 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 Guest 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 C'est la vieRetired Staff

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I got C in Chemistry 111. So....I'll let someone else field this one.

3. ### 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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5. ### 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 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 = 30Ã—10= 300 W

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

1200+300 = 1500 W for first 10 sec

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

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

120+480= 600 W for the other 10 seconds Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
#6
Last edited: Jan 13, 2009
7. ### Shai Gar Guest

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

8. ### IndigoSensor Product ObtainedRetired Staff

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I hate mechanical physics....

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9. ### Pristinegirl Well-known member

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To be more specific 2.1 kW, according to Newton's formula Yes! Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
10. ### Shai Gar Guest

so... 3 horsepower should be enough  cheers. I'm using this to make a new law.

11. ### 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 Addicted to Bagels

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*goes back to painting*

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13. ### gloomy-optimist Used to live here

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I don't know why I enjoy physics so much 8|

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14. ### Architectonic Regular Poster

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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 kgÂ·m/sÂ²
Work = force.displacement
Work is in Joules or kg.mÂ²/sÂ²
Power is in N.m/s or kg.mÂ²/s^3

Work = m.(d|v|)Â²/2 assuming linear acceleration
Average power = work/time
Instantaneous power: You can now solve your question.

#14
Last edited: Mar 21, 2009
15. ### mayflowBanned

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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 CuddlemasterRetired Staff

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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 GONE

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Exactly!

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18. ### Anders Schlander One

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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

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[/FONT][FONT=&quot]
[/FONT]

#18
Last edited: Mar 22, 2009
19. ### mayflowBanned

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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 CuddlemasterRetired Staff

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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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