WoEnPoMo

1. WoEnPoMo Work, Energy, Power, and Momentum

2. Question • Which has more energy: a tennis ball dropped from the third story of a building or one dropped from the first?

3. Work and Energy • Yesterday: decided that more energy = more ability to do stuff • bounce • displace sand • Work: doing stuff • by using energy • Energy: the ability to do work • Don’t worry about the circular definition. You have an intuitive concept of energy

4. Work • Amazing demonstration! • watch me drag this book/block/box • I’m doing work on it (because I’m giving it energy) • What affects how much work I’m doing? • force • distance of pull • W = ||F||||d|| • sound familiar?

5. Work • But what if I don’t pull horizontally? • What component of my pull does work? • vertical component is wasted “lifting” • “Real” formula (until calc class): • W = ||F||||d||cosθ • differentiates from torque (Fdsinθ) • Units: Nm (sound familiar?) • Joules (J) • Vector or scalar • scalar!

6. Practice Problem • A man picks up a 50N briefcase and walks with it to his house 2mi away. • How much work did he do on the suitcase while walking? • If he lifted the briefcase 1m, how much work did he do while lifting? • As he puts the briefcase down, how much work does he do? • What’s the total work done for the whole cycle? 0 J 50 J -50 J 0 J

7. Practice Problem • A hunter on The Oregon Trail TM pulls his 20kg deer kill on a mat sled by exerting a force of 160N through a rope a) How much work if he pulls horizontally for 10m? b) What about pulling at a 30° angle again for 10m? c) Say μk between sled & ground = .2. Find the work done by friction & net work if he pulls horizontally for 10m d) Find work done by friction and net work if he pulls at a 30° angle again for 10m (160N)(10m)cos0° = 1600J (160N)(10m)cos30° = 1386J Wfric = -400J Wnet = 1200J Wfric = -240J Wnet = 1146J

8. A quick derivation • We just learned W = FΔx = maΔx • Recall v2 = v02 + 2aΔx • Rearrange this into aΔx = (v2-v02)/2 • Substitute in • W = m(v2-v02)/2 • Expand • W = (1/2)mv2 - (1/2)mv02

9. Two important things • Define the common quantity as Kinetic Energy – the energy of motion • K = (1/2)mv2 • Notice that work = change in kinetic energy • Work energy theorem • Work causes a change in energy • if kinetic energy is changed, that’s realized as a change in speed

10. Practice • The driver of 1000kg car traveling at 35m/s stands on his brakes to avoid rear-ending the traffic at the Sagamore. This causes a friction force of 8000N. What’s the minimum distance needed to avoid an accident? • - old way: Big gun #1 • - new way: work-energy theorem

11. Practice • A 1000kg car has brakes that produce 8000N of friction. It leaves skid marks 27m long at the scene of an accident. What was the minimum speed of the car? • If he were traveling twice as fast, how long would the skid marks be? 20.8 m/s 108.2m, which is 4x more

12. Another form of energy…and more practice • You lift a 2kg book straight up 1m and hold it still. Find the work done by… • You • gravity (2kg)(10m/s2)(1m)cos0 = 20J (2kg)(10m/s2)(1m)cos180 = -20J • But work changes energy. The book started and ended at rest. So what form of energy was changed? And how can we get back the energy we put into lifting the book?

13. Gravitational Potential Energy • Gravity pulls stuff downwards (to earth) • So if you drop an object, it can do work • Since it can do work, it must have “stored up” energy ready to be released… • Gravitational Potential Energy • What factors could affect it? • Mass • Height • Acceleration due to gravity U = mgh

14. The higher you are, the more you can fall, and the more work you can do • Where is there no potential energy (U = 0)? High potential Medium potential Low potential

15. Practice • Hooligans are dropping pennies (m = 2.5g) from apartment windows. • Find the potential energies of the pennies held out the 1st story window 10m above the ground and the 2nd story window 20m above. • Now, find the work done by gravity when falling from the 2nd story to the first. Should it be positive or negative? Remember, work causes a change in energy! • Repeat the above two problems for pennies dropped from 50m and 60m (.0025kg)(10m/s2)(10m) = .25J (.0025kg)(10m/s2)(20m) = .50J .50J - .25J = .25J Positive because motion and force are down (.0025kg)(10m/s2)(50m) = 1.25J (.0025kg)(10m/s2)(60m) = 1.50J 1.50J - 1.25J = .25J It’s the same difference! (hilarious)

16. Moral of the story? • All that matters is the change in heights – NOT absolute height • Therefore, we can make ANY height to be zero potential energy – all that matters is our difference in heights. • Pro tip – make the lowest part of the problem have zero potential energy.

