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

Conservation of energy Contents: Definition Conservation of Energy Sample problem 1 Sample problem 2 Whiteboards. Conservation of Energy. Energy is neither created nor destroyed, it just moves around. The total energy of a closed system remains constant.

arthur-murray
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Conservation of energy Contents: Definition Conservation of Energy Sample problem 1

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

  2. Conservation of Energy 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) TOC

  3. Speeds you up work Slows you down work Conservation of Energy Total Energy before = Total Energy After Comes from = Goes to Assets = Expenditures Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 TOC

  4. Example 1 Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 v = 4.5 m/s 250 kg What is its velocity at the bottom? 1.75 m 0 + (250 kg)(9.8 N/kg)(1.75 m) + 1/2(250 kg)(4.5 m/s)2 + 0 = 0 + 0 + 1/2(250 kg)v2 + 0 TOC

  5. Example 2 Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 v = 6.2 m/s What is its velocity after the puddle? 890 kg 3.6 m (Puddle - Exerts 3200 N of retarding force) 1/2(890 kg)(6.2 m/s)2 =(3200 N)(3.6 m) + 1/2(890 kg)v2 TOC

  6. Whiteboards: Conservation of Energy 1 | 2 | 3 | 4 | 5 | 6 | 7 TOC

  7. What speed at the bottom? vi = 0 15 kg h = 2.15 m Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + mgh + 0 =0 + 0 + 1/2mv2 (15 kg)(9.8 N/kg)(2.15 m) = 1/2(15 kg)v2 W 6.5 m/s

  8. What speed at the bottom? vi = 5.8 m/s 15 kg h = 2.15 m Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + mgh + 1/2mv2 =0 + 0 + 1/2mv2 (15 kg)(9.8 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2 W 8.7 m/s

  9. What final velocity? vi = 4.6 m/s 350 kg Pushes with 53 N for 35 m Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 Fd + 0 + 1/2mv2 =0 + 0 + 1/2mv2 (53 N)(35 m) + 1/2(350 kg)(4.6 m/s)2 = 1/2(350 kg)v2 W 5.6 m/s

  10. v = 3.68 m/s 2.34 kg The hammer pushes in the nail 3.50 mm. (.00350 m). What force did it exert on the nail Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + 0 + 1/2mv2 =Fd + 0 + 0 1/2(2.34 kg)(3.68 m/s)2 = F(.0035 m) W 4530 N

  11. 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 Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + mgh + 0 =Fd + 0 + 0 (125 kg)(9.8 N/kg)(3.2 m + .25 m) = F(.25 m) W 17,000 N

  12. What speed at the top? vi = 8.6 m/s 15 kg h = 4.25 m h = 2.15 m Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + mgh + 1/2mv2 =0 +mgh + 1/2mv2 (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)v2 W 5.7 m/s

  13. Brakes for 4.5 m. Final speed after is 4.0 m/s. What force brakes? vi = 5.8 m/s 15 kg h = 2.15 m Fd + mgh + 1/2mv2 =Fd + mgh + 1/2mv2 0 + mgh + 1/2mv2 =Fd + 0 + 1/2mv2 (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 W 99.6 N

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