TELEMATICS 2016 GRADE 12 PHYSICAL SCIENCES CAPS - Western Cape

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TELEMATICS 2016

GRADE 12

PHYSICAL SCIENCES CAPS

QUESTIONS ANSWERS AND STUDY TIPS

Normal Force Simultaneous equations in Physics K vs f graph in Photoelectric Effect

Volume-Volume Calculations Chemical Equilibrium Application of the Mole

GRADE 12 PHYSICAL SCIENCES CAPS BROADCASTING PROGRAM 2016 GRAAD 12 FISIESE WETENSKAPPE KABV UITSAAI-PROGRAM 2016

Day

Date

Tuesday

16 February

Wednesday 17 February

Wednesday 9 March

Time

Subject

16:00 ? 17:00 Physical Sciences

Topic Normal force

16:00 ? 17:00 Fisiese Wetenskappe Normale krag

16:00 ? 17:00 Physical Sciences

Simultaneous equations in Physics

Thursday 10 March

Tuesday 5 April Wednesday 6 April Wednesday 11 May

Thursday 12 May

Thursday Tuesday

21 July 26 July

16:00 ? 17:00 Fisiese Wetenskappe Gelyktydige vergelykings in Fisika

16:00 ? 17:00 Physical Sciences 16:00 ? 17:00 Fisiese Wetenskappe

Photo-Electric effect Foto-elektriese effek

16:00 ? 17:00 Physical Sciences 16:00 ? 17:00 Fisiese Wetenskappe 16:00 ? 17:00 Physical Sciences

Volume-Volume calculations

Volume-Volume berekeninge

Chemical equilibrium

16:00 ? 17:00 Fisiese Wetenskappe Chemiese ewewig

Tuesday

6

16:00 ? 17:00 Physical Sciences

September

Wednesday 7

16:00 ? 17:00 Fisiese Wetenskappe

September

Application of the mol Toepassing van die mol

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Content and Study Tips

LESSON 1: NORMAL FORCE AND MOTION OF CONNECTED BODIES

Study Tips: The most important document that you should consult in order to prepare successfully for the 2016 PHYSICAL SCIENCES FINAL EXAMINATION is the EXAMINATION GUIDELINES (EG) dated 2014.

DEFINITION of a NORMAL FORCE: A normal force is the force or the component of a force which a surface exerts on an object with which it is in contact, and which is perpendicular to the surface. (EG, NSC (CAPS), page 7)

UNDERSTAND THE NORMAL FORCE:

The normal force is a contact force that a surface exerts on an object in contact with it in order to counter-act its weight, as illustrated below:

Block

The paper is too weak to counter-act the weight of the block

Paper

Fig. 1

The table is strong enough to counter-act the weight of the block. The force it exerts on the block is called the normal force (FN)

Fig. 2

FN (Normal force)

EXAMPLE

Write down an equation that can be used to calculate the magnitude of the normal force (FN) acting on the block in the diagram below. Numerical values are not required.

FN

Study Tip: When a body is in

equilibrium, the algebraic sum of the

upward forces equals the algebraic sum

of the downward forces.

This is the reason why FN = Fg in this

example.

This fact will be exploited to find FN in all

Fg

the examples and activities that follow. Algebraic sum means that the sum is

not a vector sum.

ANSWER: FN = Fg

OR FN = w

OR

FN = mg

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ACTIVITY 1.1

1. Write down an equation that you can use to calculate the magnitude of the normal force (FN) acting on the block in each of the following cases. Numerical values are not required.

F

1.1

F

1.2

1.3

ANSWERS

FN

Fsin

FN

F

1.1 Fg

FN F Fsin Fg

1.2

Fg Fg cos

1.3

1.1 FN = Fg - Fsin

Resolve F into its components. The block is in equilibrium. The algebraic sum of the upward forces = the algebraic sum of the downward forces.

FN + Fsin = Fg FN = Fg - Fsin

1.2 FN = Fg + Fsin

Resolve F into its components. The block is in equilibrium. The algebraic sum of the upward forces = the algebraic sum of the downward forces.

