Calculating Final Velocity of an Object with Power Input

In summary, the conversation discusses finding the final velocity of an object with a mass of M and a power input of P watts, after traveling a distance of X meters. The final velocity is found to be directly proportional to the cube root of X, the cube root of P, and M. The conversation also touches on the topic of conservation of momentum and energy in the case of a rocket with constant power input. It is noted that acceleration is not constant for constant power.
  • #1
HalfThere
18
0

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[tex]^{2}[/tex]
X = [tex]\int[/tex]V(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 = [tex]\sqrt{2E}[/tex]/M

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

Now take the equation for X
X = [tex]\int[/tex]V(t) dt

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

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

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

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

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

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?
 
Last edited:
Physics news on Phys.org
  • #2
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: [tex]3^{1/3}*p^{1/3}*x^{1/3}*m^{-1/3}[/tex] as the final velocity. Though I may have screwed up...
 
Last edited:
  • #3
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?
 
Last edited:
  • #4
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.
 
  • #5
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).
 
  • #6
HalfThere said:
V = [tex]\sqrt{2E}[/tex]/M

Here is the mistake... the M should be under the square root.
 
  • #7
Ah, you're indeed right. I see now. And it makes sense if you look at units too.

Okay, Thanks learningphysics!
 
  • #8
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...
 
  • #9
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?
 

1. How is final velocity calculated?

Final velocity can be calculated using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. What is power input in the context of calculating final velocity?

Power input is the rate at which energy is transferred to an object. In the context of calculating final velocity, it refers to the amount of energy being used to accelerate the object.

3. How does power input affect the final velocity of an object?

The higher the power input, the greater the acceleration of the object and therefore the higher the final velocity. This is because more energy is being transferred to the object, resulting in a larger change in velocity.

4. Can an object's final velocity be greater than the initial velocity?

Yes, an object's final velocity can be greater than the initial velocity if there is a positive acceleration. This means that the object is gaining speed over time.

5. Is power input the only factor that affects final velocity?

No, final velocity is also affected by other factors such as the mass of the object and any external forces acting on it. These factors can either increase or decrease the final velocity, depending on their magnitude and direction.

Similar threads

  • Introductory Physics Homework Help
Replies
16
Views
334
  • Introductory Physics Homework Help
Replies
15
Views
246
  • Introductory Physics Homework Help
Replies
3
Views
819
Replies
5
Views
334
  • Introductory Physics Homework Help
Replies
25
Views
974
  • Introductory Physics Homework Help
2
Replies
38
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
670
  • Introductory Physics Homework Help
Replies
19
Views
628
  • Introductory Physics Homework Help
Replies
10
Views
863
  • Introductory Physics Homework Help
Replies
4
Views
946
Back
Top