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SECONDARY SCHOOL IMPROVEMENT PROGRAMME 2016 GRADE 12 PHYSICAL SCIENCES (Mechanics) Gauteng department of education LEARNERS SUMMARY NOTES AND ASSESSMENT QUESTIONS
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### Transcript of SECONDARY SCHOOL IMPROVEMENT PROGRAMME 2016

Microsoft Word - LEARNER CAMP MATERIAL MEMORANDUM (2016).docxGRADE 12
CAMP MATERIAL MEMORANDUM
DEFINITIONS
1. WORK - The work done on an object by a constant force F is F Δx cos θ , where F is the magnitude of the force, Δx the magnitude of the displacement and θ the angle between the force and the displacement
2. POTENTIAL ENERGY – Energy of an object due to its position in the field,in this case the gravitational field
3. KINETIC ENERGY -The energy of an object due to its motion
4. MECHANICAL ENERGY – The sum of the potential energy and kinetic energy
5. PRINCIPLE OF CONSERVATION OF ENERGY - The total mechanical energy
(sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. (A system is isolated when the resultant/net external force acting on the system is zero.)
6. WORK 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 = Δ K = Kf - Ki.
7. CONSERVATIVE FORCES - 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 are gravitational force, the elastic force in a spring and electrostatic forces (coulomb forces).
8. NON-CONSERVATIVE 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 are frictional force, air resistance, tension in a chord, etc.
9. POWER - Power is the rate at which work is done or energy is expended
QUESTION 1
1.1 D 1.2 B 1.3 A 1.4 B 1.5 A 1.6 C Do not show the weight, w, and a component
of the weight at the same time
1.7 A Mechanical energy ( U + K ) is conserved
1.8 C ΔU = m g Δh = ( 10 )( 2 ) = 20 J ΔK ( due to net force ) = Fnet Δx = ( 15 – 10 ) x 2 = ( 5 )( 2 ) = 10 J
1.9 C Ep at top = Ek at bottom = ½ m v 2 = ½ x 2 x 12 2 = 144 J 1.10 D W = F Δx 1.11 ( C ) W = Ep + Ek = m g h + ½ m v 2
1.12 C
1.13 B Ep = m g h increases with height
1.14 D W = ΔEp = m g Δh = ( 40 )( 9,8 )( 6 ) = 2 352 J
1.15 B Total energy remains constant
1.16 C Fx < Fy as incline X is less steep and requires less force. But W = m•g•h is the same for both. 1.17 B
WNET =FNET COS ΔX
= 31.2 N
1.19 C ΔEp = m g Δh where Δh is the vertical height gained Δh = 400 – 350 = 50 m
1.20 B
QUESTION 2
2.1 Ep = m g h
= ( 40 )( 9,8 ) ( 3 ) = 1 176 J 2.2 Ek = ½ m v 2 = ½ ( 40 )( 2 )2
= 80 J 2.3 Energy is used to compensate for air friction and friction with the slide surface. Some energy appears as noise energy and thermal kinetic energy due to friction. Some energy is also used to raise the internal energy of the surroundings
QUESTION 3

3.2 Fg = m g = ( 51,02 ) ( 9,8 ) = 500 N Fnet = Fapplied + Ffriction + Fgravity
= 3 000 600 500 = 1 900 N upwards
!".!"
= 7448,06 m.s-1
∴ v f = 86,3 m•s–1 QUESTION 4 4 .1 W = Fhorizontal Δx cos θ = ( 80 cos 30° )( 4 ) ( cos 0º ) = 277,13 J
!"
= !"##
!!  !"#.!"
= 0,30 5.3
5.4 W f+ WF = Wnett (3 500)(20)cos 180 + (5 000)(20)cos360 = Wnett -70 000 + 80 901,699 = W net Wnet = 10 901,70 J
Ff                                                                                                                                            F
!""
!""!!"##
!"
= – 9 600 N 6.5 Fnet = 150 50 = 100 N Wnet = Fnet Δx cos θ = ( 100 )( 50 ) cos 0º = ( 100 )( 50 )( 1 ) = 5 000 J Wnet = ½ m ( v f2 v i2 )
!

