Physics question: Velovity and power

[calvinbiss]

how do you relate velocity and power? i have Joules/second, and i need to know meters/sec. i know there is an equation, but i can;t remember it. please please help me!

 

[BaliBabyDoc]

velocity m/s (vector)
force Nm/s
power=force per unit time Nm/s^2

 

[BaliBabyDoc]

Damn it looked right when I typed it . . .

velocity m/s
mass kg
force kg*m/s^2
work Nm
power=work per unit time J/s or watts or Nm/s

 

[calvinbiss]

ok, i really appreciate this, and we are going to get this:

the question says i have a 0.1kg mass with a Power of 10^4 Watts, and it wants to know displacement after a year in space.

thank you thank you

 

[duke]

power = ((mass x acceleration) x distance) / time

 

[b0mbrman]

One joule = 1 newton x 1 meter

 

[duke]

1 Newton = 1 kg x 1 m/s^2

 

[duke]

calvinbiss,

are you given acceleration?

 

[Mookow]

<< ok, i really appreciate this, and we are going to get this:

the question says i have a 0.1kg mass with a Power of 10^4 Watts, and it wants to know displacement after a year in space.

thank you thank you >>



You mean it has 10^4 Joules? That really doesnt make sense if its power... that means you are adding 10^4 Joules of energy to the mass every second. Now, I assume its energy added in the form of velocity, not heat or other forms of energy. Now, you would likely reach relativistic speeds pretty fast, as the mass is pretty low. So, since I dont think we are talking about Watts, but rather Joules (and energy), that means the speed of the rock is given by
KE=1/2*m*v^2
So, 10^4 J=.05 kg *v^2
10^4 J/.05 kg = v^2
v^2=2000000J/kg
v=1414.21m/s
However, since you need displacement:
x=v*t actually, x=v*t+1/2*at^2, but a=0, so the second half doesnt count
We need the number of seconds in a year... 60secs*60mins*24hours*365days=31536000
x=1414.21 meters/second * 31,536,000 seconds
x=44598638903 meters

 

[Mookow]

bump for a possible answer in my last post

 

[BaliBabyDoc]

too much help . . . how will they learn if you give them the answer?

 

[Shantanu]

Your question makes absolutely no sense. You can't "relate" power and velocity, like the way you can relate a meter and a yard.

Velocity is measured in meters/second.

Power is measured in Kilogram*m^2/s^3. That's a difference of a Kilogram*m/s^2 factor. There's no way you can just "convert" that.

 

[Lithium381]

Not directly convertable......hrm, you need to rephrase it so we know how it's applied to eachother...good luck though if i'm not around when you do.

 

[calvinbiss]

ok, i don't know if anyone is still paying attention, but this question is trying to relate Watts with a displacement. The 10^4 W is generated by a laser, which, along with its battey, weigh 0.1kg. Given that the battery life is infinite, and that we are in space, the question is how far away from the starting point will the laser/battery be in 1 year. it is essentially a rocket with light as its exhaust. i think i have an answer, unfortunatly, no with me right now, but it was a rather large number.

tell me what you think now,

thanks



<< too much help . . . how will they learn if you give them the answer? >>



i am not trying to learn, i am trying to finish

 

[Shantanu]

Basically you want to boil the "power" down to an acceleration term, then - using some calculus - incorporate the time elapsed to find the distance. I'd solve it for you, but today's my non-studying day.

Actually, I lied. I did solve it. What I did was to set .5mv^2 equal to the integral of power with respect to time, integrated to the number of seconds in a year. The velocity value I got I halved, to get the average velocity (and this where I made an assumption: constant acceleration), and the I just multiplied that by the seconds in a year. I got 3.964 E13 meters. Cheers.

 

[calvinbiss]

i am in a non-calculas/algebra based physics, so it should be solvable without any calculas!! that is my problem, i can;t find anything to relate acceleration to.

 

[Heisenberg]

1 W = 1 J/s
Multiply the number of watts by the seconds in 1 year to get the total number of Joules.
Assuming all that energy is converted to kinetic, set the number equal to 0.5*m*v^2 and solve for v (velocity).
Then divide your velocity by time to get average acceleration (assuming you start from rest).
Then use (final velocity)^2 = (initial velocity)^2 (zero I assume) + 2*accleration*distance and solve for distance.

 

[calvinbiss]

by george i think you've done it!!

thanks