F = ma (Shai Gars Second Challenge) | INFJ Forum

F = ma (Shai Gars Second Challenge)

Discussion in 'Science and Technology' started by Shai Gar, Jan 13, 2009.

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  1. Shai Gar

    Shai Gar Guest

    [​IMG]

    Work this out:

    What wattage is needed to power a contraption that weighs 30kg, with an additional 120kg (PAX + Freight) at a speed of 140kmph continuously.

    Acceleration from 0 - 100 @ ten seconds, 100 - 140 @ 10 seconds
     
  2. Satya

    Satya C'est la vie
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    I got C in Chemistry 111. So....I'll let someone else field this one.
     
  3. gloomy-optimist

    gloomy-optimist Used to live here

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    Well, maybe my logic is faulty, but if you're moving the contraption at a constant 140k/h, then your acceleration is 0, and your force is also equal to 0.
    So if it's just a object moving at a constant speed without any indication of placement (thus assuming it's on a flat, frictionless surface), there would be no force needed.
     
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  4. OP
    Shai Gar

    Shai Gar Guest

    Edited to add acceleration
     
  5. gloomy-optimist

    gloomy-optimist Used to live here

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    F = 150k x a
    a = ∆v/t = 100/10 = 10 kmph (although I'm probably cutting corners by assuming it'd be already in kmph)
    So:
    F = 150 x 10 = 1500 for the first ten seconds

    a = 40/10 = 4 kmph
    So:
    F = 150 x 4 = 600 for the second ten seconds

    I think I may have over simplified that a bit, but oh well. I do this enough for class as is...
     
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  6. Pristinegirl

    Pristinegirl Well-known member

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    Looks like someone solved it before me. More clarified:

    F= ma
    a = dt/dv

    1. m= 30kg
    2. a = 100/10= 10 km/h
    3. F = 30×10= 300 W


    4. m = 120 kg
    5. a = 100/10 = 10 km/h
    6. F = 120×10= 1200 W

    1200+300 = 1500 W for first 10 sec

    7. m= 30 kg
    8. a= 40/10= 4 km/h
    9. 150×7= 120

    10. m = 120
    11. a = 40/10 = 4
    12. F = 120×4= 480


    120+480= 600 W for the other 10 seconds

    :)
     
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    #6 Pristinegirl, Jan 13, 2009
    Last edited: Jan 13, 2009
  7. OP
    Shai Gar

    Shai Gar Guest

    so, 2.5Kw consistently could be enough power to run a 150kg contraption(weight + pax + freight) at 140kmph.
     
  8. IndigoSensor

    IndigoSensor Product Obtained
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  9. Pristinegirl

    Pristinegirl Well-known member

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    To be more specific 2.1 kW, according to Newton's formula Yes! :)
     
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  10. OP
    Shai Gar

    Shai Gar Guest

    so... 3 horsepower should be enough :D

    :) cheers. I'm using this to make a new law.
     
  11. Pristinegirl

    Pristinegirl Well-known member

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    Oh that is great Shai and yes it would!! :)
    I would love to hear it when it's done.
    I made an innovative theory myself, using verifiable justification methods, mathematics, on a totally different branch of subject thought.
     
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  12. Rosenrot

    Rosenrot Addicted to Bagels

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    *goes back to painting*
     
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  13. gloomy-optimist

    gloomy-optimist Used to live here

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  14. Architectonic

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    Pristinegirl's version is wrong and gloomy-optimist's version is wrong due to incompleteness (no units etc).

    In the real world you have to consider frictional losses and how the power is transferred (eg what sort of motor? what direction etc?)

    In the first part, the power needed is determined by the frictional losses.

    The losses are mentioned here:
    http://autoblog.xprize.org/axp/2007/10/the-car-equatio.html

    keeping in mind the first term mgCrr is missing a cos (θ)

    The second part of the question is unsolvable without stating the nature of the acceleration - eg giving a curve.
    In the real world you have to consider frictional losses and how the power is transferred (eg what sort of motor? what direction etc?)

    But you could assume linear acceleration for now.

    (bold indicates vector, | | indicates absolute value)

    d|v| = 100 km/h = 27.8 m/s

    velocity = v = ds/dt
    (where the position (s) can be in multiple dimensions)

    Absolute value of force = |F|=m.d|v|/dt

    Force is in Newtons or kg·m/s²
    Work = force.displacement
    Work is in Joules or kg.m²/s²
    Power is in N.m/s or kg.m²/s^3

    Work = m.(d|v|)²/2 assuming linear acceleration
    Average power = work/time
    Instantaneous power: [​IMG]

    You can now solve your question.
     
    #14 Architectonic, Mar 21, 2009
    Last edited: Mar 21, 2009
  15. mayflow

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    So, now that these problems have been resolved, can anyone help me with this?

    If the following is true, where:

    what is the formula to figure out E if we know the milliwatts per square
    centimeter?
     
