Acceleration Due to Gravity #6 - Questions and Answers

Question 1 You are on the roof of the physics building, 46.0 m above the ground. Your physics professor, who is 1.80 tall, is walking along the side of the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your professor's head, how far from the building should the professor be when you release the egg? Assume that the egg is in free fall. Take the free fall acceleration to be 9.80   m/s 2 9.80m/s 2 . Answer 1 Height of building ( ℎ ) = 46.0   m (h)=46.0m Height of walking person ( ℎ 1 ) = 1.2   m (h 1 ​ )=1.2m The egg's initial speed is ( 𝑣 0 𝑢 ) = 0   m/s (v 0u ​ )=0m/s. Speed of egg ( 𝑣 ) = 2 𝑔 ℎ (v)= 2gh ​ , where 𝑔 = 9.8   m/s 2 g=9.8m/s 2 : 𝑣 = 2 ( 9.8 ) ( 46 ) = 30.03   m/s . v= 2(9.8)(46) ​ =30.03m/s. Answer 2 Use the equation: 𝑆 = 𝑢 𝑡 + 1 2 𝑔 𝑡 2 . S=ut+ 2 1 ​ gt 2 . 44.2 = 0 𝑡 + 1 2 ( 9.8 ) 𝑡 2    ⟹    𝑡 2 = 3   s . 44.2=0t+ 2 1 ​ (9.8)t 2 ⟹t 2 =3s. Professor should be at a distance of 3 × 1.2 = 3.6   m 3×1.2=3.6m. Answer 3 To solve the problem, use the equation of motion for an object in free fall: ℎ = 1 2 𝑔 𝑡 2 . h= 2 1 ​ gt 2 . Rearrange to find 𝑡 t: 𝑡 = 2 ℎ 𝑔 . t= g 2h ​ ​ . Substitute ℎ = 46.0   m , 𝑔 = 9.8   m/s 2 h=46.0m,g=9.8m/s 2 : 𝑡 = 2 × 46.0 9.8 ≈ 3.06   s . t= 9.8 2×46.0 ​ ​ ≈3.06s. During this time, the professor moves 𝑑 = 𝑣 𝑡 d=vt, where 𝑣 = 1.20   m/s v=1.20m/s: 𝑑 = 1.20 × 3.06 ≈ 9.35   m . d=1.20×3.06≈9.35m. The professor should be approximately 9.35   m 9.35m away from the building. Question 2 A skier of mass 70   kg 70kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 60   m 60m up a 3 0 ∘ 30 ∘ degree slope (assumed frictionless) at a constant speed? (b) What power must a motor have to perform this task? Answer 1 Mass of skier ( 𝑚 ) = 70   kg (m)=70kg. Applied Force ( 𝐹 ) = 𝑚 𝑔 sin ⁡ 30 = 345.35   N (F)=mgsin30=345.35N. He must be pulled up 𝑑 = 60   m d=60m: 𝑊 = 𝐹 𝑑 = 𝑚 𝑔 sin ⁡ 30 ⋅ 𝑑 = 20.6   kJ . W=Fd=mgsin30⋅d=20.6kJ. The constant speed ( 𝑣 ) = 2.0   m/s (v)=2.0m/s: Power of motor   ( 𝑃 ) = 𝐹 ⋅ 𝑣 = 686.70   W . Power of motor(P)=F⋅v=686.70W. Answer 2 From work law: 𝑊 = 𝐹 𝑑 cos ⁡ 𝜃 2 , W=Fdcosθ 2 ​ , where 𝜃 2 = 0 ∘ θ 2 ​ =0 ∘ : 𝑊 = 𝑇 𝑑 = 𝑚 𝑔 sin ⁡ 30 ⋅ 𝑑 = 2.1 × 1 0 4   J . W=Td=mgsin30⋅d=2.1×10 4 J. Question 3 A weatherman carried an aneroid barometer from the ground floor to his office atop a tower. On the ground level, the barometer read 30.15   inHg 30.15inHg; topside it read 28.607   inHg 28.607inHg. Assume that the average atmospheric air density was 0.075   lb/ft 3 0.075lb/ft 3 . Estimate the height of the building. Answer 1 Let us consider the initial pressure at ground level as 𝑃 1 P 1 ​ , and the initial pressure at the top as 𝑃 2 P 2 ​ . ℎ = 𝑃 1 − 𝑃 2 𝜌 𝑔 , h= ρg P 1 ​ −P 2 ​ ​ , where 𝜌 = 0.075   lb/ft 3 ρ=0.075lb/ft 3 . Answer 2 Convert pressure: 30.15 − 28.607 = 1.543   inHg . 30.15−28.607=1.543inHg. Convert to 109.165   lb/ft 2 109.165lb/ft 2 . 109.165 = 0.075 × 𝑔 × ℎ . 109.165=0.075×g×h. Solve for ℎ h: ℎ = 109.165 0.075 ⋅ 32.2 = 1455   ft . h= 0.075⋅32.2 109.165 ​ =1455ft. Question 4 A 0.145   kg 0.145kg baseball pitched at 39.0   m/s 39.0m/s is hit on a horizontal line drive straight back toward the pitcher at 52.0   m/s 52.0m/s. If the contact time between bat and ball is 1.00 × 1 0 − 3   s 1.00×10 −3 s, calculate the average force between the bat and ball. Answer 1 Change in velocity: Δ 𝑣 = − 39.0 − 52.0 = − 91   m/s . Δv=−39.0−52.0=−91m/s. 𝐹 = 𝑚 Δ 𝑣 Δ 𝑡 = 0.145 ⋅ ( − 91 ) 1 × 1 0 − 3 = 13195   N . F= Δt mΔv ​ = 1×10 −3 0.145⋅(−91) ​ =13195N. Question 5 A student drops two metallic objects into a 120   g 120g steel container holding 150   g 150g of water at 2 5 ∘ C 25 ∘ C. One object is 240   g 240g of copper initially at 8 5 ∘ C 85 ∘ C, and the other is aluminum initially at 5 ∘ C 5 ∘ C. If the water reaches 2 5 ∘ C 25 ∘ C, find the mass of the aluminum chunk. Answer Specific heat: Copper (c) = 0.093 \, \text{cal/g} \,^\circ\text{C}, Aluminum (c) = 0.215 \, \text{cal/g} \,^\circ\text{C}. Heat lost = heat gained: 240 ⋅ 0.093 ⋅ ( 85 − 25 ) = 𝑚 ⋅ 0.215 ⋅ ( 25 − 5 ) . 240⋅0.093⋅(85−25)=m⋅0.215⋅(25−5).