Option 1 Option 2

𝑊𝐹𝑔 = 𝐹𝑔∆𝑥 cos 𝜃

𝑊𝐹𝑔 = 𝑚𝑔∆𝑥 cos 180°

𝑊𝐹𝑔 = (2)(9,8)(8)(−1)

𝑊𝐹𝑔 = −156,8 𝐽

𝑊𝐹𝑔 = −∆𝐸𝑝

𝑊𝐹𝑔 = −𝑚𝑔ℎ

𝑊𝐹𝑔 = −(2)(9,8)(8)

𝑊𝐹𝑔 = −156,8 𝐽

Option 1 Opion 2

Force required = Fg = mg = (2)(9,8) = 19,6 N

Wg= F∆xcosΘ 156,8 = F (8) (cos 00) F= 19,6 N

Option 1 Option 2

P = t

W

= (19,6)(8) 1

= 156,8 W

P = F.v v=1

8

t

x= 8 m.s-1

P= 19,6 x 8 P= 156,8


QUESTION 5:

𝑃 =𝑊𝑛𝑐

∆𝑡

40000 𝑊 =𝑊𝑛𝑐

1 𝑠

𝑊𝑛𝑐 = ∆𝐸𝐾 + ∆𝐸𝑃

𝑊𝑛𝑐 = (𝑚𝑔ℎ𝑓 − 𝑚𝑔ℎ𝑖) + (1

2𝑚𝑣𝑓

2 −1

2𝑚𝑣𝑖

2)

60000 = (200 × 9,8 × 20 − 200 × 9,8 × 0) + (1

2200𝑣𝑓

2 −1

2200 × 02)

60000 = 39200 + 100𝑣𝑓2

𝑣𝑓2 =208

𝑣𝑓 = 14.42 𝑚 ∙ 𝑠−1 Therefore, water is discharged at 14,42 m∙s-1

QUESTION 6:

Option 1 Option 2

Applying the work kinetic energy

theorem

knet EW

kifkf EExF 0180cos

22)1(

22

if

f

mvmvxF

2

)49(3

2

.3.60

2

v

5,732

.3.60

2

v

2

.3.5,7360

2v

23.25,13 vx 2327 v

29 v 1.39 smv

Applying Newton`s second law and

the equations of kinematics

maFnet

)2

(

22

x

vvmF

if

)22

7(330

22

x

v

)4

49(330

2

v

1473430 2 vx

23147430 vx 23147120 v

2

3

27v

1.39 smv

2

)7(3

2

.3.230

22

v

x


MEMO SELF-ASSESSMENT 2 MARKS: 10

QUESTION 1

1.1. A conservative force is a force for which the work done in moving an object between

two points is independent of the path taken. (2)

1.2. The total mechanical energy (sum of gravitational potential energy and kinetic

energy) in an isolated system remains constant. (2)

1.3.

98,1v m∙s-1

OR

98,1v m∙s-1 (4)

[10]

EM(A)= EM(C)

CKPAKP EEEE )()(

CA mvmghmvmgh )2

1()

2

1( 22

030,08,9( 2

2

110,08,9( v

)98,094,2(22 v

EM(D)= EM(C)

CKPDKP EEEE )()(

CD mvmghmvmgh )2

1()

2

1( 22

030,08,9( 2

2

110,08,9( v

)98,094,2(22 v

ANY ONE

ANY ONE

𝐹 𝑔

�⃗�


Lesson 8- WORK, ENERGY & POWER

Topic: Revision exercise.

Objective:


Calculate the work done on an object

Apply the work-energy theorem to objects on horizontal, vertical and inclined planes (for both frictionless and rough surfaces).


= ΔEk + ΔE

p



Perform calculations using Pave= Fvave when an object moves at a constant speed along a rough horizontal surface or a rough inclined plane.



INTRODUCTION SUMMARY

Physical quantity Work-energy theorem and the principle of conservation of energy

Law of conservation of mechanical energy

Work Kinetic energy Potential energy Mechanical energy Power

Classification of forces

Work done by a constant force is defined as the scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement.

𝑾 = 𝑭∆𝒙 𝒄𝒐𝒔𝜽 SI unit: joule [J]

Kinetic energy is the energy an object has due to its motion.

𝑬𝑲 =𝟏

𝟐𝒎𝒗𝟐

SI unit: joule [J]

Gravitational potential energy is the energy an object has due to its position in a gravitational field relative to some reference point.

