Unit 3 Lesson 3 Conservative and NonConservative Forces

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Unit 3, Lesson 3: Conservative and Non-Conservative Forces

Unit 3, Lesson 3: Conservative and Non-Conservative Forces

Conservative Forces: – The work done in moving something from A to B is

Conservative Forces: – The work done in moving something from A to B is not dependant on the path. E. g. » Gravity » Electrostatic Force » Elastic Force

Non-Conservative Forces: – The work done moving something from A to B depends on

Non-Conservative Forces: – The work done moving something from A to B depends on the path taken. E. g. • Friction • Air resistance • Work done against a non-conservative force is converted to other forms of energy, such as heat and sound. • Work done against a conservative force is stored as potential energy.

Conservation of Energy: – If there are no non-conservative forces: PE 1 + KE

Conservation of Energy: – If there are no non-conservative forces: PE 1 + KE 1 = PE 2 + KE 2 – With non-conservative forces: • W’ = work done by the nonconservative force • W’ = ΔE = ΔPE + ΔKE • PE 1 + KE 1 + W’ = PE 2 +KE 2

Example: Bri skis down a 30°slope starting from rest. After she has lost 200

Example: Bri skis down a 30°slope starting from rest. After she has lost 200 m in elevation, her speed is 150 km/h. If her mass is 60 kg, find: a) The work done by friction and air resistance (W 1) b) The average resistance force. d = 400 m 200 m 30° Note: 150 km/h = 41. 67 m/s

Note: 150 km/h = 41. 67 m/s d = 400 m 200 m 30°

Note: 150 km/h = 41. 67 m/s d = 400 m 200 m 30° Energy before + Work(-) = Energy after mgh + W’ = ½mv 2 a) W’ = ½mv 2 – mgh = ½ (60)(41. 67)2 – (60)(9. 8)(200) = 52, 100 – 117, 600 = -65, 500 J b) W’ = F • d -65, 500 = - F (400) F = 160 N

Varying Forces F d W=F • d= area under F vs. d graph F

Varying Forces F d W=F • d= area under F vs. d graph F d

Example: Loyally defending his friend Hannah, Logan fires a 10 g bullet through a

Example: Loyally defending his friend Hannah, Logan fires a 10 g bullet through a 60 cm long gun barrel. F(N) Using the graph, find: 1500 N a) The work done on the bullet. b) The velocity of the bullet 20 40 60 d(cm) as it exits the gun. a) W = area = 0. 4 m • 1500 N = 600 J Note: distance must be in meters b) W = ΔKE 600 = ½ (0. 01)v 2 v = 350 m/s

Recommended Questions: • Energy Practice Problems #1, 3, 8 -10, 14 -17, and 19

Recommended Questions: • Energy Practice Problems #1, 3, 8 -10, 14 -17, and 19 -20