Unit 3: Energy and Momentum Lesson 1: Work

Definition of Work: 1D: work = force x distance 2D: work = parallel force x distance W = F װ • d UNIT of MEASURE: Joules

W = F װ • d = 100 cos70 x 30 = 1030 J b) f = 100 cos70 = 34.2 N Example 1: Damian gets a summer job mowing lawns. His lawnmower’s handle makes an angle of 70°with the ground. A 100 N force is applied causing the 20 kg machine to cover 30 m at a constant speed. Find: A) The work done by Damian B) The work done by friction C) The work done by the normal force N W = F װ • d = 100 cos70 x 30 = 1030 J f I do – 15 min P b) f = 100 cos70 = 34.2 N W = - 34.2 x 30 = -1030 J mg c) Nװ = 0 W= 0 Note: net work done is 0!

Brain Break!

Fnet = 0 N = mg cos θ = 193 N f = uN = 77.2N P = f + mg sin θ = 111 N Example 2: Morgan pushes a 20 kg pile of textbooks 15 m up a rough ramp (u=0.4), inclined at 10 degrees, at a constant speed. Find the work done by each individual force and by the net force. N P Fnet = 0 N = mg cos θ = 193 N f = uN = 77.2N P = f + mg sin θ = 111 N f mg We do – 15 min Wnet = 0 WN = N װ • d = 0 Wf = -f • d = -1160 J Wp = P • d = 1670 J Wg = -mg sin θ • d = -510 J Interpretation Morgan does 1670 J of work. 1160 J is dissipated by friction (heat). 510 J is converted to gravitational potential energy.

Practice: Do #1 and 2 on pg. 82 10 min

Unit 3: Energy and Momentum Lesson 2: The Work-Energy Theorem

Work – Energy Theorem W = ΔKE “Work Energy Theorem” Fnet |---------------------------d--------------------------| Vo → V W = ΔKE “Work Energy Theorem” W = work done by NET force KE = ½mv2

Example 1: Famous astronaut Dane VeldBOOM is making a weekend trip to the moon. Find the thrust force (assumed constant) needed to accelerate his 600 kg rocket from 100 m/s to 150 m/s over 1.5 km (in space). W = ΔKE F • d = ½mv2 - ½mvo2 F (1500) = ½(600)(150)2 - ½(600)(100)2 F (1500) = 3.75 x 106 J F = 2500 N I do – 5 min

Video: World’s Fastest Crash Test http://www.dailymail.co.uk/news/article-2052533/Worlds-fastest-crash-test-Car-smashes-wall-120mph-speed-family-runarounds-hit.html 5 min

A) the work done by the wall on the car Example 2: A 600 kg smart car smashes into a concrete wall at 20 m/s, crumpling 0.8 m. Find: A) the work done by the wall on the car B) the average force on the car during the collision 0.8m I do – 10 min W = ΔKE = 0 - ½mv2 = 0 – ½(600)(20)2 = -120 000 J b) W = F • d F = W/d = -120 000 / 0.8 = -150 000 N

Conservation of Energy For an object under the influence of gravity, where no other forces do any work: E.g. – projectile – pendulum m V2 h2 V1 h1 Total mechanical energy = PE + KE is CONSERVED! (if no other forces do work) A famous example of this is Newton’s cradle:

Example 3: Tarzan, a 75 kg “ape-man” runs along a branch at 3 Example 3: Tarzan, a 75 kg “ape-man” runs along a branch at 3.0 m/s, grabs a vine, and swings away. The branch is 5.0 m high. Find: The speed of Tarzan as he barely misses the ground The maximum height he swings to His speed when 2.0 m above the ground 5.0m The vine is massless and unstretchable and there is no air resistance. I do – 5 min PE 1+ KE1 = PE2 + KE2 Define PE = 0 at ground level The vine does no work (always perpendicular to direction of motion) (75)(9.8)(5) +½(75)(3)2 = 0 + ½(75)v2 v2 = 2[(9.8)(5) + ½(3)2] v = 10.3 m/s

b) (75)(9.8)(5) + ½(75)(3)2 = (75)(9.8)h + 0 4012.5 = (75)(9.8)h 5.0 m b) (75)(9.8)(5) + ½(75)(3)2 = (75)(9.8)h + 0 4012.5 = (75)(9.8)h h = 5.46 m 10 min c) Etotal = ½mv2 + mgh 4012.5 = ½(75)v2 + (75)(9.8)(2) 37.5v2 = 2542.5 v = 8.2 m/s

Recommended Questions: Energy Practice Problems #2, 11-13, and 18 15 min