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

• ChetBarkley
In summary, the conversation discusses the forces acting on a rocket clamped to the top of a vertical spring, including the weight force and spring force. It also covers the calculation of the compressed length using the spring force equation, as well as the use of potential and kinetic energy equations to determine the final height of the rocket. The conversation also mentions the need to account for the work done by the rocket and the starting and ending points of the equations.

#### ChetBarkley

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".