Energy - Kwantlen Polytechnic Universityfacultyweb.kpu.ca/~mikec/P1101_Overheads/Work.pdfEnergy and...
Transcript of Energy - Kwantlen Polytechnic Universityfacultyweb.kpu.ca/~mikec/P1101_Overheads/Work.pdfEnergy and...
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Energy
• Many types
• Units 1 Joule (J) = 1 kg-m2/s2
• Kinetic Energy – K = ½mv2
– (also rotational and vibrational)
• Gravitational Potential Energy – Ug = mgh
• Elastic (Spring) Potential Energy – Us = ½kx2
• Internal or Thermal Energy – ETH
• Chemical Energy – Echem
• Mass Energy – E = mc2
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• Has Science identified all the possible energies?
• Probably not
• Dark Energy
– Causes the accelerating expansion of universe
– Not really known what it is
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W + Q = Esys
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0 = Esys
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Heat and Work
• Heat Q flows from a hotter to a cooler body (will
discuss later)
• When an external object exerts a force on a
system as it moves, work W is done.
• Internal forces occur is pairs (NIII), so no work
is done by internal forces
• Work is a scalar.
• Magnitude depends on orientation of force and
path.
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Biological Systems and Work • To get muscle to exert a continuous force takes
a constant supply of energy. Exhaust supply
and get tired.
• Skeleton is fairly rigid but can be used as a
support. Must still use muscle to keep skeleton
aligned. Less energy needed.
• Using body weight uses almost no energy.
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r
F
Displacement
r
F
cosrFrFW
Fr
F 𝑊 = 𝐹 ∙ ∆𝑟 = 𝐹𝑟∆𝑟 = 𝐹𝑦∆𝑥+𝐹𝑥∆𝑦+𝐹𝑧∆𝑧
Work in Vector format
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Let F = fk so a = 0 and velocity does not change.
Wext = E
Wext = Fℓ - fkℓ E = 0
But E = ETH increases, block gets warmer!
How can this be?
F
fk
ℓ
Work and Friction
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• Friction is a complex
force made up of the
forces between bumps
on surfaces.
• Since the surfaces are
not rigid, ℓ is not
distance these forces
all act over.
• For small distances,
good enough and ETH
0
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Using Work Energy Methods
• Use if interested in changes of height or speed
• Identify the system (often connected by ropes or
springs!)
• Identify objects exerting forces on system
• Note gravity force accounted by mgh already.
• Internal system forces do no work.
• Look at t = 0 and tfinal, how has the energy of
each member of the system changed?
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Power • Rate of change of energy or at which work is
done
• 𝑃 =∆𝐸
∆𝑡 , amount of energy transformed in time t
• 𝑃 =𝑊
∆𝑡 , amount of work done in time t
• 𝑃 =𝑊
∆𝑡=
𝐹𝑥∆𝑥
∆𝑡= 𝐹𝑥
∆𝑥
∆𝑡= 𝐹𝑣, rate of energy
transfer by F acting on object travelling at v
• 1 Watt (W) = 1 Joule/second
• 1 Horsepower (Hp) = 745 W
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Efficiency
• When a complex object
like a person or engine
does external work it
must use internal
energy (food or fuel).
• Not all that internal
energy goes to do work
• Breathing/circulation
etc also need energy
• Heat is also produced
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Efficiency
• e = Work done / Energy consumed
• Always e < 1 (Law of thermodynamics)
• for people e 25%
• Or e = Pout / Pin
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Energy and the Body - Food
• Food = chemical energy
• Food + O2 CO2 + energy
• Energy heat + ATP
• ATP is fuel molecule of cells
• metabolism slow/burning fast give same energy
– (use bomb calorimeter to measure E in food)
• Amount of O2 is proportional to energy
– (measure O2 consumption to measure body E needs)
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Energy and the Body - Food
• ATP end product
• Intermediate fuel storage as glycogen and
glucose in muscle tissue and liver
– Can be converted to ATP quickly
– ~400 g
– Endurance athletes use it all up and “hit the wall”
• Long term storage as fat
– Slower to convert
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Energy and the Body - Food
Food Energy in 1 g
(KJ)
Protein 17
Fat 38
Carbohydrates 17
Ethanol (alcohol) 29
Organic acids 13
Polyols (sugar alcohols, sweeteners) 10
Fibre 8
Source: Wikipedia
Units 1 Kcal = 4.19 KJ
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Example
3.5 g * 38 KJ/g + 12g * 17KJ/g + 4g*17KJ/g =405 KJ
405 Kj / 4.19 KJ/Kcal = 97 KCal
Food Energy in 1 g
(KJ)
Protein 17
Fat 38
Carbohydrates 17
Ethanol (alcohol) 29
Organic acids 13
Polyols (sugar alcohols, sweeteners) 10
Fibre 8
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Example
How many grams of
fat per serving?
Food Energy in 1 g
(KJ)
Protein 17
Fat 38
Carbohydrates 17
Ethanol (alcohol) 29
Organic acids 13
Polyols (sugar alcohols,
sweeteners) 10
Fibre 8
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Energy and the Body - Output
Source: Wikipedia
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Energy and the Body - Output
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People are very good at converting food to
fat (~100% efficient).
So a person who needs 2000 Kcal a day but
eats 2400 Kcal of food, coverts 400 Kcal to
fat.
(how may grams?)
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Rating Exercise
• Devices measure O2/CO2 and thus the total energy E used
• = total food need to consume
• Eout = Econsumed
• External work W done lifting and lowering legs is much less b/c most of the energy go to heat and internal work (breathing, beating heart, etc.).
• W = eE
http://siliconangl
e.com/blog/2010
/01/04/the-
suunto-t6c-is-a-
wearable-sports-
laboratory/
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Problem may tell you the energy needed for
the exertion. Efficiency not a factor.
Problem may tell you mechanical details so
you can calculate the mechanical work
output. Now use efficiency number to figure
out energy consumed.
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General Example
A person has a normal metabolic rate of 100
W. He is measuring doing 100 W of
mechanical work. Assume he is 25%
efficient.
(a) What is his metabolic rate now?
(b) At what rate does he produce heat?
(c) How much extra heat must he get rid of
by flushing, sweating or removing
clothing?
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We know Efood = Wmechanical + Qheat
Or Pfood = Pwork + Pheat
Now, before he exercises Pwork = 0 and
Pfood = Pheat = 100 W.
Now he exercises so his MR (metabolic
rate) goes up
PfoodNEW = (Pwork = 100 W) + Pheat
NEW
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Now efficiency is e = Pwork/Pfood
So PfoodNEW
= Pwork/0.25 = 400 W.
Now we can find the PheatNEW
400 W = 100 + PheatNEW
Or PheatNEW
= 400 W – 100 W = 300 W
The excess heat is the difference between
how much he produces now and when he
doesn’t exercise.
Pexcess heat = 300 W – 100 W = 200 W
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Example
Playing badminton “burns” 4 Kcal/hr per kg
of body mass. How long does a 65-kg player
need to play to burn off 150 Kcal?
• The 65-kg player burns 4 * 65 Kcal/hr =
260 Kcal/hr
• 150 Kcal / 260 Kcal/hr = 0.58 Hr = 35 min
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Weightlifter “benches” 80-kg 200 times
through 40 cm. How many KJ burned?
Assume e = 25%.
• Note applying muscle requires work up and
down if you don’t drop the weight.
• Work = 200 mgh 2 = 200 80 9.8 .4
2
= 313.6 KJ
• E = W/e = 313.6 KJ / 0.25 = 1254 KJ