How to Calculate the Speed of a Block Released from a Spring

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To calculate the speed of a block released from a spring, the initial potential energy (PE) stored in the spring must be considered, which is given by Ue = 0.5kd². The block's mass is 1.20 kg, the spring constant is 2.20 N/m, and the compression distance is 0.420 m. The total energy in the system is conserved, transitioning from potential energy to kinetic energy as the block is released. The confusion arose from incorrectly equating initial potential energy with zero, leading to an erroneous calculation. Understanding the correct application of energy conservation principles is crucial for solving this problem.
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Homework Statement


A block of mass 1.20 kg is placed in contact with a spring with a force constant of k = 2.20 N/m. When released, the spring pushes the block and expands a distance of 0.420 meters. If the floor is frictionless, the speed of the block as it leaves the spring is ? m/s.


Homework Equations


Elastic energy is Ue = .5kd2
Energy in a block-spring system is E = .5kx2 + .5mv2


The Attempt at a Solution


Well the problem is I haven't been able to get anywhere, because in my obviously incorrect thinking, if the block has yet to be released, then all the energy is potential, so E = Ue, but if E = Ue, then .5kx2 = 0, which can't be right. So, in short, I really don't know what I'm doing here. Am I using the wrong equations or something?
 
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robertmatthew said:

Homework Statement


A block of mass 1.20 kg is placed in contact with a spring with a force constant of k = 2.20 N/m. When released, the spring pushes the block and expands a distance of 0.420 meters. If the floor is frictionless, the speed of the block as it leaves the spring is ? m/s.


Homework Equations


Elastic energy is Ue = .5kd2
Energy in a block-spring system is E = .5kx2 + .5mv2


The Attempt at a Solution


Well the problem is I haven't been able to get anywhere, because in my obviously incorrect thinking, if the block has yet to be released, then all the energy is potential, so E = Ue, but if E = Ue, then .5kx2 = 0, which can't be right. So, in short, I really don't know what I'm doing here. Am I using the wrong equations or something?
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When the block is not yet released, the spring is compressed by 0.420 m. So its initial PE is not 0.
 
I just solved it, but I think my problem was that I was calculating the initial potential energy while also calculating the final kinetic energy, but still using E = PE, so I ended up with zero on one side and .5mv^2 on the other. I'm not really sure why I did Kf and Ui at the same time, I guess I'm having an off day, haha. But thank you!
 
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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