Acceleration problem

Originally posted by sonhouse
What do you think I defined? E. One megawatt. Jeez.
Getting close to c is what the fitzgerald contraction formula is all about.
Did you read my PM's?

Of course you add kinetic energy to a body under acceleration, call it a battery, so what? Ion rockets do that by accelerating at a constant but low acceleration, the total kinetic energy always goes up. That's just the way the cookie crumbles!

Watts are power, not energy.

Originally posted by joe shmo
This is explicitly solvable as stated

"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)" (You also stated how they were the same thing, that however is incorrect)

none of the others (from you OP and including up to the above statment) had solutions as stated.

Here is the solution to yo ...[text shortened]...

P*t = 1/2 *m*(a*t)^2

m= (2*P*t)/(a*t)^2

using your numbers I come up with 20,825 kg

You only have one power of t on top and t^2 on the bottom. They do not cancel out.

EDIT - Let me check:

Pt = K

(m d^2/t^3)(t) = (m d^2/2t^2)

(m d^2/t^3)(t) = 1/2 m(at)^2

2(m d^2/t^3)(t)/(at)^2 = m

2P/ta^2 = m

See? The mass depends on the time given constant power and acceleration.

Originally posted by AThousandYoung
You only have one power of t on top and t^2 on the bottom. They do not cancel out.

They cancel

Power inherently has the factor of time that you seek

P=J/s

J=kg*m/s^2*m

so the equation for mass with just units becomes


[((kg*m*m/s^2)/s)(s)]/[(m/s^2)^2*(s^2)]

=kg

Originally posted by joe shmo
They cancel

Power inherently has the factor of time that you seek

P=J/s

J=kg*m/s^2

so the equation for mass with just units becomes


[((kg*m/s^2)/s)(s)]/[(m/s^2)^2*(s^2)]

=kg

See my addition to that post.

Originally posted by joe shmo
They cancel

Power inherently has the factor of time that you seek

P=J/s

J=kg*m/s^2

so the equation for mass with just units becomes


[((kg*m/s^2)/s)(s)]/[(m/s^2)^2*(s^2)]

=kg

[((kg*m/s^2)/s)(s)]/[(m/s^2)^2*(s^2)] =kg

You have mass on both sides.

its this darn formatting

m= meters

and I forgot a factor od meters in my first post

This seems relevant:

http://www.physicsforums.com/archive/index.php/t-117271.html

Wait, maybe not...they're assuming acceleration is constant.

Originally posted by AThousandYoung
[((kg*m/s^2)/s)(s)]/[(m/s^2)^2*(s^2)] =kg

You have mass on both sides.

do you agree on this

P*t = 1/2 *M*v^2

(J/s)*s = J

v =a*t

v^2 = (a*t)^2

sub it in , solve for mass....it makes no difference

Originally posted by joe shmo
do you agree on this

P*t = 1/2 *M*v^2

(J/s)*s = J

v =a*t

v^2 = (a*t)^2

sub it in , solve for mass....it makes no difference

Those are all correct, but I don't see the relevance.

Let's start with this.

P*t = 1/2 *M*v^2

How are you going to cancel that t on the left? What is the velocity?

Originally posted by AThousandYoung
Those are all correct, but I don't see the relevance.

it shows you that

Power*time does indeed = energy

and the kinetic Energy is 1/2 *mass*velocity squared

since we know that the velocity is defined as acceleration* time

you can sub it in and see that Energy still = Energy

so if you solve the equation for mass you will indeed get a mass.

Originally posted by AThousandYoung
Let's start with this.

P*t = 1/2 *M*v^2

How are you going to cancel that t on the left? What is the velocity?

v=a*t

Originally posted by joe shmo
it shows you that

Power*time does indeed = energy

and the kinetic Energy is 1/2 *mass*velocity squared

since we know that the velocity is defined as acceleration* time

you can sub it in and see that Energy still = Energy

so if you solve the equation for mass you will indeed get a mass.

I agree with all of that, except the last sentence. The last sentence does not follow.

Originally posted by joe shmo
v=a*t

What's t?

Originally posted by AThousandYoung
What's t?

time, the same time that is on the other side of the equation