I have been working on this problem for too long! I can not get the correct answ

Question

I have been working on this problem for too long! I can not get the correct answer. This is what I have set up:

1/2mv^2 + mgh= 1/2kx^2+mgh

In this case the distance that the spring gets pushed down is relevant according to the professor.

The answer I am getting is incorrect. Any help is greatly appreciated!!! Thank you!

Question: An inclined plane of angle ? = 20.0

1/2mv^2 + mgh= 1/2kx^2+mgh. An inclined plane of angle ? = 20.0 degree has a spring of force constant k = 545 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.27 kg is placed on the plane at a distance d = 0.303 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

Explanation / Answer

Your best shot is to continue screwing around with kinetic energy and potential energy

Let
x = compression of spring when block comes to a stop

ignoring frictional losses, the potential energy of the spring will equal the sum of the initial kinetic energy and the potential energy converted in the drop

PS = KE + PE

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