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

F = ma (Shai Gars Second Challenge)

S

Shai Gar

dd7abfc26ac768881a1c393f4cb49b88.png


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
 
I got C in Chemistry 111. So....I'll let someone else field this one.
 
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.
 
Shai Gar said:
dd7abfc26ac768881a1c393f4cb49b88.png


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
Click to expand...

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...
 
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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so, 2.5Kw consistently could be enough power to run a 150kg contraption(weight + pax + freight) at 140kmph.
 
I hate mechanical physics....
 
so... 3 horsepower should be enough :D

:) cheers. I'm using this to make a new law.
 
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.
 
*goes back to painting*
 
I don't know why I enjoy physics so much 8|
 
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:
7d8f9810361e11d14e2dfa0517eec52f.png


You can now solve your question.
 
Last edited:
So, now that these problems have been resolved, can anyone help me with this?

If the following is true, where:

S = E2/377 = 377 H2

S is power density in watts/square meter
E is the electric field strength in volts/meter
H is the magnetic field strength in amperes/meter
377 is the impedance of free space in ohms.

To convert S to milliwatts per square centimeter, multiply the entire equation by 0.1, which does the meter to centimeter and watt to milliwatt conversion.
Click to expand...

what is the formula to figure out E if we know the milliwatts per square
centimeter?
 
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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VindurNótt said:
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
Click to expand...



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]
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mayflow said:
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?
Click to expand...

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.
 
Anders Schlander said:
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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]
Click to expand...
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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