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 K i + U i = K f + U f  0 + mgh = ½mv f 2 + mgy  Solving for v f v f is independent of the mass !!!

Example 8.2 Roller Coaster Speed Using Mechanical Energy Conservation (Frictionless!) ½mv f 2 + mgy f = ½mv i 2 + mgy i Only vertical differences matter! Horizontal distance doesn’t matter! Mass cancels! Find Speed at bottom? Known: y i = y = 40m, v i = 0, y f = 0, v f = ?  0 + mgy i = ½mv f  v f 2 = 2gy i = 784m 2 /s 2  v 2 = 28 m/s

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: v A = 0, y A = 0, x A = y B = 0.120m v C = 0, y C = 20m, U C = mgy C, m = 35.0g E C = E A  K C + U gC + U sC = K A + U gA + U sA ½mv C 2 + mgy C + ½kx C 2 = ½mv A 2 + mgy A + ½kx A 2  0 + mgy C + 0 = ½kx A 2  ½kx A 2 = mgy C  k = 2mgy c /x A 2 =2(0.0350kg)(9.80m/s 2 )(20.0m)/(0.120m) 2 k = 953 N/m yAyA yByB yCyC

Example 8.3 Spring-Loaded Gun, final ( B). Find v B ? Use: E B = E A K B + U gB + U SB = K A + U gA + U SA ½mv B 2 + mgy B + ½kx B 2 = ½mv A 2 + mgy A + ½kx A 2  ½mv B 2 + mgy B + 0 = ½kx A 2 v B 2 = (kx A 2 – 2mgy B )/m v B 2 =388.1m 2 /s 2  v B = 19.7 m/s yAyA yByB yCyC

Example 8.4 Ramp with Friction (Example 8.7 Text book) Problem: the 3.0 kg crate slides down the rough ramp. If: v i = 0, y i = 0.5m, y f = 0 ƒ k = 5N (A). Find speed at bottom v f At the top: E i = K i + U gi = 0 + mgy i At the bottom: E f = K f + U gf = ½ m v f Recall: If friction acts within an isolated system Δ E mech = Δ K + Δ U = E f – E i = – ƒ k d  ½ m v f 2 – mgy i = – ƒ k d Solve for v f

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 ƒ k = 5N The total ΔE mech is only kinetic (the potential energy of the system remains fixed): Δ E mech = Δ K = K f – K i = – ƒ k d  Where: K f = ½ m v f 2 = 0 K i = ½ m v i 2 = 9.68 J Then: K f – K i = 0 – 9.68 J = – (5N)d  d = (9.68/5) = 1.94 m

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. ΔE mech = ΔK + ΔU ΔE mech =(K f – K i ) + (U f – U i ) = 0 (½mv f 2 – 0) + (0 – mgh)= 0 ½mv f 2 – mgh = 0  Same result as the child is falling vertically trough a distance h!

Example 8.5 Motion on a Curve Track (Friction) If a kinetic friction of ƒ k = 2N acts 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 Δ E mech = Δ K + Δ U = – ƒ k d Δ E mech = (K f – K i ) + (U f – U i ) = – ƒ k d Δ E mech = ½mv f 2 – mgh = – ƒ k d Δ E mech = ½(20) v f 2 – 20(10)(2) = –100J

Example 8.6 Spring-Mass Collision (Example 8.8 Text book) Frictionless! K +U s = E mech remains constant (A). Assuming: m= 0.80kg v A = 1.2m/s k = 50N/m Find maximum compression of the spring after collision (x max ) E C = E A  K C + U sC = K A + U sA ½mv C 2 + ½kx max 2 = ½mv A 2 + ½kx A ½kx max 2 = ½mv A 

Example 8.6 Spring-Mass Collision, final (B). If friction is present, the energy decreases by: Δ E mech = –ƒ k d Assuming:  k = 0.50 m= 0.80kg v A = 1.2m/s k = 50N/m Find maximum compression of the spring after collision x C Δ E mech = –ƒ k x C = –  k nx C = –  k mgx C  Δ E mech = –3.92x C (1) Using: Δ E mech = E f – E i Δ E mech = (K f – U f ) + (K i – U i ) Δ E mech = 0 – ½kx C 2 + ½mv A Δ E mech = – 25x C (2) Taking: (1) = (2): – 25x C = –3.92x C Solving the quadratic equation for x C : x C = 0.092m < 0.15m (frictionless) Expected! Since friction retards the motion of the system x C = – 0.25m Does not apply since the mass must be to the right of the origin.

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 m 2 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

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  E mech =  K +  U g +  U S =  U g +  U S = U gf – U gf + U sf – U si  E mech = 0 – m 2 gh + ½kh 2 – 0  E mech = – m 2 gh + ½kh 2 (1) If friction is present, the energy decreases by: Δ E mech = –ƒ k h = –  k m 1 gh (2) Taking (1) = (2): – m 2 gh + ½kh 2 = –  k m 1 gh  k m 1 gh = m 2 gh – ½kh 2 