Maximum Height Reached by Block on Spring

AI Thread Summary
The discussion focuses on calculating the maximum height a block reaches after being released from a compressed spring. The initial energy stored in the spring is calculated as 22.4874 J, while the weight of the block is determined to be 2.45 N. The work-energy principle is applied to find the distance the block travels, resulting in a calculated height of 9.178 m. However, it is noted that the final height must account for the spring's compression, requiring a subtraction of the compression distance from the total height. The correct approach emphasizes the need to adjust for the initial compression of the spring to find the actual height above the release point.
Ry122
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A block of mass 0.250 kg is placed on top of a light vertical spring of constant k = 5200 N/m and is pushed downward so that the spring is compressed 0.093 m. After the block is released, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

my attempt:
energy in spring = (.5)5200(.093)^2=22.4874J
Weight of block = .250x9.8=2.45N
W=fxs
22.4874=2.45xs
s=9.178

What am i doing wrong?
 
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use the Energy formula involving initial and final energy.
 
what is it?
 
What makes you think you are doing these wrong? I got the same answer. Same thing with your centrifugal force problem. They both look good to me.
 
it was wrong because it asks how high does the block go above the spring so i had do minus the length of compression of the spring from the final answer.
 
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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