Conservation of Energy with Spring and Incline (Diagram Included)

AI Thread Summary
The discussion focuses on the conservation of energy involving elastic potential energy in a spring and an incline. The user attempts to apply the energy conservation equation but encounters a negative value when attempting to solve for velocity, indicating a potential error in their calculations. They express confusion over the validity of their formula and the assumption that height is negligible. The key concern is understanding why the negative value arises and whether the approach taken is correct. Clarification on the calculations and assumptions is needed to resolve the issue.
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Elastic Potential Energy with Spring and Incline (Diagram Included)

Homework Statement


04.21.2012-17.21.10.png



Homework Equations


ET = 1/2(m)(V2) + mgh + 1/2k*x2


The Attempt at a Solution


ET1 - Wf = ET2
1/2k*x2 - Wf = mgh + 1/2mV2

where, Wf = [(0.53)(2kg)(9.8m/s2)(cos67)](1.3m+0.4m)

^ I used this formula and went through by subbing in all numbers but I get error since you can't square root a negative value. Plus we can assume Height off ground is negligible since it cancels out on both sides of equation.
 
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does anyone know why I'm getting that negative error? Is my formula correct?
 
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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