𝑬𝒑 = 𝒎𝒈𝒉

SI unit: joule [J]

Total Mechanical energy is equal to the sum of the gravitational potential energy plus the kinetic energy. 𝑬𝑴 = 𝑬𝑲 + 𝑬𝒑

SI unit: joule [J]


𝑷 =𝑾

∆𝒕

If the body possesses rectilinear uniform motion (v = constant), then:

𝑷𝒂𝒗𝒆 = 𝑭𝒗𝒂𝒗𝒆 SI unit: watt [W]

A conservative force is a force for which the work done in moving an object between two points is independent of the path taken. Examples: gravitational force, electrostatic force and elastic force. A non-conservative force is a force for which the work done in moving an object between two points depends on the path taken. Examples: Frictional force, tension, etc

Work- energy theorem: The work done by the resultant (net) force acting on an object is equal to the change in the kinetic energy of the object.

𝑾𝒏𝒆𝒕 = 𝜟𝑬𝑲 𝑾𝒏𝒆𝒕 = 𝑬𝑲𝒇 − 𝑬𝑲𝒊

𝑊𝑛𝑒𝑡 = 𝑊𝑐 + 𝑊𝑛𝑐 Using the work- energy theorem: 𝑊𝑐 + 𝑊𝑛𝑐= ∆𝐸𝐾 But: 𝑊𝑐𝑜𝑛𝑠 = −∆𝐸𝑝

−∆𝐸𝑝 + 𝑊𝑛𝑐= ∆𝐸𝐾

Hence: 𝑾𝒏𝒄 = ∆𝑬𝑲 + ∆𝑬𝒑

This equation can be used to solve conservation of energy problems and is called the principle of conservation of energy.

Total mechanical energy in an isolated system remains constant (is conserved).

𝑬𝑴(𝒊𝒏𝒊𝒕𝒊𝒂𝒍) = 𝑬𝑴(𝒇𝒊𝒏𝒂𝒍)

In an isolated system, only conservative forces are acting. OR The total mechanical energy of an object remains constant as the object moves, provided that the net work done by external non-conservative forces is zero.

0 = ∆𝑬𝑲 + ∆𝑬𝒑


Development: REVISON EXERCISES QUESTION 1 Two blocks are connected by a light inextensible string that passes over a frictionless pulley as in Figure below. The system is released from rest while m2 is on the floor and m1 is a distance h above the floor.

1.1. Assuming m1˃m2, find an expression for the speed of m1 just as it reaches the floor. 1.2. Taking m1= 6.5 kg, m2=4.2 kg, and h= 3.2 m, evaluate your answer in question 1.1. 1.3. Calculate the speed of each block when m1 has fallen a distance of 1.6 m. 1.4. Calculate the maximum height reached by block 2. 1.5. Calculate the work done by the gravitational force on block 2.

Answer 1.1

𝑣 = √2𝑔ℎ(𝑚1 − 𝑚2)

𝑚1 + 𝑚2

1.2 𝑣 = 3,67 𝑚 ∙ 𝑠−1 1.3 𝑣 = 2,60 𝑚 ∙ 𝑠−1 1.4.

ℎ𝑚𝑎𝑥 = 3,89 𝑚

1.5

𝑊 = −160,11 𝐽


QUESTION 2 (FS trial 2018) adapted

2.1 A block of mass 20 kg is pulled with a force of 60 N as indicated in the

diagram below. The block slides over a distance of 20 m.

Ignore all effects of friction.

2.1.1 Define the work done on an object by a constant force in

words.

(2)

2.1.2 Calculate the work done by the applied force. (3)

2.1.3 Does the force of gravity do work on the cart? Explain. (2)

2.2 ABC is a frictionless part of a track. A is a point 1,2 m above the

ground. Block P of mass 0,3 kg slides from point A with an

initial speed of 2 m·s-1 and reaches point B as shown in the

diagram below.

.2.1 State the principle of conservation of mechanical energy in

words:

(2)

2.2.2 Calculate the speed of block P at point B. (4)

Block P then collides with a stationary block Q of mass 0,4 kg. After the

collision, the two blocks slide through BC and up the rough incline CD.

