Box-Spring Collision: How Much Compression?

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A 1.5kg box sliding at 12m/s approaches a spring with a spring constant of 2000N/m, raising the question of how far the spring will compress to stop the box. The kinetic energy of the box is calculated to be 108J, which is converted into the potential energy of the spring. The correct approach involves using the equation W=1/2kx^2 to find the compression distance. Initial calculations suggested a compression of 0.007m, but further analysis indicates it should be around 0.3m. The discussion emphasizes that gravitational acceleration does not factor into this scenario.
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Homework Statement


1.5kg box sliding on frictionless surface with a speed of 12m/s approaches a horizontal spring. The spring has a k of 2000N/m. How far will the spring be compressed in stopping the box? How far will the spring be compresses when the box's speed is reduced to half its initial speed?


Homework Equations


Fs=-kx
W=1/2kx^2


The Attempt at a Solution



Not sure if I needed this but I found the KE to be 108J. I tried to solve for x by 1.5kgx9.80m/s^2/2000N/m= .007m. Not sure!
 
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Try this - http://hyperphysics.phy-astr.gsu.edu/hbase/pespr.html

One is correct about the kinetic energy. Without dissipative forces, e.g. friction, the kinetic energy of the traveling mass will be transformed into the stored mechanical energy of the spring.

The acceleration of gravity is not part of this.
 
I put in the 2000N/m and 108J and got 0.3m?
 

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