Finding the Speed of a Box Tied to a Propeller on a Spring

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The discussion revolves around calculating the speed of a box tied to a propeller on a spring after the spring has moved a distance X. The user outlines their approach, using energy conservation principles, including initial and final kinetic energy, gravitational potential energy, and spring potential energy. They propose that the total work done by the force F equals the change in mechanical energy, leading to the equation FX = 1/2*mv^2 + MgX + 1/2*KX^2. Another participant confirms that the equation appears correct, noting that the spring is not compressed initially. The conversation emphasizes the application of energy conservation in solving the problem.
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Hello:
I was wondering if someone could check if the following is correct: A
box of mass M is tied to a propeller with an upward force of F. Ignore
weight of propeller. The box is placed on a vertical spring that is initially
at equilibrium and then the propeller is ignited. The spring constant is K.
Find speed after the spring has moved by a distance of X. So would I do:
K_{initial} = 0; U_{initial} = 0; U_{spring} = 0 initially. Kf = 1/2*mv^2
; U_{final} = MgX;U_{spring} = 1/2*kX^2.
And the total work done by the force F is F.x, or at the distance X, FX.
So since the total work done is equal to the change in mechanical energy, so
I write: FX = 1/2*mv^2 +MgX + 1/2*KX^2 and solve for v? Thank you.
 
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I see that the spring is not compressed at the start. So yes, the equation as you have written it looks right.
 
Hello:
Oh right, thanks!
 
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