Spring Ramp Problem: Maximum Distance for Compressed Spring Calculation

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The problem involves a 500-g block sliding down a frictionless track from a height of 2.00 m, striking a spring with a spring constant of 20 N/m. The energy conservation principle is applied, equating gravitational potential energy (mgh) to the elastic potential energy of the spring (1/2 kx^2). The calculation yields a maximum compression of the spring at approximately 0.99 m. The solution is confirmed as correct by other participants in the discussion. This demonstrates the effective application of energy conservation in solving spring compression problems.
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Homework Statement



A 500-g block is released from rest and slides down a frictionless track that begins 2.00 m above the horizontal as shown in the figure. At the bottom of the track, where the surface is horizontal, the block strikes and sticks to a light spring with a constant of 20 N/m.find the maximum distance the spring is compressed.

Homework Equations



I am pretty sure that this is just an MGH type problem, so I used :

mgh=1/2kx^2



The Attempt at a Solution



Using the above formula, I calculated: (.500)(9.8)(2)=.5(20)(x^2)

Solving the equation for value x i get .99 m.

Can anyone advise me to whether this is correct or not?
 
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That is correct.
 
awesome, thanks alot!
 

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