Rocket's Speed After Ignition & w/o Spring Tied Down

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kayleech
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Homework Statement


A 10.6 kg weather rocket generates a thrust of 226.0 N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 402.0 N/m, is anchored to the ground. Initially, before the engne is ignited, the rocket sits at rest on top of the spring.

A: After the engine is ignited, what is the rocket's speed when the spring has stretched 19.4 cm past its natural length?

B: What would be the rocket's speed after traveling the distance if it weren't tied down to the spring?


Homework Equations


Es=0.5kx^2
Ek=0.5mv^2
Eg=mgy


The Attempt at a Solution


For A:
I tried making a conservation of energy equation:
Es=Ek+Eg
v=square root of ((kx^2 - 2mgy)/m)

That didn't work, since I don't know y (x=0.2584 m as per another part of the question). Do I need to take the thrust force into account?
I didn't even know where to start for B after getting stuck on A.
 
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Yes, you need to take the thrust force into account. The rocket thrust provides the energy to lift the rocket and stretch the spring. Realize that before the rocket is ignited, the spring is compressed due to the rocket's weight.
 
I'm not exactly sure how I would go about adding that into the energy equation though.
 
The thrust exerts a force on the rocket. Figure out the work done by that force.
 
Yes... but I still don't know where to go from there.

I did v=square root of [((kx^2 - 2mgy + (Fthrustxy))/m]
x=0.2584m
y=0.194m

I don't think that makes sense, seeing as how it was the wrong answer.
 
For part B, the rocket is accelerating at a constant rate. A formula exists that relates distance and velocity given a constant acceleration -- what is it? Hint: The rocket travels more than 19.4 cm, as the rocket weight initially makes the spring shorter than its natural length.

Given the nature of the problem, I assume you have investigated the behavior of a mass hanging from a spring. Think of how the solution to this problem pertains to part A.
 
kayleech said:
I did v=square root of [((kx^2 - 2mgy + (Fthrustxy))/m]
Here's how the energy equation should look:
[tex]F_{thrust}y + 1/2 k x_1^2 = 1/2 k x_2^2 + mgy + 1/2mv^2[/tex]

In words that says: The energy added by the rocket engine plus the initial spring energy equals the final energy, which is the sum of spring energy, gravitational energy, and kinetic energy.
x=0.2584m
y=0.194m
Careful here. Those are the initial and final spring stretches:
[itex]x_1[/itex] = 0.2584m
[itex]x_2[/itex] = 0.194m

The total distance that the rocket rises, y, is y = [itex]x_1 + x_2[/itex].

I don't think that makes sense, seeing as how it was the wrong answer.
Give it another shot.
 
Thanks! That worked.