SECONDARY SCHOOL IMPROVEMENT PROGRAMME 2016

GRADE 12

PHYSICAL SCIENCES (Mechanics)

Gauteng department of education

LEARNERS SUMMARY NOTES AND ASSESSMENT QUESTIONS

CAMP MATERIAL MEMORANDUM

WORK ,ENERGY AND POWER

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

W = F Δx

= 40 x 8 = 320 J

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 ϴΔX

=100COS600 X 5

= 250 J

1.18 C

Fg =mg = 58.8N

Fnet = +T +(+fr) + (-Fg)

= 100 + 10 -58.8

= 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

Ep = m g h

= kg • m • s–2 • m

= kg • m 2 • s–2

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

3.1

✓✓✓

3.2 Fg = m g ✓ = ( 51,02 ) ( 9,8 ) ✓ = 500 N✓ Fnet = Fapplied + Ffriction + Fgravity

= 3 000 ‒ 600 ‒ 500 = 1 900 N upwards

Wnet = Fnet Δx cos θ✓ = ( 1 900 ) ✓ ( 100 ) cos 0º✓ = 190 000 J✓ 3.3 Wnet = ½ m ( v f2 – v i2 ) ✓ 190 000 = ½ ( 51,02 ) ( v f2 – 0 ) ✓ v f2 = !"####

!".!"

= 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✓

QUESTION 5 5.1 A non conservative force is one for which work depends on the path taken. Friction is a good example of a non conservative force. ✓✓ 5.2 In the y-direction: FNET = Fg + FN + FAPPLIED┴ ✓ 0 = 14 700 - FN - (5000sin360 ) ✓ (substitution of all in line above/ ) FN = 11 761,07 N ✓ µ k =!"

!" ✓

= !"##

!!  !"#.!"

= 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║  

QUESTION 6 6.1 The work done on an object by a net force is equal to the change in the object's kinetic energy. ✓✓ 6.2 Wnet = ΔEk✓ Fnet Δx = ½ m ( vf 2 – vi2 ) Fnet ( 200 ) ✓ = ½ ( 100 ) ( 402 – 0 ) ✓ Fnet =!"  !!!

!""

= 400 N forwards✓ 6.3 Wnet = ΔEk Fnet Δx = ½ m ( vf 2 – vi2 ) Only a negative force causes braking – 8 400 Δx✓ = ½ ( 2 600 ) ( 102 – 302 ) ✓ – 8 400 Δx = 1 300 ( 100 – 900 ) Δx =!!"##×!""

!""!!"##

=123.83 m✓ 6.4 Wnet = ΔEk Fnet Δx = ½ m ( vf 2 – vi2 ) Fnet ( 50 ) ✓= ½ ( 2 400 ) ( 0 – 202 ) ✓ FNET = !!  !""

!"

= – 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 ) ✓

5 000 = ½ ( 10 ) ( v f2 ‒ 02 ) ✓ 5 000 = 5 v f2 v f2 = !"""

!

= 1 000 ∴ v f = 31,62 m•s‒1

✓

QUESTION 7 7.1 E p = m g h✓ = ( 20 ) ( 9,8 ) ( 2 ) ✓ = 392 J✓ 7.2 ✓✓ 7.3 Upward direction is positive. Fgravity = Fg = m g✓ = ( 100 ) ( – 9,8 ) ✓ = – 980 N✓ Fnet = Fapplied + Fg = 980 ‒ 980 ✓ = 0 N Wnet = Fnet Δx cos θ✓ = ( 0 )( 10 ) cos 0º✓ = ( 0 )( 10 )( 1 ) = 0 J✓

 

7.4 The gravitational field did negative work that transferred the same amount of energy out of the system. ✓✓

 

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 = !!✓

= !"#$%

!  !  !" ✓

= 546,9 W ✓

10.3.1 NO

10.3.2 Windmill is more environmental friendly. ✓ (Or similar)  

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 + FfΔxcosϴϴ + 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

12.1

(4)

OR

12.2 OPTION 1

Weight/  w/FG  

N  FApplied    100  N  

f  

FApplied    100  N  

Weight/  w/FG  

N  

Ff  

ü  

ü  ü  

ü  

 

Work done by gravity = F ∆x Cosθ ü

WG = (10 x 9, 8) (5) Cos 1100 ü

= –167, 59 Jü

OPTION 2

Work done = ΔEp = mgh-0ü

= mg ∆x Sin 200

= 10 x 9.8 x 5 x Sin 200ü

= 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 – mg Cos 200 = 0

FN = (10) (9, 8) Cos 200 ü

= 92,1 Nü

fk = µ FN = (0,4) (92,1) = 36,8 Nü

∴ Work done by friction = fk ∆x Cos θ

W = (36, 8) (5) Cos 1800ü

= – 184 J ü

OPTION 2 Fk = µkN

= 0,4 mg Cos θü

ü  

 

= 0,4 x 10 x 9,8 Cos 200ü

= 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

12.5.2 WF = F ∆x Cosθ ü

= (100) (5) Cos00 ü

= 500 N ü

12.5 WNET = Δ EKü

WG + Wf + WN + WF = ½ mvf2 – ½ mvi

2

(–168) + (–184) + (0) + (500) ü = ½ (10) vf2 – 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

 

13.3 Wdone = Fappliedcos ϴ∆x✓

=(800)(12)(1) ✓

= 9600 J✓

13.4

13.5 tension in the cable or air friction ✓

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

 

 

14.3

 

14.4

 

14.5  

 

14.6

 

 

 

 

QUESTION 15

15.1 OPTION 1

 

OPTION  2  

 

 

 

15.2 No✓,it is not an isolated system/there is no external force acting on the

object✓

15.3 Gravitational force /Weight✓

15.4 The force is perpendicular to the displacement✓

15.5 B TO A✓

15.6 OPTION1

 

 

OPTION 2

 

 

OPTION 3

 

 

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

net = ΔK ✓

WF

+ Wf + W

w = K

f – K

i

WF

- 8,5 x 104 + (5 000)(9,8)(55)cos180°✓ = 0✓

∴WF

= 2,78 x 106 J✓

OPTION 2 W

net = ΔK ✓

WF

+ Wf -ΔE

p = K

f – K

i

WF

- 8,5 x 104 - (5 000)(9,8)(55) ✓= 0✓

WF

= 2,78 x 106 J ✓

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)

20.3 P = W/tü = (6.23 x107 J)/300üü

=2.08 x 105Wü (4)

20.4.1 Vf2 =Vi

2 + 2a∆y ü (OPTION)

Vf2 = 02 + 2(9.8)106ü

Vf = 45.58m.s-1 ü

OPTION 2

Ek at bottom = Ep at top

19.1  

19.2  

19.3  19.4  

19.5  

19.6  

 

v = √2ghü

= √(2x9,8x106) ü

= 45.58m.s-1ü

OPTION 3

Ep= mgh ü

= 2.5x9,8x106ü

= 2597J

Ek = ½ mv2

2597 = ½ x2.5xv2

v = 46.04 m.s-1ü

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)