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Elastic Energy March 22, 2013 - p. 1/9
Elastic Energy March 22, 2013 - p. 2/9
March 22, Week 9 Today: Chapter 7, Elastic Potential Energy - - PowerPoint PPT Presentation
March 22, Week 9 Today: Chapter 7, Elastic Potential Energy - - PowerPoint PPT Presentation
March 22, Week 9 Today: Chapter 7, Elastic Potential Energy Homework Assignment #6 - Due Today Mastering Physics: 9 problems from chapters 5 and 6 Written Questions: 6.73 Homework Assignment #7 - Due March 29 Mastering Physics: 6 problems from
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring.
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. s Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. s Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance Fsp = ks
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. s Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance Fsp = ks k = spring constant, Unit: N/m s = stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. s lo Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance Fsp = ks k = spring constant, Unit: N/m s = stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. s lo l Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance Fsp = ks k = spring constant, Unit: N/m s = stretching distance
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Elastic Energy March 22, 2013 - p. 2/9
Hooke’s Law
A simple example of a variable force is the force needed to stretch a spring. s lo l s = l − lo Hooke’s Law - The force needed to stretch or compress a spring increases linearly with stretching distance Fsp = ks k = spring constant, Unit: N/m s = stretching distance
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N (a) 0 m
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N (a) 0 m (b) 0.5 m
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N (a) 0 m (b) 0.5 m (c) 1 m
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N (a) 0 m (b) 0.5 m (c) 1 m (d) 1.5 m
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N (a) 0 m (b) 0.5 m (c) 1 m (d) 1.5 m (e) 2 m
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Elastic Energy March 22, 2013 - p. 3/9
Spring Exercise
Fsp = ks A horizontal 50 N force is applied to a 100 N/m spring whose unstretched length is 0.5 m. What is the spring’s length after the force has been applied? 50 N l0 = 0.5 m s = 0.5 m (a) 0 m (b) 0.5 m (c) 1 m (d) 1.5 m (e) 2 m
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? 0.3 m
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? 0.3 m (a) 30 N, Left
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? 0.3 m (a) 30 N, Left (b) 30 N, Right
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? 0.3 m (a) 30 N, Left (b) 30 N, Right (c) 20 N, Left
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? 0.3 m (a) 30 N, Left (b) 30 N, Right (c) 20 N, Left (d) 20 N, Right
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? 0.3 m (a) 30 N, Left (b) 30 N, Right (c) 20 N, Left (d) 20 N, Right (e) 50 N, Right
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Elastic Energy March 22, 2013 - p. 4/9
Spring Exercise II
Fsp = ks A 5-kg mass is attached, as shown, to a 100 N/m spring whose unstretched length is 0.5 m. If the mass is pushed 0.3 m to the left, what is the magnitude and direction of the force exerted by the spring on the mass? s = 0.3 m (a) 30 N, Left (b) 30 N, Right (c) 20 N, Left (d) 20 N, Right (e) 50 N, Right Springs can push or pull
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks s2 s1
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks s2 s1
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks s2 s1 F1 F2 W = 1
2(s2)(F2) − 1 2(s1)(F1)
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks s2 s1 F1 F2 W = 1
2(s2)(F2) − 1 2(s1)(F1)
W = 1
2(s2)(ks2) − 1 2(s1)(ks1)
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Elastic Energy March 22, 2013 - p. 5/9
Work to Stretch a Spring
s1 s2 s F Fsp = ks s2 s1 F1 F2 W = 1
2(s2)(F2) − 1 2(s1)(F1)
W = 1
2(s2)(ks2) − 1 2(s1)(ks1)
W = 1
2ks2 2 − 1 2ks2 1
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring.
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring.
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring.
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring.
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring. s1
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring. s1 s2
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring. s1 − → F
el
Fel = −ks s2
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring. s1 − → F
el
Fel = −ks s2 Wel = − 1
2ks2 2 − 1 2ks2 1
- Elastic work converted to potential energy
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Elastic Energy March 22, 2013 - p. 6/9
Elastic Potential Energy
Elastic Potential energy - Potential energy due to a spring. s1 − → F
el
Fel = −ks s2 Wel = − 1
2ks2 2 − 1 2ks2 1
- Elastic work converted to potential energy
Wel = −∆Uel ⇒ Uel = 1
2ks2
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Elastic Energy March 22, 2013 - p. 7/9
Conservation of Elastic Energy
If a spring is the only force doing work on something, E1 = E2
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Elastic Energy March 22, 2013 - p. 7/9
Conservation of Elastic Energy
If a spring is the only force doing work on something, E1 = E2 E = K + Uel = 1 2mv2 + 1 2ks2
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Elastic Energy March 22, 2013 - p. 7/9
Conservation of Elastic Energy
If a spring is the only force doing work on something, E1 = E2 E = K + Uel = 1 2mv2 + 1 2ks2 1 2mv2
1 + 1
2ks2
1 = 1
2mv2
2 + 1
2ks2
2
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Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring?
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Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring? (a) 125 J 500 N/m = 0.25 m
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Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring? (a) 125 J 500 N/m = 0.25 m (b) 50 kg · m/s 500 N/m = 0.1 m
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Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring? (a) 125 J 500 N/m = 0.25 m (b) 50 kg · m/s 500 N/m = 0.1 m (c) 125 J 250 N/m = 0.5 m
SLIDE 55
Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring? (a) 125 J 500 N/m = 0.25 m (b) 50 kg · m/s 500 N/m = 0.1 m (c) 125 J 250 N/m = 0.5 m (d)
- 125 J
250 N/m = 0.7071 m
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Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring? (a) 125 J 500 N/m = 0.25 m (b) 50 kg · m/s 500 N/m = 0.1 m (c) 125 J 250 N/m = 0.5 m (d)
- 125 J
250 N/m = 0.7071 m (e)
- 125 J
500 N/m = 0.5 m
SLIDE 57
Elastic Energy March 22, 2013 - p. 8/9
Elastic Potential Energy Exercise
A 10 kg mass slides across a frictionless, horizontal floor going 5 m/s (and therefore has 125 J of kinetic energy) when it collides with the k = 500 N/m spring shown. What is the maximum compression of the spring? (a) 125 J 500 N/m = 0.25 m (b) 50 kg · m/s 500 N/m = 0.1 m (c) 125 J 250 N/m = 0.5 m (d)
- 125 J
250 N/m = 0.7071 m (e)
- 125 J
500 N/m = 0.5 m
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work.
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work.
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
− → F
- ther
ANY other force doing work
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
− → F
- ther
ANY other force doing work Wtotal = Wg + Wel + Wother
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Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
− → F
- ther
ANY other force doing work Wtotal = Wg + Wel + Wother ∆K
SLIDE 65
Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
− → F
- ther
ANY other force doing work Wtotal = Wg + Wel + Wother ∆K −∆Ug
SLIDE 66
Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
− → F
- ther
ANY other force doing work Wtotal = Wg + Wel + Wother ∆K −∆Ug −∆Uel
SLIDE 67
Elastic Energy March 22, 2013 - p. 9/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
− → F
- ther
ANY other force doing work Wtotal = Wg + Wel + Wother ∆K −∆Ug −∆Uel 1 2mv2
i + mgyi + 1
2ks2
i + Wother = 1
2mv2
f + mgyf + 1
2ks2
f