A spring loaded gun question from my test.

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The discussion revolves around a physics problem involving a spring-loaded gun that fires a mass. The mass of 0.16 kg is launched with an elastic potential energy of 0.85 J after being compressed 0.10 m. Participants suggest using energy conservation principles, equating the spring's potential energy to gravitational potential energy at maximum height. The key formula to use is E = mgh, which simplifies the calculation of height. The conversation highlights different approaches to solving the problem, emphasizing the conversion of energy types.
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A spring loaded gun question from my test. Please Help!

Homework Statement


Suppose a .16kg mass on a spring loaded gun that has been compressed .10 m has elastic potential energy of .85 J. How high above the spring's equilibrium point can the gun fire the mass if the mass is fired straight into the air?


Homework Equations


I am not even sure which equations to use!


The Attempt at a Solution

 
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Welcome to PF, Krislynn.
The question is about energy. You are given the energy stored in the spring, so you must look up the formula for that. The spring energy is converted to the kind of energy the mass has when it is at its maximum height. You will need the formula for that kind of energy, too. The energy is entirely converted from one form to another, so you just put an equal sign between the two energy formulas, put in the numbers you know and solve for the height.
 


Thank you so much!
 


Most welcome! I now see you didn't need the spring energy formula because the energy was given. Just E = mgh should solve it.
 


or if you feel like doing it the really clunky and obtuse way (what I did for some reason), .85=(mv^2)/2 solve for v and then its a really easy kinematics problem
 
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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