Reaction time website to test reaction time http://www.gwc.maricopa.edu/class/phy101/Flash/reaction%20time.htm

Lab work and lots more http://hendrix2.uoregon.edu/~dlivelyb/phys101/Lab3.pdf

Bellringer 2/9 1. What is work? 2. Does moving something at an angle make work easier or harder? Why? 3. Rank these in order of work – low to high: B A C D

Work and energy Physics

Work A. work is done when a force causes an object to move B. eqn: W = Fd (units = Joules) C. Calculating Work 1. Constant force exerted at an angle 2. Work is done only if a force is exerted in the direction of motion 3. A force  to motion does no work 4. Be sure to use NET Force for work – in direction of motion 5. Work can be negative (opp of motion)

Work A joule represents a relatively modest amount of work. You do a joule of work when you lift a medium-sized apple through a height of 1 meter. EX: © 2014 Pearson Education, Inc.

EX: A 105 g hockey puck is sliding across the ice EX: A 105 g hockey puck is sliding across the ice. A player exerts a constant 4.50 N force over a distance of 0.150 m. How much work does the player do on the puck? W = Fd = (4.50 N) (0.150 m) W = 0.675 J

EX 2: (whiteboard?) You play Rock Hero with friends, you push the drum set to slide it into possession. You exert a horizontal force of 24N. The drum set has a force of friction of 22N. You slide the drum set 1.5m. The Weight of the drum set is 54N. What is the total work done on the drum set?

W = Fd (need force) Fnet x = FA – Ff Fnet x = 24N – 22N = 2 N W = Fd W = (2N) (1.5m) W = 3J

D. force can be applied at an angle eqn: W = Fdcos

EX: A sailor pulls a boat a distance of 30 EX: A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25.0 angle with horizontal. How much work is does the sailor do on the boat if he exerts a force of 255N on the rope? d = 30.0 m  = 25.0 F = 255 N W? W = Fdcos = (255 N) (30.0m) cos 25.0 W = 6.93 x 103 J

KINETIC ENERGY I. Kinetic Energy – energy in motion A. eqn: KE = 1/2mv2 B. Units: Joules

EX: A 3900kg truck is moving at 6.0m/s. A. What is the truck’s kinetic energy? KE = 1/2mv2 KE = (1/2) (3900kg) (6.0m/s)2 KE = 70,200J (70kJ) B. What is KE if speed is doubled? KE = (1/2) (3900kg) (12.0m/s)2 KE = 280,000 J OR 2.8x105 J

II. Work and KE are related A. Total work done on an object equals the change in its kinetic energy. B. eqn: Wtotal = ΔKE Δ = change in = final – Initial so: Wtotal = 1/2mvf2 – 1/2mvi2 C. Called the work-energy theorem

EX: How much work is required for a 74kg sprinter to accelerate from rest to a speed of 2.2m/s? Wtotal = ΔKE Wtotal = 1/2mvf2 – 1/2mvi2 Wtotal = (1/2) (74kg) (2.2m/s)2 – 0 Wtotal = 179J

Partner work – person on left The kinetic energy of a small boat is 15,000J. If the boat’s speed is 5.0m/s, what is its mass? 1200kg

Partner -person on right What is the speed of a 0.15kg baseball whose kinetic energy is 77J? 32m/s

EX: A student lifts a 4.0kg box of books vertically from rest with an upward force of 52.7N. The distance of the lift is 1.60m. Find: (a) work done by student, (b) work done by gravity, and (c) final speed of books (a) Wstudent = Fd Wstudent = (52.7N) (1.6m) Wstudent = 84.3J

(b) Wgravity = Fd Wgravity = mgd Wgravity = (4.10kg) (9.8m/s2) (1.60m) Wgravity = 64.3J (c) Wtotal = ΔKE Wtotal = 1/2mvf2 – 1/2mvi2 84.3J – 64.3J = (1/2) (4.10kg) (vf2) – 0 vf = 3.12m/s

Partner work: At t=1.0s, a 0.40kg object is falling with a speed of 6.0m/s. At t=2.0s, it has kinetic energy of 25J. (a) What is the kinetic energy of the object at t = 1.0s? (b) What is the speed of the object at t=2.0s? (c) How much work was done on the object between t=1.0s and t=2.0s?

