Calculating the Height of a Block Launched from a Compressed Spring

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SUMMARY

The discussion focuses on calculating the height a block rises when launched from a compressed spring. A block with a mass of 250 g is placed on a vertical spring with a spring constant of 100 N/m, compressed by 10 cm. The initial energy calculated using the formula E=½kx² equals 0.5 J. The correct approach involves recognizing that the block is not attached to the spring upon release, leading to the conclusion that the block rises to a height of 20 cm.

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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!
 

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