Calculating the Height of a Block Launched from a Compressed Spring

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A block of mass 250 g is launched from a compressed spring with a spring constant of 100 N/m, initially compressed by 10 cm. The initial energy calculated was 0.5 J, using the formula for elastic potential energy. The discussion clarifies that the block is not attached to the spring when it leaves, which means the elastic potential energy is fully converted to gravitational potential energy. Upon correcting the approach, the correct height the block rises is determined to be 20 cm. The resolution highlights the importance of understanding the energy transfer during the launch.
Suraj M
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Homework Statement


A block of mass 250 g is kept on a vertical spring with spring constant 100N/m fixed from below.
The spring is now compressed to have a length 10 cm shorter than its natural length and is released. How high does the block rise?g = 10m/s

Homework Equations


E=½kx²
and other basic equations

The Attempt at a Solution


So I found out initial energy imparted,
$$E_1 = ½kx² = 0.5 J$$
then equated to change in potential and new spring energy.
$$E_2 = mgh + ½k(x-h)²$$ where h is the height required
solving E₁= E₂, I get an awkward answer. the answer given is 20 cm. How?
 
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Your E2 is incorrect .

Why do you think spring would have elastic potential energy when the block leaves it ?
 
Oh so you're assuming the block is not attached to the spring, huh, didn't consider that!
Oh i get the right answer then, its just a regular launch. thank you Tanya!
 
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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