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**Question 45
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(a) Up the incline: F_{A }= mg = 0.8 x 10 = 8 N

Down the incline: F_{P}= F_{W}sin q + F_{f}= 1 x 10 x 0.57 +F_{f}= 5.7 +F_{f }Constant speed hence F_{UP}= F_{DOWN}; thus 8N = 5.7 + F_{f}or F_{f}= 2.3 N

(b) F_{N}= 1 x 10 x cos 35°

m = F_{f}/F_{N}= 3.3/(1 x 10 x cos 35° ) = 0.28

The diagram of the forces acting is as follows, where F_{W}= weight of the bob, T = tension in the string and F_{H}= the wind force.

Because the pendulum bob is at rest when held to one side, the sum of the three forces is zero; that is F_{H}+ F_{W}+ T = 0. The three vectors must form a closed triangle if this is true:

F_{W}= mg = 3 x 10^{-3}N

Using trigonometry: the wind force F_{H}= F_{W}tan 35° = 2.1 x 10^{-3}N

Using trigonometry: the tension T = F_{W}/cos37° = 3 x 10^{-3}/0.8 = 3.7 x 10^{-3}N

(a) F_{TOTAL }= F_{W}+ F

3 x 10^{4}= 2.5 x 10^{3}+ 2.5 x 10^{3}a

5000 = 2.5 x 10^{3}a

a = 2 ms^{-2}upwards

(b) s = ½ at^{2 }

F_{DOWN}= F_{W}sin q - (F_{f}+ 500 + 80 (No. people)

For one person:F = 1060 x 10 sin 30° - (0.2 x 10600 cos30° + 580)

F = 5300 - 2416 = 2884 N

a = F/m = 2884/1060 = 2.72 ms^{-2 }

For two people:F = 1120 x 10 sin 30° - (0.2 x 11200 cos30° + 640)

F = 5600 - 2580 = 3020 N

a = F/m = 3020/1120 = 2.69 ms^{-2 }

Therefore one person would go faster.

s = 290 m; m = 0.60

F_{f}= m F_{N}= m mg

F_{f}= ma

a = F_{f}/m = m mg/m = m g = 0.6 x 10 = 6 ms^{-2 }

v^{2}= u^{2}+2as

0 = u^{2}+ 2 x 6 x 290

u = 59 ms^{-1}(212 km h^{-1})

Basic relationships: F_{H}= F_{f}= F_{A}cosq ; F_{N}= F_{W}- F_{A}sinq

F_{f}= m F_{N}

F_{f}= m (F_{W}- F_{A}sinq )

F_{A}cosq = m (F_{W}- F_{A}sinq )

F_{A}x 0.74 = 0.1(750 - F_{A}0.67)

0.807 F_{A}= 75

F_{A}= 93 N

u = 20 ms^{-1}; F = -6 x 10^{3}; s = 22.0 m

a = F/m = -6 x 10^{3}/1500 = - 4.0 ms^{-2 }v^{2}= u^{2}+2as = 20^{2}+ 2 x^{-}4 x 22

v^{2}= 400 +^{-}176

v = 15 ms^{-1 }

Convert 60 kmhAt terminal velocity F^{-1}to ms^{-1}= 16.7 ms^{-1}

_{W}= ½ c r A v^{2 }½ c r A_{1}v_{1}^{2}= ½ c r A_{2}v_{2}^{2 }cancelling ½ c r from both sides

A_{1}v_{1}^{2}= A_{2}v_{2}^{2 }A_{1}v_{1}^{2}= 2A_{1}v_{2}^{2}(replacing A_{2}^{ }with 2A_{1})

v_{1}^{2}= 2 v_{2}^{2}

16.7^{2}= 2 v_{2}^{2 }278 = 2 v_{2}^{2 }v_{2}= 11.7 ms^{-1 }

m = 1.2 x 10^{6}ms^{-1}; F_{UP}= 450 x 10^{-18 }N; F_{W}^{ }= 9.11 x 10^{-31}x 10 N

F_{NET}= F_{UP}- F_{W }F_{NET}= 450 x 10^{-18 }N - 9.11 x 10^{-31}x 10 N = 4.5 x 10^{-16}N

a = F/m = 4.5 x 10^{-16}/ 9.11 x 10^{-31}= 4.9 x 10^{14}ms^{-2 }t = s/v = 30 x 10^{-3}/ 1.2 x 10^{6}= 2.5 x 10^{8}seconds

s (up) = ut + ½ at^{2}= 0 + ½ x 4.9 x 10^{14}x (2.5 x 10^{-8})^{2}= 0.153 m (15.3 cm)

The weight of the box on the incline F_{W }= mg = 14 x 10 = 140N

The weight of the hanging mass F_{W2 }= m_{2}g = 14 x 10 = 140N (also)

F_{f}+ F_{W}sinq = F_{W2 }m F_{N}+ F_{W}sinq = 140

m (140 cos 30° ) + 140 sin 30° = 140

m x 121 + 70 = 140

m = 0.58