17. Work and energy • Recall: W = (1/2)mv2 – (1/2)mv02 • W = ΔK • Positive work increses speed • What about Wgrav and U? • Start at the bottom of a ladder • As you climb, what happens to your potential energy? • Ufinal > Uinitial ΔU > 0 • What’s the work done by gravity? • Negative (force down, movement up) Wgrav = -ΔU

18. Big Gun #2 • Conservation of Energy • The total amount of energy in an isolated system is remains the same before and after any process • E1= E2 where 1 and 2 represent different locations, times, heights, etc

19. Practice • A diver (m = 50kg) drops off a platform 10m above the water’s surface. Neglecting air resistance, how fast is he moving halfway to the water? • Etop = Ebot • Ktop + Utop = Kbot + Ubot • mghtop = (1/2)mv2bot • Vbot = √2gh = 10m/s

20. Your turn • How fast is the diver going when he hits the water? • How fast does he hit the water if he jumps up off the platform at 3.5m/s? Vbot = 14.14 m/s Vbot = 14.6 m/s 10m

21. More practice • A ball is hit at a 40°angle at 30m/s. How high above the hit point does it travel? • What if it’s hit at 20°? • Use common sense! Even though energy is a scalar, the different angles will clearly cause different answers, right? h = 18.59m h = 5.26m

22. Another one…together • The Der Stuka waterslide in Orlando is 21.9m tall. If it’s frictionless, how fast is a 60kg woman going at the bottom? • Her measured speed is actually 18m/s. How much energy was lost due to friction? 20.9m/s -3420 J

23. More friction • A skier starts from rest at the top of a frictionless slope 20m tall. At the bottom is level dirt where friction exists,μ=.2. Find • The skier’s speed at the bottom of the slope • How far he travels on the dirt V= 20m/s 100m

24. Practice • A roller coaster of mass m goes through a circular loop of radius R. • what’s the minimum speed the coaster must have at the top so it doesn’t fall? • What will be the speed at the bottom? • Normal force at bottom? V =sqrt(Rg) V =sqrt(5Rg) FN =6mg

25. What about U not caused by gravity? • Springs have equilibrium positions they “rest” at. • External forces displace from equilibrium • Fspring tries to restore that equilibrium • Restoring force • Short experiment – measure the relationship between F and Δx

26. Hooke’s law • Fspring = -kx • k = spring constant • units = N/m • big = stiff; small = soft • x = displacement from equilibrium • negative because always restores

27. Spring work • Is force constant (like gravity)? • tricky, since displacement matters • you’ll learn how to deal with this in calculus • Let’s deal with average force • Favg = (F1+F2)/2 • call F1 equilibrium position • Favg = F2/2= -kx/2 Wspring = Favgx= (1/2)(-kx)x = -(1/2)kx2

28. Spring work • In general, work done by a spring: • Wspring = - ((1/2)kx2– (1/2)kx02) • Spring does negative work as its stretched (you do positive work on spring) • Just like always, work changes energy • Uspring= (1/2)kx2

29. Example • A 50kg acrobat drops 2m onto a springboard with k=8000N/m. How far does she compress the spring if it’s frictionless? .5m

30. Practice • A .5kg block rests on a horizontal frictionless surface. It’s pushed back against a spring, k=625N/m, compressing it 10cm. The block is released. • a) what distance does the block travel up a frictionless incline of 30º? • b) How fast when it’s halfway there? d = 1.25 m v = 2.5 m/s

31. Big Gun #2…IN SPACE!

32. Wait – why negative? zero potential energy here, in the flat “plane” of space M – mass warps spacetime has potential energy because in the gravity “well” but below the zero line

33. Space practice – using big gun #2 • A space probe of mass Mprobe is being launched to explore Jupiter’s moons. It’s shot up vertically from the surface of earth. What’s the minimum launch velocity so that the probe will escape earth?

