Unit 6, Lesson 1: Work is Converted to Energy

POTENTIAL ENERGY “stored energy” In physics, “work” is done when an applied force causes an object to move. The potential energy stored is equal to the work done creating it. 5 min

Gravitational Potential Energy The work done to lift (or lower) something is equal to its change in gravitational potential energy. m EP = m·g·h h 5 min h=0 (reference point)

relative to the floor in the room. Try This: Find the potential energy of a 60 kg student sitting on an 80 cm high stool… relative to the floor in the room. relative to the floor in the basement below (3.0 m lower). a) EP = mgh = 60 x 9.8 x 0.8 = 470 J b) EP = mgh 10 min = 60 x 9.8 x 3.8 = 2200 J

KINETIC ENERGY “the energy of motion” Kinetic energy = the amount of work needed to accelerate a mass “m” from rest to velocity “v” 5 min EK = ½ mv2

Brain Break!

Heat Energy Definition: The total kinetic energy of the molecules in a substance. Temperature: A measure of the average kinetic energy per molecule. Q - amount of heat energy (J) C - specific heat capacity of the substance (energy needed to raise 1 kg by 1°C) m - mass (kg) ΔT - change in temperature (°C) 10 min Q = mcΔT

Example 1: Find the heat energy needed to raise a cup of water (250 g) from room temperature (20°C) to the boiling point. Q = mcΔT = 0.25 x 4200 x (100-20) = 84000 J Example 2 (Try it!): Find the final temperature when 20000 J of heat is added to a 3.0 kg concrete block at 20°C. (c = 2900 J / kg°C for concrete) 20 min Q = mcΔT 20000 = 3.0 x 2900 x ДT ДT = 2.3°C T = 22.3 °C

Unit 6, Lesson 2: Work and Power

Work work = force x distance (Joules (J) = N x m) Example: How much work is done by a crane that lifts 120 kg up 75 m? F = mg W = F·d = m·g·d = 120 x 9.8 x 75 = 88000 J 10 min

Power power = rate of doing work power = rate of using energy power = work (Watts (W) = J/s) time Example: Find the power output of a 60 kg athlete who climbs the Grouse Grind in 40 minutes. (Elevation gain = 800 m) 10 min power = work time = (mg)d t = 60 x 9.8 x 800 2400 = 196 W

Brain Break!

Practice: Pg. 139 #1-5 and 8-10 35 min

Lesson 3: Human Power Lab Once you’ve gathered your materials, you may go make your measurements on any hill you choose – just let me know where you are going! *

Unit 6, Lesson 4: Work-Energy Theorem

Work – Energy Theorem W = ΔE The work done on an object by a net force equals the change in mechanical energy of that object. 5 min

Example 1 A 1000 kg car going 20 m/s brakes and comes to a stop. How much work is done? What happens to the lost kinetic energy? EK = ½mv2 = ½(1000)(20)2 = 200 000 J W = ΔEK = 0 – 200 000 J = – 200 000 J (lost kinetic energy) 10 min The lost kinetic energy is converted to heat due to work done by friction.

Example 2 (Try It!) How much work is done when a car of mass 1200 kg speeds up from 10 m/s to 20 m/s? If this is done in 15 seconds, find the power output. W = Δ EK W = ½mv2 - ½mvo2 W = ½(1200)(20)2 - ½(1200)(10)2 = 180 000 J 15 min P = W t = 180 000 15 = 12 000 W

Brain Break!

