Spring Potential Energy #2 - Q&A

Question 1 A sled with a rider having a combined mass of 120   kg 120kg travels over the perfectly smooth icy hill shown in the accompanying figure. How far does the sled land from the foot of the cliff (in m)? Answer 1 The rider begins with: 𝐾 𝐸 = 1 2 𝑚 𝑣 2 = 0.5 ( 120 ) ( 22.5 ) 2 = 30 , 375   J KE= 2 1 ​ mv 2 =0.5(120)(22.5) 2 =30,375J and 𝑃 𝐸 = 𝑚 𝑔 ℎ = 120 ⋅ 9.8 ⋅ 0 = 0 PE=mgh=120⋅9.8⋅0=0 So, 𝐸 𝑡 = 𝐾 𝐸 + 𝑃 𝐸 = 30 , 375   J E t ​ =KE+PE=30,375J At the top of the hill, some of the 𝐾 𝐸 KE is turned into 𝑃 𝐸 PE: 𝑃 𝐸 = 𝑚 𝑔 ℎ = 120 ⋅ 9.8 ⋅ 11.0 = 12 , 936   J PE=mgh=120⋅9.8⋅11.0=12,936J Since total energy is conserved, 𝐾 𝐸 KE becomes: 𝐾 𝐸 = 𝐸 𝑡 − 𝑃 𝐸 = 30 , 375 − 12 , 936 = 17 , 439   J KE=E t ​ −PE=30,375−12,936=17,439J Using 𝐾 𝐸 = 1 2 𝑚 𝑣 2 KE= 2 1 ​ mv 2 , solve for 𝑣 v: 𝑣 = 2 ⋅ 𝐾 𝐸 𝑚 = 2 ⋅ 17 , 439 120 = 17.05   m/s v= m 2⋅KE ​ ​ = 120 2⋅17,439 ​ ​ =17.05m/s The sled is 11.0   m 11.0m high. The time of vertical fall is given by: ℎ = 𝑣 0 𝑡 + 1 2 𝑎 𝑡 2 h=v 0 ​ t+ 2 1 ​ at 2 11 = 0 + 1 2 ( 9.8 ) 𝑡 2 11=0+ 2 1 ​ (9.8)t 2 Solve for 𝑡 t: 𝑡 = 2 ℎ 𝑎 = 2 ( 11 ) 9.8 = 1.498   s ≈ 1.5   s t= a 2h ​ ​ = 9.8 2(11) ​ ​ =1.498s≈1.5s The horizontal velocity is 𝑣 𝑥 = 17.05   m/s v x ​ =17.05m/s. The distance traveled before impact is: 𝑑 = 𝑣 𝑥 𝑡 = 17.05 ⋅ 1.5 = 25.5   m d=v x ​ t=17.05⋅1.5=25.5m Final Answer: The sled lands 25.5   m 25.5m from the foot of the cliff. Question 2 A 1000   kg 1000kg safe is 2.0   m 2.0m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses it 50   cm 50cm. What is the spring constant of the spring? Answer 1 The safe’s potential energy is used to compress the spring, which then absorbs all of the safe’s potential energy: 𝑚 𝑔 ( ℎ + 𝑥 ) = 1 2 𝑘 𝑥 2 mg(h+x)= 2 1 ​ kx 2 Substitute: 𝑘 = 2 𝑚 𝑔 ( ℎ + 𝑥 ) 𝑥 2 k= x 2 2mg(h+x) ​ 𝑘 = 2 ⋅ 1000 ⋅ 9.8 ⋅ 2.5 ( 0.5 ) 2 k= (0.5) 2 2⋅1000⋅9.8⋅2.5 ​ 𝑘 = 196 , 000   N/m k=196,000N/m Final Answer: The spring constant is 196 , 000   N/m 196,000N/m. Question 3 A point charge has a charge of 2.50 × 1 0 − 11   C 2.50×10 −11 C. At what distance from the point charge is the electric potential: a) 90.0   V 90.0V? b) 30.0   V 30.0V? Take the potential to be zero at an infinite distance from the charge. Answer 1 The distance is given by: 𝑉 = 𝑘 𝑞 𝑟    ⟹    𝑟 = 𝑘 𝑞 𝑉 V= r kq ​ ⟹r= V kq ​ For 𝑉 = 90.0   V V=90.0V: 𝑟 = ( 9 × 1 0 9 ) ( 2.50 × 1 0 − 11 ) 90.0 = 2.5   m r= 90.0 (9×10 9 )(2.50×10 −11 ) ​ =2.5m For 𝑉 = 30.0   V V=30.0V: 𝑟 = ( 9 × 1 0 9 ) ( 2.50 × 1 0 − 11 ) 30.0 = 7.5   m r= 30.0 (9×10 9 )(2.50×10 −11 ) ​ =7.5m Final Answer: (a) 𝑟 = 2.5   m r=2.5m (b) 𝑟 = 7.5   m r=7.5m Question 4 a. When the displacement of a mass on a spring is 𝑥 = 1 2 𝐴 x= 2 1 ​ A, what fraction of the mechanical energy is kinetic energy and what fraction is potential energy? b. At what displacement, as a fraction of 𝐴 A, is the energy half kinetic and half potential? Answer 1 The total energy is given by the equation: 𝐸 = 1 2 𝑘 𝐴 2 E= 2 1 ​ kA 2 For part (a): When 𝑥 = 1 2 𝐴 x= 2 1 ​ A: 𝑈 = 1 2 𝑘 𝑥 2 = 1 2 𝑘 ( 1 2 𝐴 ) 2 = 1 4 ( 1 2 𝑘 𝐴 2 ) = 1 4 𝐸 U= 2 1 ​ kx 2 = 2 1 ​ k( 2 1 ​ A) 2 = 4 1 ​ ( 2 1 ​ kA 2 )= 4 1 ​ E 𝐾 = 𝐸 − 𝑈 = 𝐸 − 1 4 𝐸 = 3 4 𝐸 K=E−U=E− 4 1 ​ E= 4 3 ​ E So, 𝑈 = 1 4 𝐸 U= 4 1 ​ E and 𝐾 = 3 4 𝐸 K= 4 3 ​ E. For part (b): When 𝑈 = 𝐾 = 1 2 𝐸 U=K= 2 1 ​ E: 1 2 𝑘 𝑥 2 = 1 2 𝐸 = 1 2 ( 1 2 𝑘 𝐴 2 ) 2 1 ​ kx 2 = 2 1 ​ E= 2 1 ​ ( 2 1 ​ kA 2 ) 𝑥 2 = 1 2 𝐴 2    ⟹    𝑥 = 𝐴 2 x 2 = 2 1 ​ A 2 ⟹x= 2 ​ A ​ Final Answer: (a) 𝑈 = 1 4 𝐸 U= 4 1 ​ E, 𝐾 = 3 4 𝐸 K= 4 3 ​ E (b) 𝑥 = 𝐴 2 x= 2 ​ A