Block falling on spring/max compression

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The discussion revolves around a physics problem involving a block released from a height above a spring. The block compresses the spring by 20.5 cm, and the work done by the block on the spring is calculated as 9.35 J, while the work done by the spring on the block is -9.35 J. The initial height (h0) is determined to be 1.158 m. The main issue arises in part (d), where the user struggles with the correct application of energy conservation principles, particularly in accounting for the spring's compression when calculating gravitational potential energy. Clarification is needed on the reference point for potential energy to accurately solve for the new height when the block is released from 2.00h0.
wilcofan3
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Homework Statement



A 700 g block is released from rest at height h0 above a vertical spring with spring constant k = 445 N/m and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring 20.5 cm.

(a) How much work is done by the block on the spring?


(b) How much work is done by the spring on the block?

(c) What is the value of h0?

(d) If the block were released from height 2.00h0 above the spring, what would be the

Homework Equations


The Attempt at a Solution



I got parts a, b, and c right, and have tried a few different things with d and have gotten it wrong.

For a and b, I got 9.35 (b was -9.35), and for c, I got 1.158 m. I went back to my equation for the spring, where mgh = .5kx^2, and solved for x but keep getting the wrong answer. I used my answer in c as the new height in mgh, is there something wrong with doing that? I ended up getting the equation...

x^2= 2mgh/k

Obviously I then just took the square root of that.
 
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I have some things in mind:

yes your final equation is wrong. The 2ho is above the spring but the spring is compressed by an amount x. This amount also contributes as Potential energy ( its under the point of zero of Potential Gravitational Energy).
 
look at your energy changes. You have an initial Gravitational Potential Energy, That energy is then transformed into kinetic energy.. But not completely. At the exact point before the object hits the spring the object has both potential and kinetic energy. And that total energy is converted into Elastic Potential Energy. Because energy is constant, you can say Gravitational Potential Energy before = Elastic Potential Energy After, but is your point of reference where your saying that your Grav. Potential Energy at the top of the spring? or the bottom of the spring? The point where Ep is zero isn't the ground because the spring could be on an airplane or x amount of height above the ground.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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