Spring constant???? Physics help, please!!!

A block of mass 528 g is pushed against the
spring (located on the left-hand side of the
track) and compresses the spring a distance
4.7 cm from its equilibrium position. The block starts from
rest, is accelerated by the compressed spring,
and slides across a frictionless horizontal track. It leaves the
track horizontally, flies through the air, and
subsequently strikes the ground.
The acceleration of gravity is 9.81 m/s2 .

H=1.9 m
distance traveled=4.41 m

What is the spring constant?
Help please....I've been trying do use the formula k=mg/x

K=(0.528*9.81)/(0.047)
k=110
but its wrong!!! don't know what to do

All right, let's first find the block's horizontal speed as it leaves the end of the track. We know it travels horizontally 4.41 meters and falls 1.9 meters during that time, so we need to find out how long it was in the air.

∆d = vo*t + 1/2at² (all displacements and velocities are vertical in this case)

1.9 m = (0 m/s)t + 1/2(9.81 m/s²)t²

1.9 m = (4.905 m/s²)t²

t² = 0.387 s²

t = 0.622 s

So the block was in the air for 0.622 s, and it traveled horizontally 4.41 meters during that time.

v = ∆d / t (displacement and velocity are horizontal this time)

v = 4.41 m / 0.622 s

v = 7.09 m/s

The block was moving at 7.09 m/s as it left the track. Given the absence of friction, we may assume that was the speed at which it left the end of the spring. Let's calculate its kinetic energy based on that speed:

K = 1/2 mv²

K = 1/2(0.528 g)(7.09 m/s)²

K = 13.3 J

Which must also have been the potential energy stored in the spring when it was compressed.

U = 1/2kx²

13.3 J = 1/2k(0.047 m)²

13.3 J = (0.0011 m²)k

k = 12000 N/m, or 1.2x10^4 N/m

I hope that helps. Good luck!

I don't remember all the equations, but I think you are looking at this too simply.