Spring Potential Energy #1 - Q&A

Question 1 A proton with an initial speed of 800 , 000   m/s 800,000m/s is brought to rest by an electric field. a. Did the proton move into a region of higher potential or lower potential? b. What was the potential difference that stopped the proton? c. What was the initial kinetic energy of the proton, in electron volts? Answer 1 Given: Initial velocity of the proton 𝑣 𝑖 = 800 , 000   m/s v i ​ =800,000m/s. (a) The potential is given by the formula: 𝑉 = 𝑞 𝑚 V= m q ​ The speed of the proton at the end is: 𝑣 𝑓 = 0 v f ​ =0 As the proton has a positive charge and slows down while traveling, its potential is lower at the beginning. Hence, the proton moves towards the region of higher potential. (b) Using the law of energy conservation: 𝐾 1 + 𝑈 1 = 𝐾 2 + 𝑈 2 K 1 ​ +U 1 ​ =K 2 ​ +U 2 ​ Rewriting: 𝑞 𝑉 𝑖 + 1 2 𝑚 𝑣 𝑖 2 = 𝑞 𝑉 𝑓 + 1 2 𝑚 𝑣 𝑓 2 qV i ​ + 2 1 ​ mv i 2 ​ =qV f ​ + 2 1 ​ mv f 2 ​ Since 𝑣 𝑓 = 0 v f ​ =0: 𝑞 ( 𝑉 𝑓 − 𝑉 𝑖 ) = 1 2 𝑚 𝑣 𝑖 2 q(V f ​ −V i ​ )= 2 1 ​ mv i 2 ​ The potential difference is: Δ 𝑉 = 1 2 𝑚 𝑣 𝑖 2 𝑞 ΔV= q 2 1 ​ mv i 2 ​ ​ Substitute values: 𝑚 𝑝 = 1.673 × 1 0 − 27   kg m p ​ =1.673×10 −27 kg (mass of proton) 𝑞 = 1.602 × 1 0 − 19   C q=1.602×10 −19 C (charge of proton) 𝑣 𝑖 = 800 , 000   m/s v i ​ =800,000m/s Δ 𝑉 = 1 2 ⋅ ( 1.673 × 1 0 − 27 ) ⋅ ( 800 , 000 ) 2 1.602 × 1 0 − 19 ΔV= 1.602×10 −19 2 1 ​ ⋅(1.673×10 −27 )⋅(800,000) 2 ​ Δ 𝑉 ≈ 3 , 341.82   V ΔV≈3,341.82V (c) The initial kinetic energy of the proton is given by: 𝐾 𝑖 = 1 2 𝑚 𝑣 𝑖 2 K i ​ = 2 1 ​ mv i 2 ​ Substitute values: 𝐾 𝑖 = 1 2 ( 1.673 × 1 0 − 27 ) ( 800 , 000 ) 2 K i ​ = 2 1 ​ (1.673×10 −27 )(800,000) 2 𝐾 𝑖 = 5.3536 × 1 0 − 14   J K i ​ =5.3536×10 −14 J Convert to electron volts: 1   eV = 1.602 × 1 0 − 19   J 1eV=1.602×10 −19 J 𝐾 𝑖 = 5.3536 × 1 0 − 14 1.602 × 1 0 − 19 K i ​ = 1.602×10 −19 5.3536×10 −14 ​ 𝐾 𝑖 ≈ 3.34   MeV K i ​ ≈3.34MeV Answer Summary: (a) Proton moves to a region of higher potential. (b) Potential difference: Δ 𝑉 ≈ 3.34   kV ΔV≈3.34kV. (c) Initial kinetic energy: 𝐾 𝑖 ≈ 3.34   MeV K i ​ ≈3.34MeV. Question 2 A 2.0-kg piece of wood slides on a surface. The curved sides are perfectly smooth, but the rough horizontal bottom is 30   m 30m long and has a kinetic friction coefficient of 𝜇 𝑘 = 0.20 μ k ​ =0.20. The wood starts from rest 4.0   m 4.0m above the rough bottom. (a) Where will this wood eventually come to rest? (b) For the motion from the initial release until the piece of wood comes to rest, what is the total amount of work done by friction? Answer 1 (a) Using energy conservation: 𝑈 1 + 𝐾 1 = 𝐾 2 + 𝑈 2 + 𝑊 friction U 1 ​ +K 1 ​ =K 2 ​ +U 2 ​ +W friction ​ At the top: 𝐾 1 = 0 K 1 ​ =0, 𝑈 1 = 𝑚 𝑔 ℎ U 1 ​ =mgh. At the bottom: 𝐾 2 = 0 K 2 ​ =0, 𝑈 2 = 0 U 2 ​ =0. 𝑚 𝑔 ℎ = 𝑊 friction mgh=W friction ​ Work done by friction: 𝑊 friction = 𝜇 𝑚 𝑔 𝑑 W friction ​ =μmgd Set equal to potential energy at height: 𝑚 𝑔 ℎ = 𝜇 𝑚 𝑔 𝑑 mgh=μmgd Cancel 𝑚 𝑔 mg: ℎ = 𝜇 𝑑 h=μd Substitute values: 4.0 = 0.20 𝑑 4.0=0.20d 𝑑 = 20   m d=20m The wood will come to rest 20   m 20m from the starting point. (b) The total work done by friction is: 𝑊 friction = 𝜇 𝑚 𝑔 𝑑 W friction ​ =μmgd Substitute values: 𝑊 friction = 0.20 ⋅ 2.0 ⋅ 9.8 ⋅ 30 W friction ​ =0.20⋅2.0⋅9.8⋅30 𝑊 friction = 117.7   J W friction ​ =117.7J Answer Summary: (a) Wood comes to rest 20   m 20m from starting point. (b) Total work done by friction: 117.7   J 117.7J. Question 3 Find the tension in each cord in the figure if the weight of the suspended object is 𝑊 W. Answer 1 (a) Using Lami’s theorem: For equilibrium at the joint: 𝑇 1 sin ⁡ 3 0 ∘ = 𝑇 2 sin ⁡ 6 0 ∘ = 𝑊 sin ⁡ 9 0 ∘ sin30 ∘ T 1 ​ ​ = sin60 ∘ T 2 ​ ​ = sin90 ∘ W ​ Solve for 𝑇 1 T 1 ​ and 𝑇 2 T 2 ​ : 𝑇 1 = 𝑊 sin ⁡ 3 0 ∘ = 0.5 𝑊 T 1 ​ =Wsin30 ∘ =0.5W 𝑇 2 = 𝑊 sin ⁡ 6 0 ∘ = 0.866 𝑊 T 2 ​ =Wsin60 ∘ =0.866W Question 4 Four point charges form a square with sides of length 𝑑 d. Part A: Potential at center: 𝑉 total = 𝑘 𝑑 ( 2 𝑞 + 5 𝑞 − 3 𝑞 ) = 5 𝑘 𝑞 𝑑 V total ​ = d k ​ (2q+5q−3q)= d 5kq ​ Part C: Total electric potential energy: 𝑈 total = − 6.71 𝑘 𝑞 2 𝑑 U total ​ =− d 6.71kq 2 ​ Question 5 Two springs are used to launch a block. What is the block’s speed now? Answer 1: With two springs: 𝑣 0 = Δ 𝑥 2 𝑘 𝑚 v 0 ​ =Δx m 2k ​ ​ The speed is 2 2 ​ times the speed with one spring.