Understanding Work and Energy: Step-by-Step Problem Solving Guide

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AI Thread Summary
The discussion revolves around a user's confusion regarding work and energy problems, specifically seeking help with a particular question involving a spring constant. Participants are encouraged to provide a step-by-step explanation, including how to approach the problem and create a free body diagram. The importance of understanding the spring constant equation and its application to the problem is emphasized. Additionally, there is a focus on setting up the correct equations over time to solve the problem effectively. Overall, the thread aims to clarify concepts related to work and energy in physics.
muna580
I was just praticing some problems based on the work/enegry. Well, I am confused, and I have no idea how to do this. Cna someone please help me?

http://img228.imageshack.us/img228/4813/untitled1tj3.jpg

Can you please give me a step by step explantion on how to do this.
 
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1986M1a looks to be pretty straightforward. How would you start this problem? What would the free body diagram look like?
 
This is the qeustion I need help with, can you please help

http://img228.imageshack.us/img228/4813/untitled1tj3.jpg
 
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muna580 said:
This is the qeustion I need help with, can you please help

http://img228.imageshack.us/img228/4813/untitled1tj3.jpg
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Sure. What is the spring constant equation? How does it apply to this problem? What equations are you setting up versus time?
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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