The frictional force acting on the blocks as they move up the incline is 0,5

N. The blocks come to rest at a distance of 0,3 m up the incline.

3 Define linear momentum in words. (2)

2.2.4 State the law of conservation of linear momentum in words. (2)

Calculate the:

2.2.5 Speed of the combination of blocks P and Q immediately

after the collision

(4)

2.2.6 Maximum height, h, reached by the combination

of blocks

(5)


2.3 Water is pumped from a 23 m deep bore hole. The requirement is 300 kg water every minute. The speed at which the water must be ejected is 5,5 m·s-1.

2.3.1

Define power in words (2)

2.3.2

Calculate the minimum power required by the electric pump to pump the water at the above mentioned rate.

(8)

[36]

- -

QUESTION 3

A block of mass 𝒎𝟏 = 2,40 𝑘𝑔 is connected to a second block of mass 𝑚2 = 1,80 𝑘𝑔 as shown

below. When the blocks are released from rest, they move through a distance 𝑑 = 0,50 𝑚 at

which point 𝑚2 hits the floor. When block 2 hits the ground the speed of the blocks is 0,95 m·s-

1.

3.1. Gravitational force is a conservative force. Define the term conservative force. ( 2) 3.2. Calculate the work done by the gravitational force on block 2. (3) 3.2. Use energy principles to calculate the coefficient of kinetic friction between

block 1 and the horizontal surface. (6)

[11]


SOLUTIONS

QUESTION 1

∆𝐸𝑀 = 0

𝐸𝑀𝑖 = 𝐸𝑀𝑓

(𝐸𝑘(1) + 𝐸𝑘(2))𝑖+ (𝐸𝑃(1) + 𝐸𝑃(2))𝑖

= (𝐸𝑘(1) + 𝐸𝑘(2))𝑓+ (𝐸𝑃(1) + 𝐸𝑃(2))𝑓

(1

2𝑚1𝑣1

2 +1

2𝑚2𝑣2

2)𝑖+ (𝑚1𝑔ℎ1 + 𝑚2𝑔ℎ2)𝑖 = (

1

2𝑚1𝑣1

2 +1

2𝑚2𝑣2

2)𝑓+ (𝑚1𝑔ℎ1 + 𝑚2𝑔ℎ2)𝑓

(1

2𝑚1(0)2 +

1

2𝑚2(0)2)

𝑖+ (𝑚1𝑔ℎ1) + 𝑚2𝑔 × 0)𝑖 = (

1

2𝑚1𝑣1

2 +1

2𝑚2𝑣2

2)𝑓+ (𝑚1𝑔 × 0 + 𝑚2𝑔ℎ2)𝑓

OR/OF

(𝑚1𝑔ℎ1)𝑖 = (1

2𝑚1𝑣1

2 +1

2𝑚2𝑣2

2)𝑓+ (𝑚2𝑔ℎ2)𝑓

𝑣1 = 𝑣2 = 𝑣

ℎ1 = ℎ2 = ℎ

(𝑚1𝑔ℎ)𝑖 = (1

2𝑚1𝑣

2 +1

2𝑚1𝑣

2)𝑓+ (𝑚2𝑔ℎ)𝑓

OR

(𝑚1𝑔ℎ)𝑖 − (𝑚2𝑔ℎ)𝑓 = 𝑣2

2(𝑚1 + 𝑚2)

OR/OF

2𝑔ℎ(𝑚1 − 𝑚2) = 𝑣2(𝑚1 + 𝑚2)

𝑣2 =2𝑔ℎ(𝑚1 − 𝑚2)

𝑚1 + 𝑚2

𝑣 = √2𝑔ℎ(𝑚1 − 𝑚2)

𝑚1 + 𝑚2


1.2

𝑣 = √2(9,8)(3,2)(6,5 − 4,2)

6,5 + 4,2

v= √144,256

10,7

𝑣 = 3,67 𝑚 ∙ 𝑠−1 1.3

√2(9,8)(1,6)(6,5 − 4,2)

