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What is the velocity of the mass as it breaks contact with the spring?

From part 1 I got:

A 1.118 kg block is on a horizontal surface with mu-k = 0.150, and is in contact with a lightweight spring with a spring constant of 715 N/m which is compressed. Upon release, the spring does 4.326 J of work while returning to its equilibrium position. Calculate the distance the spring was compressed. Answer:0.11 m

Also I got a hint:The force that acted on the mass comes from the spring and from friction. Thus the amount of work equals the kinetic energy of the spring. From the equation of work done by both forces we can calculate the velocity of the spring. (i thought i understood it, but something seems wrong.)

I thought of using the formula K=1/2(m*v^2) but it's wrong

Also from the hint o found that the force is: 39.33 N ( i divided the work done with the distance)

I also tried V^2=2*m*g*h, but i also get it wrong

What am i missing

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# Homework Help: How do I figure velocity?

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