Calculating Final Velocity of an Object with Power Input

AI Thread Summary
The discussion revolves around calculating the final velocity of an object with a given mass, power input, and distance traveled. The initial attempt to derive the final velocity led to a miscalculation regarding the placement of mass in the equations, specifically affecting the final exponent on mass. The correct final velocity formula should reflect that the mass is under the square root, leading to a different exponent. Additionally, a broader question about the implications of constant power input on acceleration, particularly for a rocket in space, highlights the complexity of momentum and energy conservation in such scenarios. The conversation emphasizes the importance of unit consistency and the relationship between power, mass, and velocity.
HalfThere
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Homework Statement


An object of mass M begins with a velocity of 0 m/s at a point. A power input of P watts goes directly to kinetic energy until the object has traveled a distance of X meters. What is the final velocity of the object?

So, we have constant variables
M = mass
X = distance that power will be input
P = power level

And also
V = final velocity, after traveling distance X (to be solved)

Homework Equations



E = 1/2MV^{2}
X = \intV(t) dt - V(t) is V as a function of time.
E = P*t

The Attempt at a Solution



Find V as a function of E (easy)
V = \sqrt{2E}/M

Find V as a funciton of time t (use equation)
V = \sqrt{2P*t}/M

Now take the equation for X
X = \intV(t) dt

And Find X as a function of t directly, knowing the V(t) function
X = \int\sqrt{2P*t}/M dt Edit: Should be, and was calculated as (\int \sqrt{2P*t} dt)/M

Integrate (remember that sqrt(2P) is a constant)
X(t) = (2/3)*\sqrt{2P}*t^{3/2}/M

Now change to t in terms of X
t(X) = ((3/2)/\sqrt{2P}*M*X)^{2/3}

And finally slide that into the V(t) equation
V(X) = \sqrt{2P}*((3/2)/\sqrt{2P}*M*X)^{1/3}/M

Simplify (whew!)
V(X) = 3^{1/3}*2^{-1/6}*P^{1/3}*X^{1/3}*M^{-2/3}

So, V correlates directly with the cube root of X, the cube root of P, and M^(-2/3), with a weird constant.


Am I right? Am I not? If not, where did I go wrong?
 
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Your general solution is fine, but I think you made some calculation errors... your m should be under the square root... so that affects your final exponent on m... also I'm getting that the 2 cancels out (ie the power of 2 is 0)...

I'm getting: 3^{1/3}*p^{1/3}*x^{1/3}*m^{-1/3} as the final velocity. Though I may have screwed up...
 
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Which specific step does the miscalculation occur? I can't quite see what you mean.

Also, a more general question about energy: What will a constant power input do to something with a velocity with no immediate context, like a rocket in space?

On the one hand, I think acceleration should remain constant, because it's change in momentum is relative only to itself, but on the other hand, a constant rate of fuel consumption should mean a a constant power input. The two don't seem to be compatible. What gives?
 
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HalfThere said:
Which specific step does the miscalculation occur? I can't quite see what you mean.

Integrate (remember that sqrt(2P) is a constant)
X(t) = (2/3)*LaTeX graphic is being generated. Reload this page in a moment.*tLaTeX graphic is being generated. Reload this page in a moment./M

Now change to t in terms of X
t(X) = ((3/2)/LaTeX graphic is being generated. Reload this page in a moment.*M*X)LaTeX graphic is being generated. Reload this page in a moment.here?

edit: the step where u isolated t.
 
I see where I went wrong with the 2^(-1/6), it should just be 3^(1/3) for the constant, but I still don't see the problem with the M. It was transformed from M to M^(2/3), then square rooted to M^(1/3), which was over M, which is M^(-2/3).
 
HalfThere said:
V = \sqrt{2E}/M

Here is the mistake... the M should be under the square root.
 
Ah, you're indeed right. I see now. And it makes sense if you look at units too.

Okay, Thanks learningphysics!
 
HalfThere said:
Which specific step does the miscalculation occur? I can't quite see what you mean.

Also, a more general question about energy: What will a constant power input do to something with a velocity with no immediate context, like a rocket in space?

On the one hand, I think acceleration should remain constant, because it's change in momentum is relative only to itself, but on the other hand, a constant rate of fuel consumption should mean a a constant power input. The two don't seem to be compatible. What gives?

Acceleration isn't constant for constant power...
 
HalfThere said:
Ah, you're indeed right. I see now. And it makes sense if you look at units too.

Okay, Thanks learningphysics!

you're welcome.
 
  • #10
Can anyone answer the third post of the topic, where I ask about conservation of momentum V. conservation of energy in the case of the rocket?
 
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