34. Practice A rocket of mass 13000kg is launched from earth. What’s the escape velocity needed? What’s the escape velocity needed if the launch pad is on the moon? Earth: 1.12x104 m/s Moon: 2.375x103 m/s

35. Amazing physics demo! • Watch me lift this book! • During which lift did I do more work? (think carefully!) • What’s the difference between the two lifts? • What’s the same between the two? • Seems counter-intuitive, doesn’t it? But thanks to work-energy theorem, the answer’s perfectly clear!

36. Power • But we definitely need a quantifiable way to contrast the two lifts • Power takes time into account • Which lift do you think was more powerful? • So less time  more power Units = J/s = Watt (W)

37. Playing around • P = W / t power = work/time • P = (FΔx)/t W = FΔx (ignoring angle) • P = F(Δx/t) thanks math! • P = Fv power = force * avg vel

38. Practice • Killer whales can be 32 ft long & over 8000kg. They can also accelerate from 0 to 30mi/hr in a few seconds. Ignoring drag, figure out the power needed to go from 0 to 12m/s in 6 s if the whale is 8000kg • Is this number and underestimate or overestimate? 9.6x104 W = 96kW Underestimate – energy lost due to drag as well

39. Pendulums! • Weight suspended from a pivot that swings back and forth. • SIMPLE pendulums • light string/rod, heavy mass • Repetitive motion back and forth • oscillation • Period (T): time it takes to go back & forth • Frequency (f): number of events in T seconds • f = 1/T (units: 1/s = Hz)

40. Pendulums • What are some factors that might affect period of oscillation? • Simple experiment • lab stations available to test • mass of bob • length of string • shape of bob • displacement angle (don’t test beyond 30°) • Share data on whiteboard

41. Simple Harmonic Motion • Pendulums are an example of SHM • SHM: motion of a simple harmonic oscillator (awesome) • SHO: experiences a single net force which acts as a restoring force • FSHO = -kx (follows Hooke’s law) • SHM: it repeats itself! • Other examples: spring oscillator, uniform circular motion

42. SHM functions • Derived on board: • Pos: x(t) = Acos(ωt + Φ) • Vel: v(t) = -A ω sin(ωt + Φ) • Acc: a(t) = -Aω2cos(ωt + Φ)

43. Energy of SHM • K = (1/2)mv(t) = (1/2)kA2sin2(ωt +Φ) • U = (1/2)kx(t) = (1/2)kA2cos2(ωt +Φ) • Etot = K + U • = (1/2)kA2(sin2(ωt +Φ)+cos2(ωt +Φ)) • Etot = (1/2)kA2 • Mechanical energy is shown in the amplitude of the oscillation

44. Momentum • A quantity of motion • p = mv • vector or scalar? • vector! • Units? • kg·m/s • no special name

45. Changing momentum • Well, how do you change momentum? • Change velocity! • which is called… • Acceleration! • which is caused by… • Forces! • F = ma = (mΔv)/Δt = Δρ/ Δt • Two important things… • …but first, the mistake Newton made

46. Important thing #1 • Application of force over time changes momentum • Define application of force over time as a new quantity! • Impulse • J = FΔt = Δρ = mv – mv0 • Impulse changes momentum! • Analogy: work changes energy

47. Practice • A 50g golf ball of radius 2cm is hit off the tee (where it’s at rest). It leaves at v=44m/s • Find magnitude of impulse • Find duration of collision and average force if the ball compresses halfway (one radius) 2.2 kgm/s t = 9.1E-4 s F = 2400N

48. Impulse practice • A 40kg boy jumps 5m off his roof into a pile of leaves. They bring him to a stop in .3s. Find the impulse and average force. • J = -400kgm/s, Favg = 1.3E3N (3.33g) • Find the impulse and average if he missed the leaves and stopped in .03s. • J = -400kgm/s, Favg = 1.3E4N (33.3g) • Applications of impulse?

49. Important thing #2 • Recall • What happens if no external force exists? • 0 =  Δp = 0 • Momentum is conserved! • MomentumtimeA = MomentumtimeB • Big gun #3

50. Practice • A hunter shoots a shotgun shell of mass .2kg horizontally at 100m/s. The gun has a mss of 10kg. How fast is the recoil of the gun if it’s held away from the body?