  16. NaeturVindur

    NaeturVindur Cuddlemaster
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    A couple of definitions before doing this problem:
    Watts is the measure of power to do something
    Power (P)=Work(W)/Time(t)
    W=Force(F)*distance (d) *cos(Angle) (for this problem I’m going to assume that the force is in the direction of travel)
    F=Mass (m) *acceleration (a, or dv/dt)
    this is two part problem, do the total Power would be the sum of the Power for each part

    Part 1
    m=150kg
    t=10s
    a=100/10=10m/s/s
    d=(1/2)*10m/s/s*(10s)^2=500m

    F=150kg*10m/s/s=1500N
    W=1500N*500m=750000J
    P=750000J/10s=75000W

    Part 2
    m=150kg
    t=10s
    a=(140-100)/10=4m/s/s
    d=(1/2)*4m/s/s*(10s)^2=200m

    F=150kg*4m/s/s=600N
    W=600N*200m=120000J
    P=120000J/10s=12000W

    Total P=12000W+75000W=87000W
     
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  17. just me

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    Exactly!
     
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  18. Anders Schlander

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    [FONT=&quot]


    Is somewhat off, part 1 = is 0-100 KM/h so 50 km/h avg.

    50km/h / 3600s = 13.88 m / s Avg speed first 10 seconds.

    13.88m/s * 10s = 138.8 meters.


    F=150kg*10.00m/s/s*=1500N
    W=1500N*138.8m=208200J
    P=208200J/10s=20820W

    P = 2,082 KW

    Next part is not 120km/h - 100km/h, it is 140+100/2 = 120km/h

    120km/h / 3.6 = 33.333 m/s avg. through the last 10 seconds...


    and thus travels 333 meters.

    F=150kg*4m/s/s=600N
    W=600N*333M = 199800J
    199800J/10s = 19980W

    = 1,99 Kw

    1,99 Kw + 2,082 Kw = 4,072 Kw

    not sure this is right, my computer uses 0.4 Kw... 10 of my computers could move a 150kg object that fast?....

    I took the formula Vindurnott, though. Just fixed the distances...


    There is a better way of calculating the straight forward kinetic energy=0.5 MV^2 that on average is beeing measured....

    0.5 * 150kg * ( 120+50/2 ) ^2 = 1576875 Joule

    1576875/20s = 78843,75 Watt

    = 78,84375 kilowatt for constant 85 km/h travel.

    [/FONT] but by calculating energy=0.5*150*(140km/h)^2 = 1470000 joule to go 140km/h

    energy=0.5*150*(100km/h)^2 = 750000 Joile to go 100 km/h

    1470000/750000 = 1,96 ratio. It nearly doubles the amount of energy needed from 100 to 140.

    while the amount of energy for the average speed ( 120 ) = 75*(120)^2 = 1080000 joule.. 1080000/750000 = 1.44

    100km/h = 100

    120km/h = 1.44

    140km/h = 1,96 thus the maximum needs to be 1,88 for average value to work, but its not. it is 0.08 higher

    so instead of 1080000 Joule it is actually 1080000*(1+0.04) = 1123200 Joule ( note that the increase is halved, because it is an average of the 2 diferrences.. )

    Doing the same thing with the 0-100 section of acceleration,

    0 = 0

    50 = 75*(50^2) = 187500 J

    100 = 75*(100^2) = 187500 = 750000J

    0 = 0
    187500J = 1

    750000 = 4,016.

    187500 * (4,016/0.50) = 376500 Joule.

    1080000*(1+0.04) = 1123200 Joule

    376500 + 1123200 Joule = 1499700 Joule

    1499700/20 = 74975 Watt

    = 74,975 Kilowatt


    Hmm...im not going to go closer than this, without drawing a graph..... but this is what i worked out

    [FONT=&quot]
    [/FONT][FONT=&quot]
    [/FONT]
     
    #18 Anders Schlander, Mar 22, 2009
    Last edited: Mar 22, 2009
  19. mayflow

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    I think I got it worked out now...



    I expressed mw/cm^2 as .001 W/.0001 M^2, which meant to me that
    1mw/cm^2 = 10W/M^2, so that meant that...

    If: V/M = the suare root of (377 * W/M^2)

    Then: V/M = the square root of (3770 * mw/cm) - since it takes 10W/M^2 to make a mw/cm2 -

    Does that make sense?

    And yes, this was a real world problem. I needed to do an A2LA certified
    calibration on a microwave leakage detection instrument which only
    measures out in mw/cm^2, and I only have software capable of telling
    me the Electric Field I am generating in V/M, so I needed to know how to
    convert.
     
  20. NaeturVindur

    NaeturVindur Cuddlemaster
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    your first part is not even close, with one little mistake, acceleration is NOT the average velocity, F=ma, not F=mv. however, reading that did make me realize a mistake I made. I mixed up kmph and mph, so the new work is thus:
    Part 1
    m=150kg
    t=10s
    a=100,000/10=10,000m/s/s
    d=(1/2)*10000m/s/s*(10s)^2=500,000m

    F=150kg*10,000m/s/s=1,500,000N
    W=1500000N*500000m=750,000,000,000J
    P=75,000,000,000J/10s=75,000,000,000W

    Part 2
    m=150kg
    t=10s
    a=(140,000-100,000)/10=4,000m/s/s
    d=(1/2)*4m/s/s*(10s)^2=200,000m

    F=150kg*4m/s/s=600,000N
    W=600,000N*200,000m=120,000,000,000J
    P=120,000,000J/10s=12,000,000,000W

    Total P=12,000,000,000W+75,000,000,000W=87,000,000,000W
    or 87GW (giga-watts)
     
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