Aim: Applications of Conservation of Energy Do Now: A bullet fired from a rifle emerges with a...
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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!
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Transcript of Aim: Applications of Conservation of Energy Do Now: A bullet fired from a rifle emerges with a...
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
vi = 2 m/s A
B
C
D
E40 m
30 m
20 m
Roller Coaster ProblemsRoller Coaster Problems
PEPE KEKE VelocityVelocity EETT
AA
BB
CC
DD
EE
mgh
(500)(9.8)(40)
196,000 J
2 m/s
(Given)
½mv2
½(500)(2)2
1000 J
PE + KE
196,000 + 1000
197,000 J
197,000 J
197,000 J
197,000 J
197,000 J
h = 0
0 J
PE + KE = ET
0 + KE = 197,000
197,000 J
KE = ½mv2
197,000 = ½(500)v2
28 m/s mgh
(500)(9.8)(30)
147,000 J
PE + KE = ET
147,000 + KE = 197,000
50,000 J
KE = ½mv2
50,000 = ½(500)v2
14.1 m/s
h = 0
0 J
PE + KE = ET
0 + KE = 197,000
197,000 J
KE = ½mv2
197,000 = ½(500)v2
28 m/s mgh
(500)(9.8)(20)
98,000 J
PE + KE = ET
98,000 + KE = 197,000
99,000 J
KE = ½mv2
99,000 = ½(500)v2
19.9 m/s
m = 600 kg
vi = 1 m/s
A
B
C
D
E50 m40 m
20 m
30 m
PEPE KEKE VelocityVelocity EETT
AA
BB
CC
DD
EE
mgh
(600)(9.8)(50)
294,000 J
1 m/s
(Given)
½mv2
½(600)(1)2
300 J
PE + KE
294,000 + 300
294,300 J
294,300 J
294,300 J
294,300 J
294,300 J
h = 0
0 J
PE + KE = ET
0 + KE = 294,300
294,300 J
KE = ½mv2
294,300 = ½(600)v2
31.3 m/s mgh
(600)(9.8)(40)
235,200 J
PE + KE = ET
235,200 + KE = 294,300
59,100 J
KE = ½mv2
59,100 = ½(600)v2
14 m/s 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 mgh
(600)(9.8)(30)
176,400 J
PE + KE = ET
176,400 + KE = 294,300
117,900 J
KE = ½mv2
117,900 = ½(600)v2
19.8 m/s
What is the centripetal force at C?
FC = maC
FC = 5,880 N
