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Combo Energy March 25, 2013 - p. 1/9
Combo Energy March 25, 2013 - p. 2/9
March 25, Week 10 Today: Chapter 7, Combination Energy Problems - - PowerPoint PPT Presentation
March 25, Week 10 Today: Chapter 7, Combination Energy Problems - - PowerPoint PPT Presentation
March 25, Week 10 Today: Chapter 7, Combination Energy Problems Homework Assignment #7 - Due March 29 Mastering Physics: 6 problems from chapter 7 Written Questions: 7.60 Help sessions with Jonathan: M: 1000-1100, RH 111 T: 1000-1100, RH 114
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SLIDE 3
Combo Energy March 25, 2013 - p. 2/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work.
SLIDE 4
Combo Energy March 25, 2013 - p. 2/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g
SLIDE 5
Combo Energy March 25, 2013 - p. 2/9
General Energy Problems
The most general problems (this term) involve gravity, springs, and other forces all doing work. m− → g − → F
el
SLIDE 6
Combo Energy March 25, 2013 - p. 2/9
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
SLIDE 7
Combo Energy March 25, 2013 - p. 2/9
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
SLIDE 8
Combo Energy March 25, 2013 - p. 2/9
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 9
Combo Energy March 25, 2013 - p. 2/9
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 10
Combo Energy March 25, 2013 - p. 2/9
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 11
Combo Energy March 25, 2013 - p. 2/9
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
1 + mgy1 + 1
2ks2
1 + Wother = 1
2mv2
2 + mgy2 + 1
2ks2
2
SLIDE 12
Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
SLIDE 13
Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(a) v1 = 0, v2 = 9 m/s y1 = 0, y2 = −d s1 = −d, s2 = 0
SLIDE 14
Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(a) v1 = 0, v2 = 9 m/s y1 = 0, y2 = −d s1 = −d, s2 = 0 (b) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = 0, s2 = −d
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Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(a) v1 = 0, v2 = 9 m/s y1 = 0, y2 = −d s1 = −d, s2 = 0 (b) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = 0, s2 = −d (c) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = d, s2 = 0
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Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(a) v1 = 0, v2 = 9 m/s y1 = 0, y2 = −d s1 = −d, s2 = 0 (b) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = 0, s2 = −d (c) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = d, s2 = 0 (d) v1 = 9 m/s, v2 = 0 y1 = 0, y2 = −d s1 = 0, s2 = −d
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Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(a) v1 = 0, v2 = 9 m/s y1 = 0, y2 = −d s1 = −d, s2 = 0 (b) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = 0, s2 = −d (c) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = d, s2 = 0 (d) v1 = 9 m/s, v2 = 0 y1 = 0, y2 = −d s1 = 0, s2 = −d (e) Both (b) and (d) would work
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Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(a) v1 = 0, v2 = 9 m/s y1 = 0, y2 = −d s1 = −d, s2 = 0 (b) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = 0, s2 = −d (c) v1 = 9 m/s, v2 = 0 y1 = d, y2 = 0 s1 = d, s2 = 0 (d) v1 = 9 m/s, v2 = 0 y1 = 0, y2 = −d s1 = 0, s2 = −d (e) Both (b) and (d) would work
SLIDE 19
Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(d) v1 = 9 m/s, v2 = 0 y1 = 0, y2 = −d s1 = 0, s2 = −d 1 2(80 kg)(9 m/s)2 + 0 + 0 = 0 + (80 kg)(9.8 m/s2)(−d) + 1 2(1800 N/m)d2 (900 N/m)d2 − (784 N/m)d − 3240 J = 0 ← Quadratic equation d = 0.645 m or d = −0.558 m
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Combo Energy March 25, 2013 - p. 3/9
Combination Exercise I
An 80 kg man jumps onto a spring platform (k = 18000 N/m) going 9 m/s. Ignoring friction, which of the following is the correct set of variables to find the maximum compression of the spring, d? 1 2mv2
1 + mgy1 + 1
2ks2
1 = 1
2mv2
2 + mgy2 + 1
2ks2
2
(d) v1 = 9 m/s, v2 = 0 y1 = 0, y2 = −d s1 = 0, s2 = −d 1 2(80 kg)(9 m/s)2 + 0 + 0 = 0 + (80 kg)(9.8 m/s2)(−d) + 1 2(1800 N/m)d2 (900 N/m)d2 − (784 N/m)d − 3240 J = 0 ← Quadratic equation d = 0.645 m
- r d = −0.558 m Already put negative into equation.
