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Year 12 Physics Gradstart

Year 12 Physics Gradstart. 2.1 Basic Vector Revision/ Progress Test. You have 20 minutes to work in a group to answer the questions on the Basic Vector Revision/Progress Test Worksheet. 2 .2 Vector Components Part 2. 2.2 Vector Components Part 2.

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Year 12 Physics Gradstart

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  1. Year 12 Physics Gradstart

  2. 2.1 Basic Vector Revision/Progress Test You have 20 minutes to work in a group to answer the questions on the Basic Vector Revision/Progress Test Worksheet

  3. 2.2 Vector Components Part 2

  4. 2.2 Vector Components Part 2 1a) What is the northerly and easterly component of the velocity below E vN =35sin28 = 16.43150ms-1 vE =35cos28 = 30.90317ms-1

  5. 2.2 Vector Components Part 2 1b) What is the northerly and easterly component of the velocity below vE =– 12sin15 = – 3.10583ms-1 E vN =12cos15 = 11.59111ms-1

  6. 2.2 Vector Components Part 2 2. Work out the horizontal and vertical components of the force below: x y Fy =6sin30 = 3.0N Fx =6cos30 = 5.19615N

  7. 2.2 Vector Components Part 2 3. Work out the horizontal and vertical velocity components of the golfball below: x y vy =30sin60 = 25.98076ms-1 vx =30cos60 = 15ms-1

  8. 2.2 Vector Components Part 2 4(a) Work out the weight into the slope and down the slope. x y Wy =15cos22 = 13.90776N 22o W=mg = 15N Wx =15sin22 = 5.61910N

  9. 2.2 Vector Components Part 2 4(b) What is the acceleration down the slope if it is frictionless. x y In x direction a = ? Fnet = 5.61910N m = 1.5kg Fnet = a = a = a= 2.80955 a 2.8 ms-2 Wy =15cos22 = 13.90776N W=mg = 15N 22o Wx =15sin22 = 5.61910N

  10. 2.2 Vector Components Part 2 4(c) What is the normal reaction force on the mass. x N y In y direction N = Wy N = 13.90776N N  14N Wy =15cos22 = 13.90776N W=mg = 15N 22o Wx =15sin22 = 5.61910N

  11. 2.2 Vector Components Part 2 5(a) Work out an expression for the net force down the frictionless slope below. x y Fx =mgsin  W=mg Wx =mgsin

  12. 2.2 Vector Components Part 2 5(b) What is the formula for the acceleration of a mass down a frictionless slope? x y In x direction a = ? Fnet = mgsin m = m Fnet = a = a= g sin Fx =mgsin  W=mg Wx =mgsin

  13. 2.2 Vector Components Part 2 6If the surface below is frictionless, how long will it take the mass to reach the base of the slope if it starts from rest? t = ? u = 0 x = 0.4ma = g sin20o = 10sin20o = 3.40201 x = ut + ½ at2 0.4 = ½ × 3.40201 × t2 0.233904 = t2 0.483637 = t t = 0.48m A Save in memory A of your calculator

  14. 2.2 Vector Components Part 2 7 If Frmax = 4.0Nand a 15N is applied to the 3.0kg block at 30o to the horizontal (a) What is the acceleration on the mass? x y It is not worth saving 7.5 in the memory since it is so easy to enter in the calculator Fy =15sin30 = 7.5N Frmax = 4.0N In x direction Fnet = Fx - Fr ma = 12.99038 - 4 3a = 8.99038 a = 2.99679 a  3.0 ms-2 Fx =15cos30 = 12.99038N B

  15. 2.2 Vector Components Part 2 7 If Frmax = 4.0Nand a 15N is applied to the 3.0kg block at 30o to the horizontal (b) What is the normal reaction on the mass? x y N Fy =15sin30 = 7.5N Frmax = 4.0N In y direction N + Fy= W N + 7.5 = 30 N = 22.5N N  23N Fx =15cos30 = 12.99038N W =3 × 10 = 30N

  16. 2.2 Vector Components Part 2 8The 2.0kg mass below is held at rest on the 40o slope below by friction. What is the magnitude of the friction holding the mass? x y In x direction Fr = Wx Fr = 12.85575 N Fr 13N Fr 40o Wx =20sin40 = 12.85575 N W=mg = 20N