(a) KE = 1/2mv2 KE = (1/2) (0.40kg) (6.0m/s)2 KE = 7.2J (b) KE = 1/2mv2 25J = (1/2) (0.40kg) (v2) v = 11m/s (c) W = ΔKE W = 25J – 7.2J W = 17.8J

Bellringer 2/12 A two-man bobsled has a mass of 390 kg. Starting from rest, the two racers push the sled for the first 50 m with a net force of 270 N. Neglecting friction, what is the sled’s speed at the end of the 50 m? W = ΔK = Kf - Ki K = 1/2mv2 Kf = Ki + W 1/2mvf2 = 0 + Fd (1/2) (390kg) vf2 = (270N) (50m) vf = 8.3m/s (18mph)

A. stored energy – based on position B. 2 types of PE 10.4 Potential Energy (PE) A. stored energy – based on position B. 2 types of PE 1. gravitational – from gravity – distance from gravity – so height. 2. spring – how much “potential” does a spring have

D. Elastic (or spring) Potential energy (PEs) 1. eqn: *PEs = 1/2kx2 C. Gravitational potential energy (PEg) – higher it is the more energy it can store So height important (h) (sometimes “y”) Eqn: *PEg = mgh D. Elastic (or spring) Potential energy (PEs) 1. eqn: *PEs = 1/2kx2 k = constant of spring x = distance spring moves

E. EX: You lift a 7.30 kg bowling ball from the storage rack and hold it up to your shoulder. The storage rack is 0.610 m above the floor and your shoulder is 1.12 m above the floor. a. When the ball is at your shoulder, what is the bowling ball’s PE relative to floor? PE = mgh (7.3 kg)(9.8 m/s2)(1.12 m) PE = 80.1 J

b. When the ball is at your shoulder, what is the bowling ball’s PE relative to rack? PE = mgh h = 1.12m – 0.610 m = 0.51 m (7.30 kg)(9.8 m/s2)(0.51m) PE = 36.5 J c. How much work was done by gravity as you lifted the ball from the rack to shoulder level? W = Fd F = mg W = mgd (7.3 kg)(9.8m/s2)(0.51m) W = 36.5 J So W = ΔE!

F. EX: A spring with a constant of 120N/m is compressed a distance of 2.3cm. How much potential energy is stored in the spring? PEs = 1/2kx2 PEs = ½(120N/m)(0.023m)2 PEs = 0.032J

Partner work A 875kg 4 wheeler is initially moving at 22m/s and then accelerates to 44.0m/s. What is the work done during this acceleration? W = ΔKE KE = 1/2mv2 KEi = ½ (875kg) (22m/s)2 KEi = 2.12x105J KEf = (1/2) (875kg) (44m/s)2 KEf = 8.47x105J W = 8.47x105J - 2.12x105J W 6.35x105J

Other partner

Bellringer 2/17 Kahoot on phones – GOOGLE “KAHOOT.IT” Potential and kinetic energy – know eqns https://play.kahoot.it/#/?quizId=ad7e09be-9726-42b0-872b-849f079fb871

Using the Law of Conservation of Energy I. In an isolated system energy is conserved, final energy = initial energy E total initial = E total final Kf + (PEg)f + (PEs)f + = Ki + (PEg)i + (PEs)i II. Energy bar charts (LOL charts) A. Show energy before and after event B. bar graphs

C. EX: A car on a frictionless roller coaster track, launched by a huge spring, makes it to the top of the loop. Draw energy bar graph.

Now we tackle the same problem, but we put the spring outside of the system. Energy must be transferred into the system by the object outside of the system. We call this idea of transferring energy into or out of the system “doing work”.

ON WHITE BOARD: person on left Make a energy bar chart for this: An object is launched upwards using a compressed spring.

ON WHITE BOARD: person on right Make a energy bar chart for this:

ON WHITE BOARD: person on right Make a energy bar chart for this:

$ scenario State similarities and differences in these scenarios: I have $100 in my wallet I have $100 in my savings account I have $100 in a savings bond I have $100 in a mutual fund I give you $100 I give my mom $100

How does money idea relate to energy? All deal with money but location, access and availability may change (storage). How it is transferred may change This relates to energy too! Different forms of energy change into other forms but all energy is same

Note in the US – we seem to just make more money when we need it (hence the national debt) In nature, energy cannot be created when we need more – it is conserved: The amount we have today is all we get so we need to use it wisely and convert it wisely 

III. Quantitative analysis of energy problems: A. Make energy charts (LOL) B. Eqn: essentially initial energy = final energy (every situation dif) C. solve

Example Speed at the bottom of a water slide While at the county fair, Katie tries the water slide, whose shape is shown in the figure. The starting point is 9.0 m above the ground. She pushes off with an initial speed of 2.0 m/s. If the slide is frictionless, how fast will Katie be traveling at the bottom? © 2015 Pearson Education, Inc.

LOL Chart: Eqn? © 2015 Pearson Education, Inc.

Example 10.11 Speed at the bottom of a water slide (cont.) prepare Table 10.2 showed that the system consisting of Katie and the earth is isolated because the normal force of the slide is perpendicular to Katie’s motion and does no work. If we assume the slide is frictionless, we can use the conservation of mechanical energy equation. solve Conservation of mechanical energy gives Kf + (Ug)f = Ki + (Ug)i or © 2015 Pearson Education, Inc.