6,5 + 4,2

v= √72,128

10,7

𝑣 = 2,60 𝑚 ∙ 𝑠−1 1.4

∆𝐸𝑀 = 0

𝐸𝑀𝑖 = 𝐸𝑀𝑓

(𝐸𝐾(2) + 𝐸𝑃(2))𝑖= (𝐸𝐾(2) + 𝐸𝑃(2))𝑓

(1

2𝑚2𝑣2

2+𝑚2𝑔ℎ2)𝑖= (

1

2𝑚2𝑣2

2+𝑚2𝑔ℎ2)𝑓

(1

2𝑚2𝑣2

2+𝑚2𝑔ℎ2)𝑖= (0+𝑚2𝑔ℎ𝑚𝑎𝑥)𝑓

OR

(1

2𝑚2𝑣2

2+𝑚2𝑔ℎ2)𝑖= 𝑚2𝑔ℎ𝑚𝑎𝑥

OR

(1

2𝑣2

2 + 𝑔ℎ2)𝑖= 𝑔ℎ𝑚𝑎𝑥

1

2(3,67)2 + (9,8 × 3,2 = (9,8)ℎ𝑚𝑎𝑥

ℎ𝑚𝑎𝑥 = 3,89 𝑚

1.5

𝑊 = 𝐹∆𝑥𝑐𝑜𝑠𝜃

OF

ℎ𝑚𝑎𝑥 =

𝑣2

2 + 𝑔ℎ𝐴

𝑔

ℎ𝑚𝑎𝑥 =

(3,67)2

2 + (9,8)(3,2)

9,8

ℎ𝑚𝑎𝑥 = 3,89 𝑚


𝑊 = 𝑚𝑔∆𝑥𝑐𝑜𝑠𝜃

𝑊 = 4,2 × 9,8 × 3,89 × 𝑐𝑜𝑠180𝑜

𝑊 = −160,11 𝐽

OR

𝑊 = −∆𝐸𝑝

𝑊 = −(𝐸𝑃(𝑓) − 𝐸𝑃(𝑖))

𝑊 = −(𝑚𝑔ℎ𝑓 − 𝑚𝑔ℎ𝑖)

𝑊 = −(4,2 × 9,8 × 3,89 − 4,2 × 9,8 × 0)

𝑊 = −160,11 𝐽

QUESTION 2

2.1.1

Work done by a constant force is defined as the scalar quantity equal to the product of the force multiplied by the displacement of the point of application of the force by the cosine of the angle formed by the line of action of the force and the direction of the displacement. .

(2)

2.1.2

(3)

2.1.3

No The force of gravity acts at right angles (is perpendicular) to the Displacement of the block.

(2)

2.2.1

The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant.

(2)


2.2.2

(4)

2.2.3

Linear momentum is the product of an object's mass and its velocity.

(2)

2.2.4

The total linear momentum of an isolated system remains constant (is conserved). Die totale lineêre momentum in 'n geslote sisteem bly konstant (behoue).

(2)

2.2.5


2.2.6

(5)

2.3.1


(2)

2.3.2

OPTION 1

PKnc EEW

OR/OF

)(2

1 22

ifnc vvmW )( if hhmg

)05,5)(300(2

1 2 ncW )023)(8,9(300 J5,15772

t

WP

60

5,72157P

WP 63,2021

OPTION 2


t

WP

P=Fv

t

xFP

Fnet=ma OR amF net

F-Fg=ma F =m(a+g) F=300(0,66+9,8) F=3137,283 N

60

23283,3137 xP

P=1202,63 W

OPTION 3

t

WP

P=Fv

t

xFP

Fnet=ma OR amF net

F-Fg=ma

F =m

g

x

vv if

2

22

F =300

8,9

232

0)5,5( 22

F=300(0,66+9,8) F=3137,283 N

60

23283,3137 xP

P=1202,63 W

(8)

xavv if 222

)23(20)5,5( 25 a

a = 0,65761 m∙s-2

ANY ONE

ANY ONE

ANY ONE


QUESTION 3 3.1A conservative force as a force for which the work done in moving an object between two points is independent of the path taken. (2) 3.2. 𝑊𝑔 = 𝐹𝑔∆𝑥𝑐𝑜𝑠0𝑜

𝑊𝑔 = 𝑚𝑔∆𝑥𝑐𝑜𝑠0𝑜

𝑊𝑔 = (1,8)(9,8)(0,5)(1)

𝑊𝑔 = 8,82 𝐽 (3)

3.3 𝑊𝑛𝑐 = ∆𝐸𝑘 + ∆𝐸𝑝

𝑊𝑓 + 𝑊𝑇1 + 𝑊𝑇2 = (𝐸𝑘𝑓 − 𝐸𝑘𝑖) + (𝐸𝑝𝑓 − 𝐸𝑝𝑖)