Example 3 (Try It!) A 22 kg sled at rest is pushed 12 m with a force of 30 N against a friction force of 10 N. Find: a) the work done by the pusher b) the work done by friction c) the kinetic energy gained by the sled a) W = F·d b) W = F·d = 30x12 = -10x12 15 min = 360 J = -120 J c) ΔEK = W (done by net force) = 20x12 = 240 J

Example 4 (Try It!) How much work is done to slow down a 1200 kg vehicle from 80 km/h to 50 km/h? W = ΔEK W = ½mv2 - ½mvo2 10 min W = ½(1200)(13.889)2 - ½(1200)(22.222)2 = -181 000 J

Unit 6, Lesson 5: Efficiency

Efficiency Power Output Energy Output Efficiency = x 100% = x 100% Power Input Energy Input Example 1: Find the efficiency of a 1200 W kettle that heats 1 cup (250 g) of water from 20 to 70°C in 1 minute. Output: Q = mcДT = (0.25)(4200)(70-20) = 52500 J 15 min Input: E = Pt = (1200)(60) = 72000 J Eff = (52500/72000)x100 = 73%

Example 2 (Try it!): A crane operates at 15 kW and lifts a 700 kg mass 50 m in 2 minutes. Find the crane’s efficiency. Input: E = Pt = (15000)(120) = 1 800 000 J Output: E = EP = mgh = (700)(9.8)(50) = 343 000 J 15 min Eff = (343000) x100 = 19% 1800000

Brain Break!

Example 3 (Try it!): A 6.5 hp forklift is 15% efficient when lifting a 220 kg box. How long does it take to lift the box 1.5 m? (1 hp = 746 W) Output Efficiency = 0.15 = Input Pin = 6.5 hp x 746 W/hp = 4849 W 15 min EP/t = mgh 0.15 = Pout/Pin = = (220)(9.8)(1.5) Pin t Pin t (4849) t = 4.5 s

Practice: Work and Power Problems #1 10 min

Unit 6, Lesson 6: Efficiency Practice

Example 4 (Try it!): An 800 kg 100-hp car is 22% efficient at transferring energy into forward motion. How long does it take to go from 0 to 108 km/h? (1 hp = 746 W) Pout = 16400 W t = E/P = ½ mv2 / Pout = ½ (800)(30)2 / 16400 = 22 s Output Efficiency = 0.22 = Input Pout_______ 0.22 = Pout/Pin = (100 hp)(746 W/hp) 15 min

Brain Break!

Practice: Work and Power Problems #2-5 40 min

Unit 6, Lesson 7: Conservation of Energy

Conservation of Mechanical Energy Total Mechanical Energy of an object moving in a gravitational field: E = EP + EK E = mgh + ½mv2 Conservation of Energy Law: If no other force (besides gravity) does work on an object, the total mechanical energy is conserved. 5 min Eo = Efinal

Example 1 (Try it!): Matt kicks a soccer ball of mass 0.6 kg with an initial speed of 15 m/s. Find its total mechanical energy kinetic energy at height 4.0 m speed when at 4.0 m height when the speed is 5.0 m/s. v h 15m/s 15 min a) E = mgh + ½mv2 = 0 + ½ (0.6)(15)2 = 67.5 J

b) c) EK = ½mv2 E = EP + EK E = mgh + EK EK = ½(0.6)v2 b) kinetic energy at height 4.0 m c) speed when at 4.0 m d) height when the speed is 5.0 m/s m = 0.6 kg ET = 67.5 J b) E = EP + EK c) EK = ½mv2 E = mgh + EK EK = ½(0.6)v2 67.5 = (0.6)(9.8)(4) + EK v = 12 m/s EK = 44 J d) 10 min E = mgh + ½mv2 67.5 = (0.6)(9.8) h + ½ (0.6)(5)2 67.5 = 5.88h + 7.5 h = 10.2 m

Brain Break!

Example 2 (Try it!): Christine (75 kg including her gear) skis straight down an icy slope, from rest. After she has lost 250 m of elevation, a) find her final velocity assuming no effects from air-resistance or friction. b) if her final velocity is 210 km/h, find the total amount of energy lost due to friction forces. What happens to the “lost” energy? a) Etop = Ebottom 0m/s mgh = ½ mv2 15 min gh = ½ v2 2gh = v2 2(9.8)(250) = v2 250m v = 70 m/s (252 km/h) h = 0

v = 210 = 58.3 m/s 3.6 ΔE = Ebottom – Etop = ½ mv2 - mgh b) if her final velocity is 210 km/h, find the total amount of energy lost due to friction forces. What happens to the “lost” energy? v = 210 = 58.3 m/s 3.6 ΔE = Ebottom – Etop = ½ mv2 - mgh = ½ (75)(58.3)2 - (75)(9.8)(250) 5 min = - 56 000 J (converts to heat)