SLIDE 21
Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
SLIDE 22
Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
(a) −392000 J
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Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
(a) −392000 J (b) −640 J
SLIDE 24
Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
(a) −392000 J (b) −640 J (c) 392000 J − 640 J = 391360 J
SLIDE 25
Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
(a) −392000 J (b) −640 J (c) 392000 J − 640 J = 391360 J (d) 640 J − 392000 J = −391360 J
SLIDE 26
Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
(a) −392000 J (b) −640 J (c) 392000 J − 640 J = 391360 J (d) 640 J − 392000 J = −391360 J (e) 640 J + 392000 J = 392640 J
SLIDE 27
Combo Energy March 25, 2013 - p. 4/9
Combination Exercise II
An 80 kg man skydives from a plane 500 m above the ground. If he lands with a speed of 4 m/s (and was essentially at rest when he jumped), how much work did his parachute do? Hint: (80)(9.8)(500) = 392000, 1 2(80)(4)2 = 640. 1 2mv2
1 + mgy1 + Wchute = 1
2mv2
2 + mgy2
(a) −392000 J (b) −640 J (c) 392000 J − 640 J = 391360 J (d) 640 J − 392000 J = −391360 J (e) 640 J + 392000 J = 392640 J
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Combo Energy March 25, 2013 - p. 5/9
Thermal Energy
The work done by friction is changed into Thermal Energy, Eth
SLIDE 29
Combo Energy March 25, 2013 - p. 5/9
Thermal Energy
The work done by friction is changed into Thermal Energy, Eth The work done by friction: Wf = −∆Eth
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Combo Energy March 25, 2013 - p. 5/9
Thermal Energy
The work done by friction is changed into Thermal Energy, Eth The work done by friction: Wf = −∆Eth We can recover conservation of energy by including thermal energy: 1 2mv2
i +mgyi + 1
2ks2
i +Wother +Wf = 1
2mv2
f +mgyf + 1
2ks2
f ⇒
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Combo Energy March 25, 2013 - p. 5/9
Thermal Energy
The work done by friction is changed into Thermal Energy, Eth The work done by friction: Wf = −∆Eth We can recover conservation of energy by including thermal energy: 1 2mv2
i +mgyi + 1
2ks2
i +Wother +Wf = 1
2mv2
f +mgyf + 1
2ks2
f ⇒
1 2mv2
i +mgyi+ 1
2ks2
i +Wother−∆Eth = 1
2mv2
f +mgyf + 1
2ks2
f ⇒
SLIDE 32
Combo Energy March 25, 2013 - p. 5/9
Thermal Energy
The work done by friction is changed into Thermal Energy, Eth The work done by friction: Wf = −∆Eth We can recover conservation of energy by including thermal energy: 1 2mv2
i +mgyi + 1
2ks2
i +Wother +Wf = 1
2mv2
f +mgyf + 1
2ks2
f ⇒
1 2mv2
i +mgyi+ 1
2ks2
i +Wother−∆Eth = 1
2mv2
f +mgyf + 1
2ks2
f ⇒
1 2mv2
i + mgyi + 1
2ks2
i + Wother = 1
2mv2
f + mgyf + 1
2ks2
f + ∆Eth
SLIDE 33
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s
SLIDE 34
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s (a) 10 J
SLIDE 35
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s (a) 10 J (b) 15 J
SLIDE 36
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s (a) 10 J (b) 15 J (c) 30 J
SLIDE 37
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s (a) 10 J (b) 15 J (c) 30 J (d) 45 J
SLIDE 38
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s (a) 10 J (b) 15 J (c) 30 J (d) 45 J (e) 90 J
SLIDE 39
Combo Energy March 25, 2013 - p. 6/9
Thermal Energy Exercise
A 10 kg mass sliding to the right, initially with speed 3 m/s, is stopped by friction. How much thermal energy will be created by this process? 3 m/s
No springs or gravity ⇒ 1 2mv2
i = 1
2mv2
f + ∆Eth
vi = 3 m/s, vf = 0 ⇒ 45 J = 0 + ∆Eth
(a) 10 J (b) 15 J (c) 30 J (d) 45 J (e) 90 J
SLIDE 40
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a
SLIDE 41
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a ⇒ − → F = md− → v dt
SLIDE 42
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a ⇒ − → F = md− → v dt ⇒ − → F = d(m− → v ) dt
SLIDE 43
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a ⇒ − → F = md− → v dt ⇒ − → F = d(m− → v ) dt Newton’s Second Law: − → F = d− → p dt
SLIDE 44
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a ⇒ − → F = md− → v dt ⇒ − → F = d(m− → v ) dt Newton’s Second Law: − → F = d− → p dt Momentum: − → p = m− → v Momentum measures how “hard" it is to change the velocity of an object.
