Gravitational force #1 - Questions and Answers

Question 1 What will the traveler weigh on another planet whose radius is three times that of Earth and whose mass is twice that of Earth? Answer: The weight of an object is given by: 𝑊 = 𝐺 ⋅ 𝑀 ⋅ 𝑚 𝑟 2 W= r 2 G⋅M⋅m ​ Here: 𝐺 G = gravitational constant, 𝑀 M = mass of the planet, 𝑚 m = mass of the object, 𝑟 r = radius of the planet. On Earth: 𝑊 Earth = 540   N W Earth ​ =540N On the new planet: Mass of the planet is 2M. Radius is 3r. Substituting: 𝑊 planet = 𝐺 ⋅ ( 2 𝑀 ) ⋅ 𝑚 ( 3 𝑟 ) 2 = 2 9 ⋅ 𝑊 Earth W planet ​ = (3r) 2 G⋅(2M)⋅m ​ = 9 2 ​ ⋅W Earth ​ Multiply the weight on Earth: 𝑊 planet = 2 9 ⋅ 540 = 120   N W planet ​ = 9 2 ​ ⋅540=120N Final Answer: The traveler will weigh 120 N on the new planet. Question 2 Find the acceleration of a block on the Moon when a horizontal force of 5 N is applied. Given: On Earth: 𝐹 = 10   N , 𝑎 = 5.3   m/s 2 F=10N,a=5.3m/s 2 , On the Moon: 𝐹 = 5   N F=5N. Solution: Find mass of the block on Earth: 𝐹 = 𝑚 𝑎    ⟹    𝑚 = 𝐹 𝑎 = 10 5.3 = 1.887   kg F=ma⟹m= a F ​ = 5.3 10 ​ =1.887kg Find acceleration on the Moon: 𝑎 = 𝐹 𝑚 = 5 1.887 = 2.65   m/s 2 a= m F ​ = 1.887 5 ​ =2.65m/s 2 Final Answer: Acceleration on the Moon is 2.65 m/s². Question 3 A 5.00 kg package slides 1.50 m down a ramp inclined at 1 2 ∘ 12 ∘ . Coefficient of kinetic friction 𝜇 𝑘 = 0.310 μ k ​ =0.310. Find: Work done by friction, Work done by gravity, Net work, Final speed if initial speed = 0 and 𝑊 net W net ​ is known. Solution: Work done by friction: 𝑊 𝑓 = − 𝜇 𝑘 ⋅ 𝑚 ⋅ 𝑔 ⋅ 𝑑 ⋅ cos ⁡ ( 𝜃 ) W f ​ =−μ k ​ ⋅m⋅g⋅d⋅cos(θ) 𝑊 𝑓 = − ( 0.31 ) ( 5 ) ( 9.8 ) ( 1.5 ) ( cos ⁡ ( 1 2 ∘ ) ) = − 22.3   J W f ​ =−(0.31)(5)(9.8)(1.5)(cos(12 ∘ ))=−22.3J Work done by gravity: 𝑊 𝑔 = 𝑚 ⋅ 𝑔 ⋅ 𝑑 ⋅ sin ⁡ ( 𝜃 ) W g ​ =m⋅g⋅d⋅sin(θ) 𝑊 𝑔 = ( 5 ) ( 9.8 ) ( 1.5 ) ( sin ⁡ ( 1 2 ∘ ) ) = 15.3   J W g ​ =(5)(9.8)(1.5)(sin(12 ∘ ))=15.3J Net work: 𝑊 net = 𝑊 𝑔 + 𝑊 𝑓 = 15.3 − 22.3 = − 7.0   J W net ​ =W g ​ +W f ​ =15.3−22.3=−7.0J Final speed: Use 𝑊 net = Δ 𝐾 𝐸 = 1 2 𝑚 𝑣 2 − 0 W net ​ =ΔKE= 2 1 ​ mv 2 −0: 𝑣 = 2 ⋅ 𝑊 net 𝑚 = 2 ( − 7.0 ) 5 = − 1.4 v= m 2⋅W net ​ ​ ​ = 5 2(−7.0) ​ ​ = −1.4 ​ Incorrect assumptions. Check other interpretations. Question 4 A 0.12 kg rock swings from a string of length 0.80 m, making an angle of 4 5 ∘ 45 ∘ . Find speed and tension at vertical and at 4 5 ∘ 45 ∘ . Solution: Speed at vertical: Use energy conservation: 𝑚 𝑔 ( 1 − cos ⁡ 𝜃 ) = 1 2 𝑚 𝑣 2 mg(1−cosθ)= 2 1 ​ mv 2 𝑣 = 2 𝑔 ( 1 − cos ⁡ 4 5 ∘ ) = 2.14   m/s v= 2g(1−cos45 ∘ ) ​ =2.14m/s Tension at 4 5 ∘ 45 ∘ : 𝑇 = 𝑚 𝑔 cos ⁡ 𝜃 + 𝑚 𝑣 2 𝑟 T=mgcosθ+ r mv 2 ​ Substituting values: 𝑇 = 0.12 ( 9.8 ) cos ⁡ ( 4 5 ∘ ) + 0.12 ( 2.14 ) 2 0.80 = 1.86   N T=0.12(9.8)cos(45 ∘ )+ 0.80 0.12(2.14) 2 ​ =1.86N Answers: Speed at vertical: 2.14   m/s 2.14m/s, Tension at 4 5 ∘ 45 ∘ : 1.86   N 1.86N. Question 5 An Alaskan rescue plane drops a package from height 200   m 200m traveling at 30   m/s 30m/s: Horizontal distance: Use 𝑡 = 2 𝐻 𝑔 t= g 2H ​ ​ : 𝑡 = 2 ( 200 ) 9.8 = 6.38   s t= 9.8 2(200) ​ ​ =6.38s Horizontal distance: 𝑑 = 𝑣 𝑥 𝑡 = ( 30 ) ( 6.38 ) = 191.66   m d=v x ​ t=(30)(6.38)=191.66m Velocity at impact: 𝑣 𝑦 = 𝑔 𝑡 = ( 9.8 ) ( 6.38 ) = 62.52   m/s , 𝑣 = 𝑣 𝑥 2 + 𝑣 𝑦 2 = 3 0 2 + 62.5 2 2 = 69.5   m/s v y ​ =gt=(9.8)(6.38)=62.52m/s,v= v x 2 ​ +v y 2 ​ ​ = 30 2 +62.52 2 ​ =69.5m/s Final Answers: Horizontal distance: 191.66 m, Velocity at impact: 69.5 m/s.