University:
Florida Gateway College
Course:
MAC 2312 | Calculus II
Academic year:

2024
Views:

93

Pages:

5
Author:

Ruby Bates

Question 1 Projectile motion of a quarter tossed into a dish: (a) Height of the shelf above the point where the quarter leaves your hand: Given: 𝑢 𝑦 = 6.4 m/s ⋅ sin ⁡ ( 6 0 ∘ ) = 5.5 m/s u y =6.4m/s⋅sin(60 ∘ )=5.5m/s, 𝑔 = 9.8 m/s 2 g=9.8m/s 2 , 𝑡 = 0.66 s t=0.66s. Vertical displacement equation: 𝑦 = 𝑢 𝑦 𝑡 − 1 2 𝑔 𝑡 2 y=u y t− 2 1 gt 2 𝑦 = ( 5.5 ) ( 0.66 ) − 1 2 ( 9.8 ) ( 0.66 ) 2 y=(5.5)(0.66)− 2 1 (9.8)(0.66) 2 𝑦 = 3.63 − 2.13 = 1.5 m y=3.63−2.13=1.5m Height of the shelf above the point of release: 1.5 m 1.5m. (b) Vertical component of velocity just before landing: Equation: 𝑣 𝑦 = 𝑢 𝑦 − 𝑔 𝑡 v y =u y −gt 𝑣 𝑦 = 5.5 − ( 9.8 ) ( 0.66 ) = 5.5 − 6.47 = − 0.93 m/s v y =5.5−(9.8)(0.66)=5.5−6.47=−0.93m/s Answer: Height: 1.5 m 1.5m, Vertical velocity: − 0.93 m/s −0.93m/s. Question 2 Work done by gravity on a book: Given: Mass 𝑚 = 2.0 kg m=2.0kg, Initial height ℎ 1 = 0.75 m h 1 =0.75m, Final height ℎ 2 = 2.3 m h 2 =2.3m, Gravitational force 𝐹 = 𝑚 𝑔 = ( 2.0 ) ( 9.8 ) = 19.62 N F=mg=(2.0)(9.8)=19.62N. Work done: 𝑊 = 𝐹 ⋅ 𝑑 ⋅ cos ⁡ 𝜃 W=F⋅d⋅cosθ Here, 𝑑 = ℎ 2 − ℎ 1 = 2.3 − 0.75 = 1.55 m d=h 2 −h 1 =2.3−0.75=1.55m, and cos ⁡ ( 18 0 ∘ ) = − 1 cos(180 ∘ )=−1: 𝑊 = ( 19.62 ) ( 1.55 ) ( − 1 ) = − 30.41 J W=(19.62)(1.55)(−1)=−30.41J Answer: Work done by gravity: − 30.41 J −30.41J. Question 3 Tension in the rope for an adventurer crossing between cliffs: (a) Find tension at 𝜃 = 1 3 ∘ θ=13 ∘ : Vertical force: 2 𝑇 sin ⁡ 𝜃 = 𝑀 𝑔 2Tsinθ=Mg 𝑇 = 𝑀 𝑔 2 sin ⁡ 𝜃 = ( 86.5 ) ( 9.8 ) 2 sin ⁡ ( 1 3 ∘ ) T= 2sinθ Mg = 2sin(13 ∘ ) (86.5)(9.8) 𝑇 = 847.7 0.45 = 1884 N T= 0.45 847.7 =1884N (b) Smallest angle 𝜃 θ before rope breaks: Breaking tension: 𝑇 max = 4500 N T max =4500N. Using: 𝑇 = 𝑀 𝑔 2 sin ⁡ 𝜃 , sin ⁡ 𝜃 = 𝑀 𝑔 2 𝑇 T= 2sinθ Mg ,sinθ= 2T Mg sin ⁡ 𝜃 = ( 86.5 ) ( 9.8 ) 2 ( 4500 ) = 847.7 9000 = 0.0942 sinθ= 2(4500) (86.5)(9.8) = 9000 847.7 =0.0942 𝜃 = arcsin ⁡ ( 0.0942 ) = 5.4 2 ∘ θ=arcsin(0.0942)=5.42 ∘ Answer: 𝑇 = 1884 N T=1884N, Minimum 𝜃 = 5.4 2 ∘ θ=5.42 ∘ . Question 4 Tension in the wire of a model airplane: Given: 𝑚 = 0.75 kg m=0.75kg, 𝑣 = 35.0 m/s v=35.0m/s, 𝑟 = 60.0 m r=60.0m, 𝜃 = 2 0 ∘ θ=20 ∘ . Step 1: Vertical forces: 𝑇 sin ⁡ 𝜃 = 𝑚 𝑔 ⟹ 𝑇 = 𝑚 𝑔 sin ⁡ 𝜃 Tsinθ=mg⟹T= sinθ mg 𝑇 = ( 0.75 ) ( 9.8 ) sin ⁡ ( 2 0 ∘ ) = 21.55 N T= sin(20 ∘ ) (0.75)(9.8) =21.55N Step 2: Horizontal forces: 𝑇 cos ⁡ 𝜃 = 𝑚 𝑣 2 𝑟 Tcosθ= r mv 2 𝑇 = 𝑚 𝑣 2 𝑟 cos ⁡ 𝜃 = ( 0.75 ) ( 35.0 ) 2 60.0 cos ⁡ ( 2 0 ∘ ) T= cosθ r mv 2 = cos(20 ∘ ) 60.0 (0.75)(35.0) 2 𝑇 = 15.31 0.94 = 16.28 N T= 0.94 15.31 =16.28N Answer: Tension in the wire: ≈ 45.22 N ≈45.22N. Question 5 Package dropped from a plane: (a) Horizontal distance traveled: 𝑡 = 2 𝐻 𝑔 = 2 ( 200 ) 9.8 = 6.38 s t= g 2H = 9.8 2(200) =6.38s 𝑑 = 𝑣 𝑥 𝑡 = ( 30 ) ( 6.38 ) = 191.4 m d=v x t=(30)(6.38)=191.4m (b) Final velocity: 𝑣 𝑦 = 𝑔 𝑡 = ( 9.8 ) ( 6.38 ) = 62.52 m/s , 𝑣 = 𝑣 𝑥 2 + 𝑣 𝑦 2 v y =gt=(9.8)(6.38)=62.52m/s,v= v x 2 +v y 2 𝑣 = ( 30 ) 2 + ( 62.52 ) 2 = 69.5 m/s v= (30) 2 +(62.52) 2 =69.5m/s Answer: Horizontal distance: 191.4 m 191.4m, Velocity at impact: 69.5 m/s 69.5m/s.