𝑓𝑓∆𝑥𝑐𝑜𝑠1800 + 𝑇1∆𝑥𝑐𝑜𝑠00 + 𝑇2∆𝑥𝑐𝑜𝑠1800 = [1

2(𝑚1 + 𝑚2)] 𝑣𝑓

2 − 0) + (0 − 𝑚1𝑔ℎ𝑖)

𝑓𝑓∆𝑥𝑐𝑜𝑠1800 = [1

2(𝑚1 + 𝑚1)] 𝑣𝑓

2 − 0) + (0 − 𝑚2𝑔ℎ𝑖)

𝜇𝑘𝑁∆𝑥𝑐𝑜𝑠1800 = [1

2(𝑚1 + 𝑚1)] 𝑣𝑓

2 − 0) + (0 − 𝑚2𝑔ℎ𝑖)

𝜇𝑘𝑚1𝑔∆𝑥𝑐𝑜𝑠1800 = (1

2𝑚1 + 𝑚1) 𝑣𝑓

2 − 0) + (0 − 𝑚2𝑔ℎ𝑖)

[(𝜇𝑘)(2,40)(9,8)(0,50)(−1)] = [1

2(2,40 + 1,80)] (0,95)2

− (1,80)(9,8)(0,5)

(𝜇𝑘)(−11,76) = 1,90 − 8,82

(𝜇𝑘)(−11,76) = −6,92

(𝜇𝑘) = 0,59 Option 2

𝑊𝑛𝑒𝑡 = ∆𝐸𝑘 𝑊𝑓 + 𝑊𝑇1 + 𝑊𝑇2 + 𝑊𝑔 = (𝐸𝑘𝑓 − 𝐸𝑘𝑖)

𝑓𝑓1∆𝑥𝑐𝑜𝑠1800+𝑓𝑔2∆𝑥𝑐𝑜𝑠00 = (1

2𝑚1 + 𝑚2) 𝑣𝑓

2 − 0)

𝑓𝑓∆𝑥𝑐𝑜𝑠1800 + 𝑚2𝑔∆𝑥 = [1

2(𝑚1 + 𝑚1)] 𝑣𝑓

2 − 0)

𝜇𝑁∆𝑥𝑐𝑜𝑠1800 + 𝑚2𝑔∆𝑥 = [1

2(𝑚1 + 𝑚1)] 𝑣𝑓

2 − 0)

𝜇𝑚1𝑔∆𝑥(−1) + 𝑚2𝑔𝑑 = [1

2(𝑚1 + 𝑚1)] 𝑣𝑓

2 − 0)

[(𝜇𝑘)(2,40)(9,8)(0,50)(−1)]+ 8,82 = [1

2(2,40 + 1,80)] (0,95)2

(𝜇𝑘)(−11,76) = 1,90 − 8,82 (𝜇𝑘)(−11,76) = −6,92 (𝜇𝑘) = 0,59 (6)

[11]

Anyone

Any one

Any one

Any

one

Anyone



LESSON 9: Self-assessment on law of conservation of mechanical energy.

QUESTION 1

Two blocks A and B of mass 8 kg and 24 kg respectively are connected by a light inextensible string that passes over a frictionless pulley as shown below. The masses of the pulley and the string may be ignored.

The system is released from rest while block A is on the floor and B is a distance of 1,6 m above the floor.

1.1 State the law (principle) of conservation of mechanical energy in words. (2)

1.2 Draw a labelled free-body diagram for block B while it moves

downwards.

(2)

1.3 Classify the forces acting on block B into conservative and non-

conservative. Give a reason for each answer.

(4)

1.4 Calculate the work done by the gravitational force on block B. (3)

1.5 Use the principle of conservation of mechanical energy to calculate

the:

1.5.1 speed of block B when as it reaches the floor. (6)

1.5.2 maximum height reached by block A. (5)

[22]


MEMO

1.1

The total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant.

(2)

1.2

(2)

1.3 Gravitational force (𝐹 g) is conservative. It is a force for which the work done in moving an object between two points is independent of the path taken.

Tension force (�⃗� ) is non-conservative. It is a force for which the work done in moving an object between two points depends on the path taken.

