Conservation of energy Contents Definition Conservation of Energy

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

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

Example 1 Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 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 + 1/2(250 kg)v 2 + 0

Example 2 Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 v = 6. 2 m/s 890 kg What is its velocity after the puddle? 3. 6 m (Puddle - Exerts 3200 N of retarding force) 1/ 2 = (3200 N)(3. 6 m) + 1/ (890 kg)v 2 (890 kg)(6. 2 m/s) 2 2

Example 3 Fd + mgh + 1/2 mv 2 = Fd + mgh + 1/2 mv 2 v = 4. 5 m/s What is its velocity after the puddle at the bottom? 250 kg 1. 75 m Puddle exerts 1200 N of slowing force for 2. 30 m 0 + (250 kg)(9. 8 N/kg)(1. 75 m) + 1/2(250 kg)(4. 5 m/s)2 + 0 = (1200 N)(2. 30 m) + 0 + 1/2(250 kg)v 2 + 0 5. 70 m/s

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

15 kg u=0 What speed at the bottom? h = 2. 15 m Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 0 + mgh + 0 = 0 + 1/2 mv 2 + (15 kg)(9. 81 N/kg)(2. 15 m) = 1/2(15 kg)v 2 v = 6. 4948… 6. 5 m/s 0

vi = 8. 7 m/s 15 kg What speed at the top? h = 2. 15 m Fd + mgh + 1/2 mv 2 = Fd + mgh + 1/2 mv 2 0 + mgh + 1/2 mv 2 = 0 + 1/2 mv 2 + 1/2(15 kg)(8. 7 m/s)2 = (15 kg)(9. 8 N/kg)(2. 15 m) + 1/2(15 kg)v 2 5. 8 m/s

What speed at the top? u = 8. 6 m/s 15 kg h = 4. 25 m h = 2. 15 m Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 0 + mgh + 1/2 mv 2 + 0 = 0 +mgh + 1/2 mv 2 + 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)v v = 5. 723… 5. 7 m/s

What final velocity? u = 4. 6 m/s 350 kg Pushes with 53 N for 35 m Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 Fs + 0 + 1/2 mv 2 + 0 = 0 + 1/2 mv 2 + (53 N)(35 m) + 1/2(350 kg)(4. 6 m/s)2 = 1/2(350 kg)v 2 5. 6 m/s 0

450. kg u = 5. 80 m/s h = 1. 75 m A 140. gram steel marble compresses a 35. 0 N/m spring a distance of 12. 0 cm. What is the greatest height the marble reaches above its lowest position? (18. 3 cm) What is the speed of the marble when it has risen only 15. 0 cm? (0. 811 m/s)

450. kg u = 5. 80 m/s What distance will the 15000. h = 1. 75 m N/m spring be compressed when the cart has been slowed to 4. 20 m/s? Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 0 + mgh + 1/2 mv 2 + 0 = 0 + 1/2 mv 2 + 1/2 kx 2 (450 kg)(9. 81 N/kg)(1. 75 m) + 1/2(450 kg)(5. 8 m/s)2 = 1/2 (450)(4. 2)2 + 1/2(15000 N/m)x 2 x = 1. 23… 1. 23 m

What speed at the top? vi = 7. 1 m/s Accelerator exerts 48. 0 N for 2. 10 m 15 kg h = 2. 15 m Fd + mgh + 1/2 mv 2 = Fd + mgh + 1/2 mv 2 Fd + 1/2 mv 2 = 0 + mgh + 1/2 mv 2 4. 66 m/s

15 kg vi = 5. 8 m/s Brakes for 4. 5 m. Final speed after is 4. 0 m/s. What force brakes? h = 2. 15 m Fd + mgh + 1/2 mv 2 = Fd + mgh + 1/2 mv 2 0 + mgh + 1/2 mv 2 = Fd + 0 + 1/2 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 99. 6 N

Harry Potter exerts a force of 180. N for a distance of 32. 0 m on the level speeding up a 980. kg car initially at rest. The car then rolls up an incline. How much elevation will the car gain before it stops? (Neglect friction) 0. 599 m

Harry Potter exerts a force of 180. N for a distance of 32. 0 m on the level speeding up a 980. kg car initially at rest. The car then rolls up an incline. How much elevation has the car gained when it is still going 2. 50 m/s? 0. 281 m

A 380. kg roller coaster car initially at rest is launched from the top of a 4. 20 m tall hill by a 1920 N/m spring compressed a distance of 6. 30 m. What is the speed of the car when it is later at the top of a 11. 0 m tall hill? (Neglect friction) 8. 19 m/s

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 + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 0 + 1/2 mv 2 + 1/ 4530 N 2(2. 34 0 = Fs + 0 + kg)(3. 68 m/s)2 = F(. 0035 m) 0 + 0

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 + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 0 + mgh + 0 = Fs + 0 + (125 kg)(9. 81 N/kg)(18. 0 m + 1. 50 m) = F(1. 50 m) F = 15, 941. 25 15, 900 N 0

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 =. 115 kg Fs + mgh + 1/2 mv 2 + 1/2 kx 2 = Fs + mgh + 1/2 mv 2 + 1/2 kx 2 0 + 0 1/ 1550 N/m 2 k(. 0315 +1/2 kx 2 = 0 + mgh + 1/2 mv 2 + 0 m)2 = 1/2(. 115 kg)(2. 13 m/s)2 + (. 115 kg)(9. 81 N/kg)(. 452 m) k = 1553. 63…