Conservation of energy Contents: Definition Conservation of Energy Sample problem 1 Sample problem 2 Whiteboards.

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Presentation transcript:

Conservation of energy Contents: Definition Conservation of Energy Sample problem 1 Sample problem 2 Whiteboards

Conservation of Energy TOC Energy is neither created nor destroyed, it just moves around. The total energy of a closed system remains constant. Total Energy before = Total Energy After (Pendulum of death)

Conservation of Energy TOC Total Energy before = Total Energy After Comes from = Goes to Assets = Expenditures Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv 2 Speeds you up workSlows you down work

TOC Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv m v = 4.5 m/s What is its velocity at the bottom? 0 + (250 kg)(9.8 N/kg)(1.75 m) + 1 / 2 (250 kg)(4.5 m/s) = / 2 (250 kg)v kg Example 1

TOC v = 6.2 m/s What is its velocity after the puddle? 1 / 2 (890 kg)(6.2 m/s) 2 = (3200 N)(3.6 m) + 1 / 2 (890 kg)v kg (Puddle - Exerts 3200 N of retarding force) 3.6 m Example 2 Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv 2

Whiteboards: Conservation of Energy 11 | 2 | 3 | 4 | 5 | 6 | TOC

6.5 m/s W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv mgh + 0 = / 2 mv 2 (15 kg)(9.8 N/kg)(2.15 m) = 1 / 2 (15 kg)v 2 v i = 0 h = 2.15 m 15 kg What speed at the bottom?

8.7 m/s W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv mgh + 1 / 2 mv 2 = / 2 mv 2 (15 kg)(9.8 N/kg)(2.15 m) + 1 / 2 (15 kg)(5.8 m/s) 2 = 1 / 2 (15 kg)v 2 v i = 5.8 m/s h = 2.15 m 15 kg What speed at the bottom?

5.6 m/s W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv 2 Fd / 2 mv 2 = / 2 mv 2 (53 N)(35 m) + 1 / 2 (350 kg)(4.6 m/s) 2 = 1 / 2 (350 kg)v 2 v i = 4.6 m/s 350 kg What final velocity? Pushes with 53 N for 35 m

4530 N W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv / 2 mv 2 = Fd / 2 (2.34 kg)(3.68 m/s) 2 = F(.0035 m) The hammer pushes in the nail 3.50 mm. ( m). What force did it exert on the nail 2.34 kg v = 3.68 m/s

17,000 N W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv mgh + 0 = Fd (125 kg)(9.8 N/kg)(3.2 m +.25 m) = F(.25 m) The 125 kg pile driver drives the piling in.25 m with one stroke. What force does it exert? (careful what you use as the change in height) 3.2 m

5.7 m/s W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv mgh + 1 / 2 mv 2 = 0 +mgh + 1 / 2 mv 2 (15 kg)(9.8 N/kg)(2.15 m) + 1 / 2 (15 kg)(8.6 m/s) 2 =(15 kg)(9.8 N/kg)(4.25 m) 1 / 2 (15 kg)v 2 v i = 8.6 m/s h = 2.15 m 15 kg What speed at the top? h = 4.25 m

99.6 N W Fd + mgh + 1 / 2 mv 2 = Fd + mgh + 1 / 2 mv mgh + 1 / 2 mv 2 = Fd / 2 mv 2 (15 kg)(9.8 N/kg)(2.15 m) + 1 / 2 (15 kg)(5.8 m/s) 2 = F(4.5 m) + 1 / 2 (15 kg)(4.0 m/s) 2 v i = 5.8 m/s h = 2.15 m 15 kg Brakes for 4.5 m. Final speed after is 4.0 m/s. What force brakes?