A 145 kg block is released at a 4:6 m height?

Question

A 145 kg block is released at a 4.6 m height
as shown. The track is frictionless. The block
travels down the track, hits a spring of force
constant k = 1109 N/m:
The acceleration of gravity is 9.8 m/s2 :
Determine the compression of the spring x
from its equilibrium position before coming to rest momentarily. Answer in units of m

Answer 1

..."as shown"...Shown where? Darn now I have to guess...

Is the track an incline that makes an angle A with the horizontal? Actually who cares?!

Pe (at the top)=Ke(at the bottom )

And all that good energy gets transfered to a spring
Ps= (1/2) k x^2

x=sqrt( 2Pe/k)
x=sqrt(2 mgh /k)
x=sqrt( 2 x 145 x 9.8 x 4.6/ 1109)
x=3.43 m

Answer 2

To answerer above:
As shown in the invisible link.