QUESTION 8 8.1 Ep = m g h = ( 60 ) ( 9,8 ) ( 80 ) = 47 040 J 8.2 Σ MEi = Σ MEf m g hi + ½ m vi2 = m g hf + ½ m vf2 ( 60 )( 9,8 )( 80 ) + ½ ( 60 )( 02 ) =( 60 )( 9,8 )( 0 ) + ½ ( 60 ) vf2
47 040 + 0 = 0 + 30 vf2 vf2 = !"#!#
!"
= 1 568 Vf= 39,6 m•s–1
8.3 Ep + Ek = a constant = 47 040 J 8.4 47 040 J 8.5 Ep at 35 m = m g h = ( 60 )( 9,8 )( 35 ) = 20 580 J Ep + Ek = 47 040 Ek = 47 040 – 20 580 = 26 460 J 26 460 = ½ ( 60 ) v 2
V2 =!"#"\$
=882 V = 882 =29,7 m•s –1
8.6 W = F Δx cos θ = F Δx cos 180º Friction force acts against the motion. 47 040 = Ffriction ( 0,96 ) ( – 1 )
Ffriction = !!"#!# !.!"
= – 49 000 N Friction force is negative as it opposes the motion.
QUESTION 9 9.1 Friction is present, thus it is not an isolated system. 9.2 Wfriction = Ffriction Δx cos θ = ( 2,5 ) ( 6 ) ( cos 180º ) = ( 2,5 ) ( 6 ) ( –1 ) = – 15 J 9.3 + 15 J 9.4 Useful work changes speed or height Wnc = ΔEp + ΔEk = ( m g h f – m g h i ) +( ½ m v f 2 – ½ m v i2 ) = ( 4 ) ( 9,8 ) ( 3 ) – 0 + ½ ( 4 ) ( 2 ) 2 – ½ ( 4 ) ( 5 ) 2 = 117,6 + 8 – 50 = 75,6 J 9.5 The car must do the real work required to change its height and speed as well as compensating for friction. Wtotal = 75,6 + 15
= 90,6 J
QUESTION 10
10.1 W = ΔK + ΔU = ½m(vf² – vi²) + mg(h2 – h1) = ½(90)(2² – 0²) + (90)(9,8)(37) = 32 814 J
10.2 P = ! !
QUESTION 11
11.1
11.2 The net (total) work done (on an object) is equal to the change in kinetic energy (of the object). 11.3
OPTION 1 WNET = ΔK Any one WFg+ WFT + WFf + WFN = ΔK FgΔxcos + FfΔxcos + FTΔxcos + 0 = 450 (50)(9,8)Δxcos60° + 50Δxcos180° + 300Δxcos180° = 450 (245 – 50 – 300) Δx = 450 Δx = 4,29 m OPTION 2 WNET = ΔK WFg//+ WFf + WFT + WFN + WFg = ΔK Any one/Enige een Fg//Δxcos + FfΔxcos + FTΔxcos + 0 + 0 = 450 (50)(9,8)sin30°Δxcos0° + 50Δxcos180° + 300Δxcos180° = 450 (245 – 50 – 300) Δx = 450 Δx = 4,29 m OPTION 3 Fnet = FT + Ff + W// = 300 + 50 – mg sin 30° = 300 + 50 – (50)(9,8) sin30° Both formulae = 105 Fnet Δxcos = ΔK 105 Δxcos0° = 450
FT
Ff
F
ü
ü
ü
ü
Δx= 4,29 m 11.4 fk = k N =µ k (mg cos ) Any ONE 50 = µ k (50)(9,8)cos 30° µ k = 0,12
QUESTION 12
WG = (10 x 9, 8) (5) Cos 1100 ü
= –167, 59 Jü
= 167, 59 J ü
12.3 Is a force for which the work done in moving an object between two points depends on the pathtaken. üü E.g friction, air resistance, tension in a chord (any one) ü
12.4 Is not conserved, because there is friction üü
12.5.1 OPTION 1
Σ Fy = 0:
N + FG = 0ü
FN = (10) (9, 8) Cos 200 ü
= 92,1 Nü
∴ Work done by friction = fk x Cos θ
W = (36, 8) (5) Cos 1800ü
= – 184 J ü
= 36,8N ü
∴ Work done by friction/ arbeid verrig deur wrywing = fk x Cos θü
W = (36, 8) (5) Cos 1800 ü
= – 184 J ü
= (100) (5) Cos00 ü
2
(–168) + (–184) + (0) + (500) ü = ½ (10) vf 2 – ½ (10) ((1, 5)2ü
vf = 5,64 m.s–1ü
QUESTION 13
13.1 The net (total) work done (on an object) is equal to the change in kinetic energy (of the object).
13.2
=(800)(12)(1)
QUESTION 14
14.1 The net (total) work done (on an object) is equal to the change in kinetic energy (of the object).
14.2

15.2 No,it is not an isolated system/there is no external force acting on the
object
15.5 B TO A
QUESTION 16
16.1 contact force – work done is dependent on the path taken.
16.2 force of air friction. Tension of the cable 16.3
[one mark for each force]
16.4 The net work done by an object is equal to the change in its kinetic energy OR The work done by the net force is equal to the change in the objects kinetic Energy 16.5
Wnet = Ek Fnet xcosØ = ½mvf2- ½mvi2 Fnet(20)cos1800 = ½(50)(02) - ½(50)(22) Fnet(- 20) = -100 N ma = + 5 (50) a = 5 a = 0,1 m·s-2 upwards
QUESTION 17
17.1 A force for which the work done in moving an object between two points depends on the path taken/is not independent of the path taken. 17.2 0J
17.3 Fg// - ( f + F) = 0 (Accept other correct symbols) OR F = mg sinθ – fk OR F = mgsinθ – 266 F = [100(9,8) sin 25 o] – 266 = 414,167- 266 F = 148,17 N NOTENo mark for diagram 1 mark for use of any of the three formulae
17.4
QUESTION 18
18.1 The rate at which work is done. / Work done per unit time. 18.2
18.3 OPTION 1 W
∴W F
OPTION 2 W
W F
18.4 P = !
=4.63 X 104W
18.5 Smaller than Weight / gravitational force does positive work on the truck
QUESTION 19
QUESTION 20
20.1 Power is the rate of doing work or the rate at which energy is expended. üü (2)
20.2 Ep = mgh ü = (6x 104 x 9,8 x 106) üü
= 6.23 x107 Jü (4)
=2.08 x 105Wü (4)
20.4.1 Vf 2 =Vi
Vf 2 = 02 + 2(9.8)106ü
19.1
19.2
v = 46.04 ms-1 x 3.6 = 165.76 Km.h-1 üü (5)
20.4.2 The work done by net (resultant) force on an object is equal to the change in its kinetic energy. üü (2)
20.4.3 Ek= ½ mv2 (at ground level) Ek = Wnetü = ½ x2.5x (46.04)2 ü Ekf- Eki = Fnet x
= 2650Jü 0-2650 = Fnet x0.4ü ü
Fnet = - 6625Nü (6)