Work and Potential/Kinetic Energy (spring problem)

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The discussion revolves around a physics problem involving a ball dropped onto a vertical spring, requiring the calculation of the ball's speed when the spring is compressed and the maximum compression distance. The conservation of energy principle is key, as it combines gravitational potential energy and spring potential energy. The participant initially struggled with the acceleration due to gravity affecting the ball's motion but clarified their understanding by calculating the potential energy changes. They confirmed that the remaining energy after spring compression translates into kinetic energy. The conversation concludes with a clearer grasp of the energy transformations involved in the problem.
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Homework Statement


A)
A ball with a mass of .360kg is dropped from a height of 1.20 meters above the top of a fixed vertical spring, whose force constant is 350 N/m. What is the speed of the ball when the spring has been compressed 5.0cm?
(Ignore the mass of the spring; also, notice that this problem contains both forms of potential energy (gravitational and spring pot'l En.)

B)
What is the maximum distance the spring is compressed by the ball?

Homework Equations


Work=F\cdotD
PE:
U=1/2K(xf-xi)2
\DeltaU=mg\cdot\DeltaY
Ki+Ui=Kf+Uf

m=.360 Kg
H=1.2m
\DeltaX=5.0 cm
K=350 Nm


The Attempt at a Solution


We did a problem similar to this on a frictionless horizontal plane. We used the conservation of energy to solve it. In the picture we drew we found where the Pot. and Kin. energies were zero which left the equation with only one unknown left. In that problem there was constant velocity but in the problem I just posted we have an acceleration of mg down which leaves me confused on what the right way to solve this is. I also know for part B the velocity will be 0.
 
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Figure your changes in potential energy.

m*g*h = .36*9.8*(1.2 + .5)

Now how much potential energy is in the spring?

1/2*k*x2

Whatever is left over then must be kinetic energy right?
 
Wow that makes much more sense now, thanks alot!
 
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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