FN = Fg + Fsin

1.3 FN = Fgcos

Resolve Fg into its components. The block is in equilibrium. The algebraic sum of the upward forces = the algebraic sum of the downward forces on the inclined plane.

FN = Fgcos

2. Calculate the magnitude of the normal force (FN) using the derived equations in 1.1, 1.2 and 1,3 of ACTIVITY 1.1, if m = 5 kg, F = 40 N and = 30o.

ANSWERS to ACTIVITY 1.1, question 2.

1.1: 29 N

1.2: 69 N

1.3: 42,43 N or 42,44 N

ACTIVITY 1.2

Application of FN, the normal force, to the motion of connected bodies:

You already know that frictional forces oppose motion of objects. In Grade 11 you learnt about static and kinetic friction.

The normal force FN is used in the equation to determine both static and kinetic friction.

NOTES: Instead of FN, N is used in the formula for static and kinetic friction. N is the preferred symbol and FN is the alternative symbol for normal force. Refer to page 25 of the EG.

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Content and Study Tips

Static friction

When static friction is a maximum Fs = ?sN

Where:

Fs is the static frictional force ?s is the co-efficient of static friction N is the normal force

Kinetic friction

When kinetic friction is a maximum Fk = ?kN

Where:

Fk is the kinetic frictional force ?k is the co-efficient of kinetic friction N is the normal force

In the following examples, we apply static and kinetic frictional forces using some of the normal forces derived in ACTIVITY 1.1.

EXAMPLES

1.1 A force of 10 N acts on a block at an angle of 30o to the horizontal and the block does not move. The co-efficient of static friction between the block and the surface is 0,1. Calculate the mass of the block.

10 N 30o

ANSWER Study Tips: You must know how to obtain the mass of the block from the given data before you can solve the problem. Follow the steps below in a problem solving strategy:

PROBLEM SOLVING STRATEGY: Find FN in terms of m using the upward and downward forces acting on the block. Then use the horizontal forces acting on the block to find the mass.

STEP 1:

Understand the question and the context: The block is in equilibrium. This means: (1) The algebraic sum of the upward forces equals the algebraic sum of the downward forces. (2) The algebraic sum of the horizontal forces pulling the block to the right equals the algebraic sum of the horizontal forces pulling the block to the left.

STEP 2:

Draw a diagram to show all the forces acting vertically and horizontally on the block.

Study Tips: What is a free body diagram? It is a diagram which is used to show all the forces that are acting on a body. The body is shown as a dot.

Fsin

FN

F

Fs

Fcos

Fg

NOTES: In a free body diagram and a force diagram, the components of forces are not shown.

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FN

F

Fs

Fg

Free body diagram for STEP 2

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Content and Study Tips

FN

F

Fs

Fg Force diagram for STEP 2

Study Tips: The components of F are not shown in a free body diagram or a force diagram. The diagram drawn in STEP 2 is not a force diagram or a free body diagram. Study Tips: The forces shown in the free body- and force diagrams above are vectors. Vectors are shown as arrows. The head of an arrow shows its direction and its length its relative magnitude. The tails of each arrow touch the dot or the block.

Study Tips: N in ?sN is another way of writing FN. They have the same meaning.

Study Tips: 1. If you do not put in the arrow heads you will lose the marks for each force. 2. If you do not label each force you will lose the mark for the forces. 3. If you have additional forces that should not be included one mark is deducted per additional force.

STEP 3: STEP 3:

Write down equations for the vertical and horizontal forces acting on the block, substitute the numerical values into them, and simplify:

Vertical forces: FN = Fg ? Fsin30o = m(9,8) ? (10)(0,5) = 9,8m ? 5 ... (1)

Horizontal forces: Fs = sN = Fcos30o

... (2)

Substitute FN = 9,8 ? 5 into (2): (0,1)(9,8m ? 5) = (10)(0,866) = 8,66 ... (3)

Now use equation (3) to calculate the mass m:

(0,1)(9,8m ? 5) = (10)(0,866) = 8,66

both sides by 0,1: 9,8m ? 5 = 86,6

EXAMPLES

m = 91,6 = 9,35 kg 9,8

2 kg A

1.2 A block of mass 2 kg is released at point A at the top of an inclined plane and it travels to point B at the bottom. The length of AB is 5 m. If the coefficient of kinetic friction between the surface of the block and the inclined plane is 0,05, calculate the velocity of the block when it reached B.