Practice: Conservation of Energy Problems #1 5 min

Unit 6, Lesson 8: Practice with Conservation Law

Find his velocity when he swings past ground level. Practice #1: Tarzan, a 75 kg ape-man, swings from a branch 3.0 m above the ground with an initial speed of 5.0 m/s. Find his velocity when he swings past ground level. Find the maximum height he swings to. What should his initial speed be if he is to just reach a branch 4.0 m high? b) & c) 20 min 3.0 m a) h = 0 Note: In the physics jungle there is no air resistance and the vines have no mass and are unstretchable!

a) mgho + ½mvo2 = mgh + ½mv2 m cancels out; h = 0 gho + ½vo2 = 0 + ½v2 v = √(vo2 + 2gho) v = √((5.0)2 + 2(9.8)(3.0)) v = 9.2 m/s horizontal b) gho + ½vo2 = 0 + gh v = 0 h = vo2 + ho 2g 5 min h = 4.3 m c) gho + ½vo2 = 0 + gh v = 0 vo = √(2gh - 2gho) vo = 4.4 m/s

Brain Break!

Practice #2: George of the Jungle swings from the same 3 Practice #2: George of the Jungle swings from the same 3.0 m high branch with an initial speed of 3.5 m/s. Unfortunately, he hits a tree when he is1.5 m above the ground. Find his speed just before impact and the work done by the tree. His mass is 90 kg. 15 min 3.0 m 1.5 m

mgho + ½mvo2 = mgh + ½mv2 m cancels out gho + ½vo2 = gh + ½v2 v = √(vo2 + 2gho – 2gh) v = √((3.5)2 + 2(9.8)(3.0) – 2(9.8)(1.5)) v = 6.5 m/s W = ΔKE = KE - KEo W = ½mv2 – ½mvo2 5 min W = 0 – ½(90)(6.5)2 W = -1900 J

Practice: Conservation of Energy Problems #2 and 3 10 min

Unit 6, Lesson 9: More Practice with Conservation Law Jacob Clare Travis McKenna Chase Balance Carli Nathan Taylor Anna Aidan Finley Amira Ashley Kally Annie

Heat = Work done by friction W = Ef - Eo W = ½ mv2 - mgh Practice #3: How much heat energy is produced when a speed skier loses 300 m in elevation if he starts from rest and finishes going 40 m/s? His mass (including gear) is 90 kg. Heat = Work done by friction 0m/s W = Ef - Eo W = ½ mv2 - mgh W = ½(90)(40)2 – (90)(9.8)(300) W = -1.9 x 105 J 300m 15 min 40m/s 190 000 J of heat are produced h = 0

Heat energy gained by the pool = KE lost by the meteor Practice #4: A 5 kg meteorite lands in a swimming pool with 100 000 L of water. The temperature rises 5°C. Find the velocity of the meteor right before it hits the water. Q = mcΔT = 100000 x 4200 x 5 = 2.1 x 109 J Heat energy gained by the pool = KE lost by the meteor Q = KE = ½mv2 2.1 x 109 = ½(5)v2 v = 29000 m/s 15 min

Brain Break!

Practice #5: A 20 kg meteor crashes into a 1.5m-deep pool of dimensions 4m X 6m at 17 km/s. Find the final temperature if the initial temp is 22C and all of the meteor’s energy heats the water. (Density of H2O = 1000 kg/m3.) Q = KE = (0.5)(20)(17 000)2 = 2.89x109 J mc∆T = Q [(1.5 x 4 x 6) x 1000] x 4200 x ∆T = 2.89x109 36000 x 4200 x ∆T = 2.89x109 ∆T = 19C ∆T = 22+19 = 41C 20 min

Practice: Conservation of Energy Problems #4 and 5 10 min