Mass cancels (1/2) vf2 + 0 = (1/2) (2.0m/s)2 + (9.8m/s2) (9.0m) Vf = 13.4m/s

question A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Answer: D © 2015 Pearson Education, Inc. 49

QuickCheck 10.18 A spring-loaded gun shoots a plastic ball with a launch speed of 2.0 m/s. If the spring is compressed twice as far, the ball’s launch speed will be 1.0 m/s 2.0 m/s 2.8 m/s 4.0 m/s 16.0 m/s Conservation of energy: Double x  double v © 2015 Pearson Education, Inc. 50

EX 2 with work: Monica pulls her daughter Jessie in a bike trailer. The trailer and Jessie together have a mass of 25 kg. Monica starts up a 100-m-long slope that’s 4.0 m high. On the slope, Monica’s bike pulls on the trailer with a constant force of 8.0 N. They start out at the bottom of the slope with a speed of 5.3 m/s. What is their speed at the top of the slope? Diagram? Energy bar chart?

Kf + (Ug)f = Ki + (Ug)i + W

(1/2) (25kg) vf2 + (25kg) (9.8m/s2) (4.0m) = (1/2) (25kg) (5.3m/s)2 + 0 + (8.0N) (100m) vf = 3.7m/s

Partner work: person on right A spring-loaded toy gun is used to launch a 0.01kg plastic ball. The spring, which has a spring constant of 10N/m, is compressed by 0.10m as the ball is pushed into the barrel. When the trigger is pulled, the spring is released and shoots the ball back out. What is the ball’s speed as it leaves the barrel? Assume friction is negligible. Draw energy bar chart

Kf + (Us)f = Ki + (Us)i 1/2mvf 2 + 1/2kxf2 = 1/2mvi 2 + 1/2kxi2 ½(0.01kg) (vf 2) + 0 = 0 + ½(10N/m) (-0.10m)2 vf = 3.2m/s

A 2.0 x 103 kg car is pulled 345 m up a hill that makes an angle of 15 with the horizontal. a. What is the potential energy of the car at the top of the hill? 1.75x106 J b. If the car rolls down the hill, what will its speed be if we neglect friction? 41.8m/s

BR 2/23 What is final velocity of a 34kg mass that rolls down a 12m hill from rest? Make an energy chart (LOL) solve

defn – rate at which work is done eqn: P = W/t Power defn – rate at which work is done eqn: P = W/t C. unit = Watt 1 W = 1J/s kilowatt (kW) is what power companies measure energy usage in D. Another eqn: P = F/v

d = 9.00m, t = 15.0s F = 1.20x104N P? P = W/t W = Fd EX: An electric motor lifts an elevator 9.00 m in 15.0 s by exerting an upward force of 1.20x104N. What power does the motor produce in kW? d = 9.00m, t = 15.0s F = 1.20x104N P? P = W/t W = Fd = (1.20 x104 N)(9.00m) W = 1.08 x105 J P = 1.08x105 J / 15.0 s P = 7.20 x 103 Watts = 7.20 Watts

EX: Ex. A student uses 140 N to push a block up a ramp at a constant velocity of 2.2 m/s. What is their power output? P = Fv P = (140N) (2.2m/s) P = 308 W

Quick lab on power 1. Who is the most powerful student to run up a flight of stairs? Design a lab to test this – on whiteboard Each student needs to be tested – what do you need? Record initials and power on board 2. who is the most powerful student to lift 5lbs? The 5lb weight is tied to a dowel – roll it up the 1m (on tape of rope) and time how long it takes. Each person does it Record initial and power of lifting on board

Partner Work – person on left Ex. Lover’s Leap is a 122 m vertical climb. The record time of 4 min 25 s was achieved by Dan Osman (65 kg). What was his average power output during the climb? P = ΔE / t E = Peg 4min 60s +25s = 265s 1 min P = mgh/t P = (65kg) (9.8m/s2) (122m) / 265s P = 293 W

Partner work = person on right Ex. A 1.00x103 kg car accelerates from rest to a velocity of 15.0 m/s in 4.00 s. Calculate the power output of the car. Ignore friction. P = W/t = ΔE/t = KE/t = 1/2mv2 /t P = (1/2)(1.0x103 kg) (15.0m/s)2 / (4.0s) P = 2.81x104 W

Bellringer 2/25 1. What are the three major types of energy we are studying? 2. If there are no types of energy for initial, what must be added to the initial side? 3. What is the Law of Conservation of energy? What does it really mean? 4. What is the difference between work and power? 5. list 3 types of machines

Efficiency I. Efficiency efficiency (e) of a machine is the ratio of output work to input work A measure of how much of the energy that goes into a machine actually gets used Machines are useful because they allow us to use less force over a longer distance to do the same work

Types of machines

E. Work in (Wi): total energy supplied TO a machine D. eqn: Efficiency = Wout / Win (100) E. Work in (Wi): total energy supplied TO a machine F. Work out (Wo): amount of energy actually used G. Whenever work is done, some heat is created, so Wi > Wo H. Can also calculate efficiency with power: Eff = Pout / Pin (100)

Win = Fd = (75N) (1.0m) Win = 75J Wout = ΔE = mgh = (50.0kg) (9.8m/s2) (0.1m) Wout = 49J

Efficiency = Wout / Win (100) Eff = 49J / 75J (100) Eff = 65%