30o B

Study Tips: Follow a similar procedure as in Example 1.1. HINT: Use the Work-Energy Theorem.

ANSWER: 6,69 m.s-1

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Content and Study Tips

ACTIVITY 1.3 Solve the following problem and discuss your answers in class with your teacher:

1. A 5 kg block, resting on a rough horizontal

table, is connected by a light inextensible string passing over a light

60 N

frictionless pulley to another block of mass 2 kg. The 2 kg block hangs vertically

10o

as shown in the diagram below.

A force of 60 N is applied to the 5 kg block at an angle of 10o to the horizontal,

causing the block to accelerate to the left.

5 kg

2 kg

The coefficient of kinetic friction between the 5 kg block and the surface of the table is 0,5. Ignore the effects of air friction.

1.1

Draw a labelled free-body diagram showing ALL the forces acting on the 5 kg

block.

(5)

1.2

Calculate the magnitude of the:

1.2.1 Vertical component of the 60 N force

(2)

1.2.2 Horizontal component of the 60 N force

(2)

1.3

State Newton's Second Law of Motion in words.

(2)

Calculate the magnitude of the:

1.4

Normal force acting on the 5 kg block

(2)

1.5

Tension in the string connecting the two blocks

(7)

[20]

ANSWERS

1.1

1.2.1: 10,42 N 1.2.2: 59,09 N 1.3: See EG pg. 7

Fapp

FN f T

Fg

1.4: 38,58 N 1.5: 25,37 N

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Content and Study Tips

ACTIVITY 1.4

Application of the normal force FN in the lift problem

There are two cases:

FN

Case 1

Case 2 FN

Fg

Fg

In both cases the object experiences a normal force FN that acts on it. Case 1 is the same as the example on page 3. In case 2, the bathroom scale registers a reading that is a reaction force to the weight of the object. According to Newton's 3rd Law, this reaction force has the same magnitude as the weight but is opposite in direction. This means the reaction force is FN.

Case 1: An object is in direct contact with a lift floor

Case 2: An object is placed on a bathroom scale that is in direct contact with a lift floor

NOTES: The object can be a person standing in a lift. In ACTIVITY 1.4 we will use case 1 to derive the various equations. For case 2, the equations will be the same. FRes and Fnet have the same meaning. We apply Newton's 2nd Law viz. FRes = ma where applicable.

Study Tips: If you accelerate upwards in a lift you feel heavier. If you accelerate downwards you feel lighter. If the lift cable breaks you will feel weightless.

Case 1.1: Lift is stationary. FN = Fg

(Algebraic sum of upward forces equals algebraic sum of downward forces.)

Case 1.2: Lift moves upwards at constant velocity.

FN = Fg Proof: FRes = FN ? Fg = ma But a = 0 because v is constant. Thus FN ? Fg = 0 FN = Fg

Case 1.3: Lift moves upwards at constant acceleration.

FN = ma + Fg Proof: FRes = ma. FN > Fg

FN ? Fg = ma FN = ma + Fg

Case 1.4: Lift moves downwards at constant velocity.

FN = Fg Proof: FRes = Fg ? FN = ma But a = 0 because v is constant.

Thus Fg ? FN = 0 Fg = FN

Case 1.5: Lift moves downwards at constant acceleration.

FN = Fg - ma Proof: FRes = ma. Fg > FN

Fg ? FN = ma FN = Fg - ma

Case 1.6 Lift slows down as it moves downwards.

FN = Fg + ma Proof: FRes = ma, Fg > FN, a Fg a < 0

FN ? Fg = m(-a) FN = Fg - ma

Case 1.8: Lift is free falling FN = 0 N

Proof: FRes = Fg = mg Lift and object are in free fall. Lift floor does not exert an upward force on the object.

FN = 0 N

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