Energy Balance, Units & Proble Solving: Mechanical Energy Balance ABET Course Outcoes: 1. solve and docuent the solution of probles involving eleents or configurations not previously encountered (e) (e.g. a new geoetric arrangeent, a new ter to include in an analysis, a new type of starting condition). solve probles using ultiple approaches including varied analytic approaches, diagras, foral solution steps or siple coputer progras (e) By the end of today you should be able to: Calculate the force of a spring and coplete spring force balances Calculate elastic potential energies Solve probles balancing kinetic, gravitational potential, and elastic potential energies Class Plan Handout: eg fro slides 5 8 and 11 14 worksheet (possibly) Lecture Outline: 1. Elastic Force a) Background: Equations and Balances b) Exaple : Deterining k (force balance) c) Proble: Final position of a bungee cord. Elastic Potential Energy a) Background: Equations and Balances b) Proble 1: Mass stopped by spring (finding k) c) Exaple : Deterining ax height of a spring launched ass d) Exaple 3: Roller coaster with KE, PE & Spring E 3. Wrap/Review a) Energy Balance Suary b) Proble Solving Lessons/Concepts (ephasize proble solving approach as well conservation balances.) Slides will be posted (& therefore these exaples). F s Δx Spring at rest: x = 0, where F s = 0 Elastic Force Elastic Force: Soe Other Forces Hooke s Law Constant k [=] force/length Kineatic: F = a Weight: F g = g (also frictional, drag, agnetic ) F s = k Δx A Force Balance: Elastic & Gravitational Force: Force Exaple: Deterining Spring Constant An 8 kg weight copresses a spring 10 c. What is the Hook s Law Constant (k)? 8 kg 10 c Proble Setup Diagra: Given: = 8 kg x = 10 c (copression, fro rest) Find: k Known: F (spring) = k Δx F (gravity) = a = g g = 9.8 /s Balance: g = k x 8 9.8 / 0.1 784 /
Elastic Force Balance: Force Proble: Bungee Juping Proble (Hooke s Law) A 100 kg bungee juper jups fro a bridge 00 above the river. His bungee cord has a Hooke s Law Constant of 00 N/ and is 100 long at rest (i.e., with no load). When he stops otion (no ore bouncing) how high above the river is he? Hint: Balance the Hooke s Law force & the gravitational force Unweighted 100 Weighted, = 100 kg L f 00 Have students work on Bungee proble on previous slide Ask for setup, Then solve proble (worksheet?) Length at Rest Proble Setup Given: = 100 kg, k = 00 N/ L o (no load) = 100 Find: Final Loaded Length, L f Known: F s = k(l f L o ) F g = ( g) F s = F g (when otion stops) Assuptions: no losses (friction and drag) Solution: k(l f L o ) = g + L o + 100 = 104.9 Distance above river = 00 104.9 = 95.1. / / Diagra: Unweighted 100 Weighted, = 100 kg L f =? Gravitational Potential Energy & Kinetic Energy (Review): Equations and a Siple Energy Balance Gravitational Potential Energy (U ) Kinetic Energy (K) K = ½ v v g U g = (g) h In a closed syste the total energy change is zero: E = 0 K + U g = 0 ½ (v ) + g h = 0 or ½ v + g h = constant h Previous Slide: Quick Review of KE & PE. 1. Forula: Review forula for KE and PE (but on board?). Units Work K = ½ v [=] kg = kg /s Energy units Ug = (g) h [=] kg = kg /s Energy units 3. Signs: need to ake sure signs ake sense 4. Balance Equations: Review 5. Syste: this is a balance on an object (the syste) Now add spring energy (next slide)
