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Aim: Applications of Conservation of Energy

Aim: Applications of Conservation of Energy. Do Now: A bullet fired from a rifle emerges with a kinetic energy of 2,400 J. If the barrel of the rifle is 0.5 m long, what is the average force on the bullet while in the barrel?. Δ KE = W Δ KE = Fd 2,400 J = F(0.5 m) F = 4,800 N.

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Aim: Applications of Conservation of Energy

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  1. Aim: Applications of Conservation of Energy Do Now: A bullet fired from a rifle emerges with a kinetic energy of 2,400 J. If the barrel of the rifle is 0.5 m long, what is the average force on the bullet while in the barrel? ΔKE = W ΔKE = Fd 2,400 J = F(0.5 m) F = 4,800 N ΔKE = ΔPE = W Choose the combination that works!

  2. Roller Coaster Problems m = 500 kg vi = 2 m/s A C E 40 m 30 m 20 m B D

  3. mgh (500)(9.8)(40) 196,000 J ½mv2 ½(500)(2)2 1000 J PE + KE 196,000 + 1000 197,000 J 2 m/s (Given) PE + KE = ET 0 + KE = 197,000 197,000 J KE = ½mv2 197,000 = ½(500)v2 28 m/s h = 0 0 J 197,000 J KE = ½mv2 50,000 = ½(500)v2 14.1 m/s mgh (500)(9.8)(30) 147,000 J PE + KE = ET 147,000 + KE = 197,000 50,000 J 197,000 J PE + KE = ET 0 + KE = 197,000 197,000 J KE = ½mv2 197,000 = ½(500)v2 28 m/s h = 0 0 J 197,000 J mgh (500)(9.8)(20) 98,000 J KE = ½mv2 99,000 = ½(500)v2 19.9 m/s PE + KE = ET 98,000 + KE = 197,000 99,000 J 197,000 J

  4. m = 600 kg vi = 1 m/s A C 50 m E 40 m D 30 m 20 m B

  5. mgh (600)(9.8)(50) 294,000 J ½mv2 ½(600)(1)2 300 J PE + KE 294,000 + 300 294,300 J 1 m/s (Given) PE + KE = ET 0 + KE = 294,300 294,300 J KE = ½mv2 294,300 = ½(600)v2 31.3 m/s h = 0 0 J 294,300 J KE = ½mv2 59,100 = ½(600)v2 14 m/s mgh (600)(9.8)(40) 235,200 J PE + KE = ET 235,200 + KE = 294,300 59,100 J 294,300 J mgh (600)(9.8)(20) 117,600 J PE + KE = ET 117,600 + KE = 294,300 176,700 J KE = ½mv2 176,700 = ½(600)v2 24.3 m/s 294,300 J mgh (600)(9.8)(30) 176,400 J KE = ½mv2 117,900 = ½(600)v2 19.8 m/s PE + KE = ET 176,400 + KE = 294,300 117,900 J 294,300 J

  6. What is the centripetal force at C? FC = maC FC = 5,880 N

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