  17. 2.2 Vector Components Part 2 9What will be the acceleration of the 1.6kg block on the 30o slope below if Frmax = 3.0N? x y In x direction Fnet= Wx– Fr ma = 8 – 3 1.6a= 5 a = 3.125 a 3.1 ms-2 Frmax = 3.0N 40o Wx =16sin30 = 8 N W=mg = 16N

  18. 2.2 Vector Components Part 2 10 If Frmax = 2.0Nand a 12N is applied to the 1.0kg block at 30o to the horizontal (a) What is the acceleration on the mass? x y In x direction Fnet = 12 – Wx – Fr ma = 12 – 2 – 5 1a = 5 a = 5.0 ms-2 Frmax = 2.0N Wy =10cos30 = 8.66025N A 30o Wx =10sin30 = 5 N W=mg = 10N

  19. 2.2 Vector Components Part 2 10 If Frmax = 2.0Nand a 12N is applied to the 1.0kg block at 30o to the horizontal (b) What is the normal reaction on the mass? x y N In y direction N = 8.66025N N  8.7N Frmax = 2.0N Wy =10cos30 = 8.66025N A 30o Wx =10sin30 = 5 N W=mg = 10N

  20. 2.2 Vector Components Part 2 11 If Frmax = 8.0Nand a 20N is applied to the 3.0kg block at 30o to the horizontal What is the acceleration on the mass? x y In x direction The force up the slope is not enough to overcome friction and the weight force down the slope (max = 23N) so the mass will be stationary. Frmax = 8.0N 30o Wx =30sin30 = 15 N W=mg = 30N

  21. 2.2 Vector Components Part 2 12 A 55kg cyclist is riding up a 4.0o slope at a constant speed of 3.0ms-1? (a) What is the net force on the cyclist? Since the cyclist is travelling at constant speed the net force is zero.

  22. 2.2 Vector Components Part 2 12A 55kg cyclist is riding up a 4.0o slope at a constant speed of 3.0ms-1 ? (b) What is the horizontal force of the rear wheel propelling the cyclist up the hill if there is no rolling resistance (friction)? x y F In x direction since Fnet= 0 F = 38.36606 N F 38N 4o Wx =550sin4 = 38.36606 N W=mg = 55 × 10 = 550N

  23. 2.5 Classic Lift Problems

  24. 2.5 Classic Lift Problems 1If the lift below is moving vertically at a constant speed of 3.0 ms-1: (a) Draw a force diagram for the 50kg mass. T kg W = 500N

  25. 2.5 Classic Lift Problems 1If the lift below is moving vertically at a constant speed of 3.0 ms-1: (b) What is the Normal Reaction on the mass? Since the lift is travelling at constant speed the net force is zeroand so T = W T= 500N T kg W = 500N

  26. 2.5 Classic Lift Problems 1If the lift below is moving vertically at a constant speed of 3.0 ms-1: (c) How much would the mass register on a set of scales? Since apparent weight = T Apparent Mass = Apparent Mass = Apparent Mass = 50kg T kg W = 500N

  27. 2.5 Classic Lift Problems 2If the lift is accelerating upwards at 2.0ms-2: (a) Draw a force diagram for the 50kg mass. T kg W = 500N

  28. 2.5 Classic Lift Problems 2 If the lift is accelerating upwards at 2.0ms-2 : (b) What is the Normal Reaction on the mass? a = 2.0ms-2 + Fnet = T – W ma = T - 500 50 × 2 = T – 500 100 = T – 500 600 = T T = 600N T kg W = 500N

  29. 2.5 Classic Lift Problems 2 If the lift is accelerating upwards at 2.0ms-2 : (c) How much would the mass register on a set of scales? a = 2.0ms-2 + Since apparent weight = T Apparent Mass = Apparent Mass = Apparent Mass= 60kg T kg W = 500N

  30. 2.5 Classic Lift Problems 3If the lift is moving upwards and decelerates at 3.0ms-2: (a) Draw a force diagram for the 50kg mass. T kg W = 500N

  31. 2.5 Classic Lift Problems 3 If the lift is moving upwards and decelerates at 3.0ms-2 : (b) What is the Normal Reaction on the mass? + Fnet = T – W ma = T - 500 50 × -3 = T – 500 -150 = T – 500 450 = T T = 450N T a = 3.0ms-2 kg W = 500N

  32. 2.5 Classic Lift Problems 3 If the lift is moving upwards and decelerates at 3.0ms-2 : (c) How much would the mass register on a set of scales? + Since apparent weight = T Apparent Mass = Apparent Mass = Apparent Mass= 45kg T a = 3.0ms-2 kg W = 500N

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