F s x Elastic Potential Energy x = 0, where F s = 0 Hook s Law Constant Elastic PE: U s = ½ k Δx Adding this into the Energy Balance K + U g + U s = 0 ½ (v ) + g h + ½ k (x) = 0 or ½ v + g h + ½ k x = constant Elastic Potential Energy: Energy Exaple 1: Ball launcher (an energy balance) The earlier spring/ass syste (k = 784 ) is copressed 40 c and then released. What is the resulting axiu height of the ass? Proble Setup Diagra: Given: = 8 kg Δx = 40 c (copression, fro rest) k = 784 N/ (previous calc.) Find: Maxiu height of the ass Known: U s = ½ k Δx U g = ( g) h Copress Release 8 kg 10 c 8 kg 30 c h =? Elastic Potential Energy Exaple 4: Spring Launcher Energy at the beginning (E 0 ) Energy at axiu height (E ax ) Elastic Potential Energy Exaple 4: Spring Launcher Calculation Spring copressed x = 40 c = 0.4 Spring relaxed (x = 0) U s,ax = 0 No otion K o = 0 Take this to be h = 0 U go = 0 No otion K ax = 0 Find: Height =? h ax = /.. / E o = U so = ½ k x o E ax = U g = gh ax Conservation of Energy: E o = E ax ½ k x o = g h ax
Elastic Potential Energy: Energy Exaple : Stopping a Sliding Block A.5 kg block slides on a frictionless surface into a spring. It coes to rest when the spring has been copressed 7.5 c. The spring constant is 30 N/. What was the velocity of the block before it contacted the spring? Hint: Use a before & after energy balance. Energy Exaple 1: Stopping a Sliding Block Proble Setup Diagra: on previous page Given: =.5 kg, k = 30 N/ x f = 7.5 c = 0.075 Find: V 0 =? Known: U sf = ½ k x f K 0 = ½ v 0 K 0 = U sf Assuptions: frictionless, no drag Have students try this proble Ask the for proble setup Possible steps Equation developed The answer V 0 =? k = 30 N/ Δx f = 7.5 c Solution: ½ v 0 = ½ k x f v 0 = k x f /.5 kg.5 kg / 30 0.075.5 0.85 / Exaple 5: Launched Roller Coaster 3 Spring Constant k = 5000 N/ Cart Mass, = 1000 kg 5 Work with neighbors 1. Select the datu (h =0). Describe Energy coponents at tie 0 (t 0 ) 3. Proble Restateent (diagra, given, find) 4. Solution V =? 1 Exaple 5: Launched Roller Coaster Questions (possible answers) have groups discuss, then ask class 1. Select the datu (h = 0) could be any places: At ground would be logical At height of position B siplifies calc.. Describe Energy coponents at t 0 Need to define t 0 t 0 = when spring is copressed Syste: entire roller coaster (cart & spring) Energy Coponents of Cart: Es = Us = ½ k x U g = g h (with h = 5 1 = 13 ) 3. Proble Restateent Given: Cart Mass = 1000 kg Spring Const., k = 5000 N/ x = 3 h 1 = 5, h = 1 (h = 13 ) Find: V B =? Diagra (on slide) 4. Proble Steps 1. Calculate Initial Energy (E 0 = U s0 ). Conservation of Energy E B = E 0 3. Calculate Velocity at B E B = K + Ug + Us h = 0, no spring Ug and Us = 0, E B = K B = ½ v
Exaple 5: Launched Roller Coaster Calculations Given: Cart Mass = 1000 kg Spring Const., k = 5000 N/ x = 3 h 1 = 5, h = 1 (h = 13 ) Find: V B =? Diagra: (below 3 Datu (h = 0) 5 V B =? 1 1. Calculate Initial Energy (= U g + U s ) U so = ½ k x = ½ 5000 N/ (3 ) =,500 N or J U go = g h = 1000 kg (9.8 /s) (13 ) = 17,400 kg /s or J E 0 = U so + U go = 149,900 J. Conservation of Energy E B = E 0 3. Calculate Velocity at B E B = Ug + K + Us = 0 + ½ V + 0 E 0 =E B = ½ V conservation of E v = ( E o /) ½ = ( 150 kj/1000 kg) ½ v = 17.3 /s Units [=] / = ( /s ) ½ = /s Could define spring as out of the syste in which case it would act on the syste (work) Energy in a Closed Syste Syste (of fixed ass) Energy = a property Previous Slide Suarize energy balances in general, this proble in particular Note energy crossed a boundary (springs can also be inside a syste (e.g. a wind up toy) Heat will be treated later. Next Slide Quickly note that this sections key goals have been proble solving. If tie, ask students in sall groups coe up with a list of key considerations for solving probles particularly involving energy Review the next slide as appropriate. Soe Proble Solving Keys Understand Circustances (Principles, Assuptions, Units ) conservation! Break into parts/steps (setup, initial conditions, solution steps ) Units: Pay attention throughout (ake sure they work, use to understand) Nubers & Signs: Do they ake sense at each step? Discipline: Take tie and care throughout (easy to ake accounting errors) Reread/reevaluate the proble any ties