Calculating Spring Constant for an 8kg Stone Resting on a Compressed Spring

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To calculate the spring constant for an 8kg stone resting on a compressed spring, the relationship between gravitational potential energy (Wg) and elastic potential energy (Ws) is crucial, where Wg = -Ws. The spring's compression of 10cm can be used in the formula Ws = -0.5kx^2 to derive the spring constant k. When the stone is pushed down an additional 30cm, the elastic potential energy just before release is calculated as U = 0.5kd^2, yielding a value of 6.27 x 10^-3 J. To find the change in gravitational potential energy as the stone moves to its maximum height, the formula delta(U) = mg[delta(y)] is applied, but the maximum height needs to be determined first.
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An 8kg stone rests on a spring. the spring is compressed 10cm by the stone.

What is the spriong constant?

im not really sure how to start...Ws = -0.5kx^2

im guessing that there must be a relationship between Wg and Ws obviously
-Wg = Ws... but I am not sure how to solve for Ws

Can anyone help?
 
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F = mg = -kx
 
b)the stone is pushed down an extra 30 cm. What is the elastic potential energy of the compressed spring just before the release?

U = 0.5kd^2 d=distance compressed

...U = 6.27 x 10^-3 J

ok...the next part asks for...

what is the change in the gravitational potential energy of the stone-earth system when the stone moves from the release point to its max height

so delta(U) = mg[delta(y)]

how do i get the max height?
 
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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