A rocket on a spring, related to potential/kinetic energy

AI Thread Summary
The discussion focuses on analyzing the forces acting on a rocket attached to a spring, highlighting the balance between the spring force and the weight of the rocket. The calculations show that the weight of the rocket is 117.6N and the spring compresses by 0.214m under this force. The conversation shifts to energy conservation, emphasizing that the initial kinetic energy is zero, and only the potential energy of the spring is relevant at the start. Participants point out that the energy equation overlooks the work done by the rocket and question the assumptions about the starting and ending points of the energy calculations. Clarification is sought on how to determine the final height of the rocket without time, indicating a need for a more comprehensive approach to the problem.
ChetBarkley
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Homework Statement
A 12kg weather rocket generates a thrust of 200N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is 550 N/m, is anchored to the ground.
A) Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed.
B) After the engine is ignited, what is the rocket's speed when the spring has stretched 40cm?
Relevant Equations
F[sub]spring[\sub] = -k#\delta x#
U = mgy
K = .5mv^2
Part A) So from a force diagram we can see that the only two forces acting in our system are the spring force(positive y axis) and the weight of the rocket(negative y axis), which means the spring force is equal and opposite to the weight force.

The weight is simple enough ##12* 9.8=117.6N##
and then using the spring force equation we get the compressed length to be
##117.6N = -550(\delta x)##
##\delta x = \frac{117.6}{-550} = -0.214##m

Part B) Using the potential and kinetic energy equations and know that energy must be conserved we can use the following equation
##U spring, 0 +K spring,0+U rocket,0+ K rocket,0 = Uspring, 1 +Kspring,1+Urocket,1
+Krocket,1##

From this we know that the K.E. of the rocket and the spring initially, is zero(v=0 and the spring isn't stretched), meaning that the only term on the left side of our equation is the P.E. of the spring. On the right side, the P.E. of the spring finally is zero(spring is not compressed) and so we only have the P.E of the rocket and the K.E. of the spring and the rocket.
##\frac{1}{2}k(-\delta x)^2 = mgyf, rocket + \frac{1}{2} k (\delta x)^2 + \frac{1}{2} m (vf, rocket)^2##

Seeing this I'm not sure how to find the final height of the rocket as I wasn't given a time, nor am I sure where in my problem I could calculate time.
 
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The initial compression is not negative. The initial extension is.

Your energy equation neglects the work done by the rocket, and you wrongly state there is no final elastic PE.
But you cannot answer it just using energy. What can you do instead?
 
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Are the starting and ending points of my equations from when the engines are ignited to when the rocket is no longer in contact with the spring?
 
ChetBarkley said:
Are the starting and ending points of my equations from when the engines are ignited to when the rocket is no longer in contact with the spring?
It says "clamped to the top of a vertical spring".
 
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