S

#### Shai Gar

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

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S

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

- MBTI
- INxJ

- Enneagram
- 4w3

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.

S

Edited to add acceleration

- MBTI
- INxJ

- Enneagram
- 4w3

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

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...

- MBTI
- ANFP

F= ma

a = dt/dv

1. m= 30kg

2. a = 100/10= 10 km/h

3.

4. m = 120 kg

5. a = 100/10 = 10 km/h

6.

7. m= 30 kg

8. a= 40/10= 4 km/h

9.

10. m = 120

11. a = 40/10 = 4

12. F = 120

Last edited:

S

- MBTI
- INFJ

- Enneagram
- 1w2 sx/so/sp

I hate mechanical physics....

- MBTI
- ANFP

To be more specific 2.1 kW, according to Newton's formula Yes!

S

so... 3 horsepower should be enough

cheers. I'm using this to make a new law.

cheers. I'm using this to make a new law.

- MBTI
- ANFP

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.

- MBTI
- INxJ

- Enneagram
- 4w3

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

- MBTI
- no comment

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.

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

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|

(where the position (s) can be in multiple dimensions)

Absolute value of force = |

Force is in Newtons or kgÂ·m/sÂ²

Work =

Work is in Joules or kg.mÂ²/sÂ²

Power is in N.m/s or kg.mÂ²/s^3

Work = m.(d|

Average power = work/time

Instantaneous power:

You can now solve your question.

Last edited:

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.

what is the formula to figure out E if we know the milliwatts per square

centimeter?

- MBTI
- iNfj

- Enneagram
- 9w1

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=

- MBTI
- infj

- MBTI
- INTP

[FONT="]

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="]

[/FONT][FONT="]

[/FONT]

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

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="]

[/FONT][FONT="]

[/FONT]

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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?

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.

- MBTI
- iNfj

- Enneagram
- 9w1

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:[FONT="]

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="]

[/FONT][FONT="]

[/FONT]

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)