How Is Kinetic Energy Converted in a Spring System with Friction?

[R.H.2010]

Homework Statement

A block of mass m=2.5 kg slides head on into a spring of spring constant 320 N/m. When the block stops, it has compressed the spring by 7.5 cm. The coefficient of kinetic friction between the block and the floor is .25. while the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force (B) the increase in the thermal energy of the block-floor system? (c) what is the block's speed as it reaches the spring?

Homework Equations

U=.5kx^2
Ksubi + Wsuball = Ksubf
K=.5mv^2

The Attempt at a Solution

i used conservation of energy and tried to solve for Work from the spring, however I don't know how to get the initial velocity.

.5(2.5)v^2 + .5(320)(.075)^2 + mg(mu)cos180 + Wsubs = 0 + .5(320)(.075)^@

I think the problem here is that .5kx^2 is not supposed to get reduced so the above set up is wrong. Any suggestions, please? Thank you.

 

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We know that the initial kinetic energy (defined to be the moment the block comes in contact with the spring) is equal to the final potential energy of the spring due to compression of the spring by x plus frictional losses accrued while traveling that same distance x.