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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. Speeds you up work. Slows you down work. Conservation of Energy. Total Energy before = Total Energy After Comes from = Goes to Assets = Expenditures

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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. Speeds you up work Slows you down work Conservation of Energy Total Energy before = Total Energy After Comes from = Goes to Assets = Expenditures Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 TOC

  3. Example 1 Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 v = 4.5 m/s 250 kg What is its velocity at the bottom? 1.75 m 0 + (250 kg)(9.81 N/kg)(1.75 m) + 1/2(250 kg)(4.5 m/s)2 + 0 = 0 + 0 + 1/2(250 kg)v2 + 0 TOC

  4. Example 2 Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 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

  5. Whiteboards: Conservation of Energy 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 TOC

  6. What speed at the bottom? u = 0 15 kg h = 2.15 m Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + mgh + 0 + 0 =0 + 0 + 1/2mv2 + 0 (15 kg)(9.81 N/kg)(2.15 m) = 1/2(15 kg)v2 v = 6.4948… W 6.5 m/s

  7. What speed at the bottom? u = 5.8 m/s 15 kg h = 2.15 m Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + mgh + 1/2mv2 + 0 =0 + 0 + 1/2mv2 + 0 (15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(5.8 m/s)2= 1/2(15 kg)v2 v = 8.7076… W 8.7 m/s

  8. What speed at the top? u = 8.6 m/s 15 kg h = 4.25 m h = 2.15 m Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + mgh + 1/2mv2 + 0 =0 +mgh + 1/2mv2 + 0 (15 kg)(9.81 N/kg)(2.15 m) + 1/2(15 kg)(8.6 m/s)2 =(15 kg)(9.81 N/kg)(4.25 m) 1/2(15 kg)v2 v = 5.723… W 5.7 m/s

  9. What final velocity? u = 4.6 m/s 350 kg Pushes with 53 N for 35 m Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 Fs + 0 + 1/2mv2 + 0 =0 + 0 + 1/2mv2 + 0 (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 Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + 0 + 1/2mv2 + 0 =Fs + 0 + 0 + 0 1/2(2.34 kg)(3.68 m/s)2 = F(.0035 m) W 4530 N

  11. A 125 kg experiment falls 18.0 m and has its velocity arrested in 1.50 m by an air bag. What force does the air bag exert on the experiment in stopping the experiment? 18 m Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + mgh + 0 + 0 =Fs + 0 + 0 + 0 (125 kg)(9.81 N/kg)(18.0 m + 1.50 m) = F(1.50 m) F = 15,941.25 W 15,900 N

  12. What distance will it compress the 15000 N/m spring in stopping the cart? u = 5.8 m/s 450 kg h = 1.75 m Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + mgh + 1/2mv2 + 0 =0 + 0 + 0 + 1/2kx2 (450 kg)(9.81 N/kg)(1.75 m) + 1/2(450 kg)(5.8 m/s)2 = 1/2(15000 N/m)x2 x = 1.428… W 1.4 m

  13. What must be the spring constant to give the 115 g marble a velocity of 2.13 m/s on top of the hill if the spring is compressed 3.15 cm? h = .452 m x = .0315 m, m = .115kg Fs + mgh + 1/2mv2 + 1/2kx2 =Fs + mgh + 1/2mv2 + 1/2kx2 0 + 0 + 0 +1/2kx2 =0 + mgh + 1/2mv2 + 0 1/2k(.0315 m)2 = 1/2(.115 kg)(2.13 m/s)2 + (.115 kg)(9.81 N/kg)(.452 m) k = 1553.63… W 1550 N/m

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