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Chapter 8 Conservation of Energy EXAMPLES

Chapter 8 Conservation of Energy EXAMPLES. Example 8.1 Free Fall (Example 8.1 Text book). Determine the speed of the ball at y above the ground The sum K + U remains constant At h : U i = mgh K i = 0 At y: K f = ½mv f 2 U f = mgy In general: Conservation of Energy

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Chapter 8 Conservation of Energy EXAMPLES

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  1. Chapter 8Conservation of Energy EXAMPLES

  2. Example 8.1 Free Fall(Example 8.1 Text book) • Determine the speed of the ball at y above the ground • The sumK + Uremainsconstant • At h: Ui = mghKi = 0 • At y: Kf = ½mvf2 Uf= mgy • In general: Conservation of Energy Ki + Ui = Kf + Uf  0 + mgh = ½mvf2 + mgy • Solving for vf vfis independent of the mass !!!

  3. Example 8.2 Roller Coaster Speed • UsingMechanical Energy Conservation (Frictionless!) ½mvf2 + mgyf = ½mvi2 + mgyi • Only vertical differences matter! • Horizontal distance doesn’t matter! • Mass cancels! • Find Speed at bottom? Known:yi = y = 40m, vi = 0, yf = 0,vf = ?  0 + mgyi = ½mvf2 + 0  vf2= 2gyi = 784m2/s2 v2 = 28 m/s

  4. Example 8.3 Spring-Loaded Gun(Example 8.3 Text book) • Choose point A as the initial point and C as the final point • (A). Find the Spring Constant k ? Known: vA = 0, yA = 0 , xA = yB = 0.120m vC = 0 , yC = 20m, UC = mgyC , m = 35.0g EC = EA  KC + UgC + UsC = KA + UgA + UsA ½mvC2 + mgyC + ½kxC2 = ½mvA2 + mgyA + ½kxA2  0 + mgyC + 0 = 0 + 0 + ½kxA2  ½kxA2 = mgyC  k = 2mgyc/xA2 =2(0.0350kg)(9.80m/s2)(20.0m)/(0.120m)2 k = 953 N/m yC yB yA

  5. Example 8.3 Spring-Loaded Gun, final • (B). Find vB ? Use: EB = EA KB + UgB + USB = KA + UgA + USA ½mvB2 + mgyB + ½kxB2 = ½mvA2 + mgyA + ½kxA2 ½mvB2 + mgyB + 0 = 0 + 0 + ½kxA2 vB2 = (kxA2 – 2mgyB)/m vB2 =388.1m2/s2  vB = 19.7 m/s yC yB yA

  6. Example 8.4 Ramp with Friction(Example 8.7 Text book) • Problem: the 3.0 kg crate slides down the rough ramp. If:vi = 0, yi = 0.5m , yf = 0 ƒk = 5N • (A). Find speed at bottom vf • At the top: Ei = Ki + Ugi = 0 + mgyi • At the bottom: Ef = Kf + Ugf = ½ m vf2 + 0 • Recall: If friction acts within an isolated system ΔEmech = ΔK + ΔU = Ef – Ei =– ƒk d  ½ m vf2 – mgyi =– ƒk d • Solve for vf

  7. Example 8.4 Ramp with Friction, final • (B). How far does the crate slide on the horizontal floor if it continues to experience the same friction force ƒk = 5N • The total ΔEmech is only kinetic (the potential energy of the system remains fixed): ΔEmech = ΔK = Kf – Ki = – ƒkd • Where: Kf = ½ m vf2 = 0 Ki = ½ m vi2 = 9.68 J Then: Kf – Ki = 0 – 9.68 J = – (5N)d d = (9.68/5) = 1.94 m

  8. Example 8.5 Motion on a Curve Track (Frictionless) • A child of mass m = 20 kg starts sliding from rest. Frictionless! • Find speed v at the bottom. • ΔEmech = ΔK + ΔU ΔEmech =(Kf – Ki) + (Uf – Ui) = 0 (½mvf2 – 0) + (0 – mgh)= 0 ½mvf2 – mgh = 0  • Same result as the child is falling vertically trough a distance h!

  9. Example 8.5 Motion on a Curve Track (Friction) • If a kinetic friction of ƒk = 2Nacts on the child and the length of the curve track is 50 m, find speed v at the bottom. • If friction acts within an isolated system ΔEmech = ΔK + ΔU =– ƒk d ΔEmech = (Kf – Ki) + (Uf – Ui) = – ƒk d ΔEmech = ½mvf2 – mgh = – ƒk d ΔEmech = ½(20) vf2 – 20(10)(2)= –100J

  10. Example 8.6 Spring-Mass Collision(Example 8.8 Text book) • Frictionless! • K +Us = Emech remains constant • (A). Assuming: m= 0.80kg vA = 1.2m/s k = 50N/m • Find maximum compression of the spring after collision (xmax) EC = EA  KC + UsC = KA + UsA ½mvC2 + ½kxmax2 = ½mvA2 + ½kxA2 0 + ½kxmax2 = ½mvA2 + 0 

  11. Example 8.6 Spring-Mass Collision, final • (B). If friction is present, the energy decreases by: ΔEmech = –ƒkd • Assuming: k= 0.50 m= 0.80kg vA = 1.2m/s k = 50N/m Find maximum compression of the spring after collision xC ΔEmech = –ƒk xC = –knxC = –kmgxC ΔEmech = –3.92xC (1) • Using: ΔEmech = Ef – Ei ΔEmech = (Kf – Uf) + (Ki – Ui) ΔEmech = 0 – ½kxC2 + ½mvA2 + 0 ΔEmech = – 25xC2 + 0.576 (2) • Taking:(1) = (2): – 25xC2 + 0.576 = –3.92xC • Solving the quadratic equation for xC : xC = 0.092m < 0.15m (frictionless) • Expected! Since friction retards the motion of the system • xC = – 0.25m Does not apply since the mass must be to the right of the origin.

  12. Example 8.7 Connected Blocks in Motion(Example 8.9 Text book) • The system consists of the two blocks, the spring, and Earth. Gravitational and potential energies are involved • System is released from rest when spring is unstretched. • Mass m2 falls a distance h before coming to rest. • The kinetic energy is zero if our initial and final configurations are at rest K= 0 • Find k

  13. Example 8.7 Connected Blocks in Motion, final • Block 2 undergoes a change in gravitational potential energy • The spring undergoes a change in elastic potential energy Emech= K+ Ug + US= Ug + US= Ugf – Ugf+ Usf – Usi Emech= 0– m2gh+ ½kh2 – 0 Emech= – m2gh+ ½kh2 (1) • If friction is present, the energy decreases by: ΔEmech= –ƒkh = – km1gh(2) • Taking (1) = (2): – m2gh+ ½kh2 = – km1gh km1gh = m2gh– ½kh2 

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