SLIDE 45
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a ⇒ − → F = md− → v dt ⇒ − → F = d(m− → v ) dt Newton’s Second Law: − → F = d− → p dt Momentum: − → p = m− → v Momentum measures how “hard" it is to change the velocity of an object. Unit: kg · m/s
SLIDE 46
Combo Energy March 25, 2013 - p. 7/9
Momentum
We can rewrite Newton’s Second law in a slightly different way: − → F = m− → a ⇒ − → F = md− → v dt ⇒ − → F = d(m− → v ) dt Newton’s Second Law: − → F = d− → p dt Momentum: − → p = m− → v Momentum measures how “hard" it is to change the velocity of an object. Unit: kg · m/s ← No fancy name!
SLIDE 47
Combo Energy March 25, 2013 - p. 8/9
Impulse
Related to Momentum is Impulse, − → J :
SLIDE 48
Combo Energy March 25, 2013 - p. 8/9
Impulse
Related to Momentum is Impulse, − → J : In terms of the average net force, − → F
av:
− → J = − → F
av∆t
Unit: N · s = kg · m/s
SLIDE 49
Combo Energy March 25, 2013 - p. 8/9
Impulse
Related to Momentum is Impulse, − → J : In terms of the average net force, − → F
av:
− → J = − → F
av∆t
Unit: N · s = kg · m/s Impulse-Momentum Theorem (Average Force Version):
SLIDE 50
Combo Energy March 25, 2013 - p. 8/9
Impulse
Related to Momentum is Impulse, − → J : In terms of the average net force, − → F
av:
− → J = − → F
av∆t
Unit: N · s = kg · m/s Impulse-Momentum Theorem (Average Force Version): − → F = d− → p dt ⇒ − → F
av = ∆−
→ p ∆t ⇒ − → J = − → F
av∆t = ∆−
→ p
SLIDE 51
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE
SLIDE 52
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE AFTER
SLIDE 53
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE AFTER (a) 60 kg · m/s, ←
SLIDE 54
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE AFTER (a) 60 kg · m/s, ← (b) 60 kg · m/s, →
SLIDE 55
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE AFTER (a) 60 kg · m/s, ← (b) 60 kg · m/s, → (c) 30 kg · m/s, ←
SLIDE 56
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE AFTER (a) 60 kg · m/s, ← (b) 60 kg · m/s, → (c) 30 kg · m/s, ← (d) 30 kg · m/s, →
SLIDE 57
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−
→ p A 5.0-kg block is sliding on a frictionless, horizontal surface going 6.0 m/s to the right when it hits a wall and stops. What impulse is imparted to the block? 6 m/s BEFORE AFTER (a) 60 kg · m/s, ← (b) 60 kg · m/s, → (c) 30 kg · m/s, ← (d) 30 kg · m/s, → (e) 15 kg · m/s, ←
SLIDE 58
Combo Energy March 25, 2013 - p. 9/9
Impulse-Momentum Exercise I
− → J = − → F
av∆t = ∆−