(4)

1.4 𝑊 = 𝐹∆𝑥𝑐𝑜𝑠𝜃 𝑊 = 𝑚𝑔∆𝑥𝑐𝑜𝑠𝜃 𝑊 = 24 × 9,8 × 1,6 × 𝑐𝑜𝑠 0𝑜

𝑊 = 376,32 𝐽 OR 𝑊 = −∆𝐸𝑝

𝑊 = −(𝐸𝑃(𝑓) − 𝐸𝑃(𝑖))

𝑊 = −(𝑚𝑔ℎ𝑓 − 𝑚𝑔ℎ𝑖)

𝑊 = −(24 × 9,8 × 0 − 24 × 9,8 × 1,6) 𝑊 = 376,32 𝐽

(3)

1.5

1.5.1 ∆EM=0 EMi=EMf

(Ek(A)+Ek(B))i+(EP(A)+EP(B))i

= (Ek(A)+Ek(B))f+(EP(A)+EP(B))f

(1

2mAvA

2 +1

2mBvB

2)i

+(mAghA+mBghB)i=

(1

2mAvA

2 +1

2mBvB

2)f

+(mAghA+mBghB)f

(1

2mA(0)

2+

1

2mB(0)

2)i

+(mAg×0)+mBghB)i= (1

2mAvA

2 +1

2mBvB

2)f

+(mAghA+mBg×0)f

OR/OF

�⃗�

𝐹 𝑔

Any oner

Any one

Any one/


(𝑚𝐵𝑔ℎ𝐵)𝑖 = (1

2𝑚𝐴𝑣𝐴

2 +1

2𝑚𝐵𝑣𝐵

2)𝑓+ (𝑚𝐴𝑔ℎ𝐴)𝑓

𝑣𝐴 = 𝑣𝐵 = 𝑣 ℎ𝐴 = ℎ𝐵 = ℎ

(𝑚𝐵𝑔ℎ𝐵)𝑖 = (1

2𝑚𝐴𝑣

2 +1

2𝑚𝐵𝑣2)

𝑓+ (𝑚𝐴𝑔ℎ)𝑓

OR

(𝑚𝐵𝑔ℎ𝐵)𝑖 − (𝑚𝐴𝑔ℎ)𝑓 = 𝑣2

2(𝑚𝐴 + 𝑚𝐵)

OR

2𝑔ℎ(𝑚𝐵 − 𝑚𝐴) = 𝑣2(𝑚𝐴 + 𝑚𝐵) 2 × 9,8 × 1,6(24 − 8) = 𝑣2(8 + 24) 501,76 = 𝑣2(32)

𝑣2 =501.76

32

𝑣 = 3,96 𝑚 ∙ 𝑠−1

(5)

OR

𝑣2 =2𝑔ℎ(𝑚𝐵−𝑚𝐴)

𝑚𝐴+𝑚𝐵

𝑣2 =2 × 9,8 × 1,6(24 − 8)

(8 + 24)

𝑣2 =501,76

32

𝑣 = 3,96 𝑚 ∙ 𝑠−1

Any one


1.5.2 ∆𝐸𝑀 = 0 𝐸𝑀𝑖 = 𝐸𝑀𝑓 (𝐸𝐾𝐴 + 𝐸𝑃𝐴)𝑖 = (𝐸𝐾𝐴 + 𝐸𝑃𝐴)𝑓

(1

2𝑚𝐴𝑣𝐴

2+𝑚𝐴𝑔ℎ𝐴)𝑖= (

1

2𝑚𝐴𝑣𝐴

2+𝑚𝐴𝑔ℎ𝐴)𝑓

(1

2𝑚𝐴𝑣𝐴

2+𝑚𝐴𝑔ℎ𝐴)𝑖= (0+𝑚𝐴𝑔ℎ𝑚𝑎𝑥𝐴)𝑓

OR/OF

(1

2𝑚𝐴𝑣𝐴

2+𝑚𝐴𝑔ℎ𝐴)𝑖= 𝑚𝐴𝑔ℎ𝑚𝑎𝑥

OR/OF

(1

2𝑣𝐴

2 + 𝑔ℎ𝐴)𝑖= 𝑔ℎ𝑚𝑎𝑥

1

2(3,96)2

+ (9,8 × 1,6 = (9,8)ℎ𝑚𝑎𝑥

ℎ𝑚𝑎𝑥 = 2,40 𝑚

(5)

[22]

OR/OF

ℎ𝑚𝑎𝑥 =𝑣2

2+𝑔ℎ𝐴

𝑔

ℎ𝑚𝑎𝑥

=

(3,96)2

2 + (9,8)(1,6)

9,8

ℎ𝑚𝑎𝑥 = 2,40 𝑚

Any one

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