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.

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 Conservation of energy means energy is not lost or gained  Important equations: E total = KE + GPE KE = ½ mv 2 GPE = ma grav h KE i + GPE i = Ke.

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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!

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

PEKEVelocity ETETETET A B C D E mgh (500)(9.8)(40) 196,000 J 2 m/s (Given) ½mv 2 ½(500)(2) J PE + KE 196, ,000 J h = 0 0 J PE + KE = E T 0 + KE = 197, ,000 J KE = ½mv 2 197,000 = ½(500)v 2 28 m/s mgh (500)(9.8)(30) 147,000 J PE + KE = E T 147,000 + KE = 197,000 50,000 J KE = ½mv 2 50,000 = ½(500)v m/s h = 0 0 J PE + KE = E T 0 + KE = 197, ,000 J KE = ½mv 2 197,000 = ½(500)v 2 28 m/s mgh (500)(9.8)(20) 98,000 J PE + KE = E T 98,000 + KE = 197,000 99,000 J KE = ½mv 2 99,000 = ½(500)v m/s

m = 600 kg v i = 1 m/s A B C D E 50 m 40 m 20 m 30 m

PEKEVelocity ETETETET A B C D E mgh (600)(9.8)(50) 294,000 J 1 m/s (Given) ½mv 2 ½(600)(1) J PE + KE 294, ,300 J h = 0 0 J PE + KE = E T 0 + KE = 294, ,300 J KE = ½mv 2 294,300 = ½(600)v m/s mgh (600)(9.8)(40) 235,200 J PE + KE = E T 235,200 + KE = 294,300 59,100 J KE = ½mv 2 59,100 = ½(600)v 2 14 m/s mgh (600)(9.8)(20) 117,600 J PE + KE = E T 117,600 + KE = 294, ,700 J KE = ½mv 2 176,700 = ½(600)v m/s mgh (600)(9.8)(30) 176,400 J PE + KE = E T 176,400 + KE = 294, ,900 J KE = ½mv 2 117,900 = ½(600)v m/s

What is the centripetal force at C? F C = ma C F C = 5,880 N