Block falls on a spring?

Question

A 9.3 kg block is dropped onto a spring of spring constant 1086 N/m from a height of 800 cm. When the block is momentarily at rest, the spring has been compressed by 50 cm. Find the speed of the block when the compression of the spring is 10 cm. The acceleration of gravity is 9.81 m/s^2. Answer in units of m/s.

Answer 1

Potential energy Pe is transferred to kinetic energy Ke (just before it starts to compress the spring) and then Ke is transformed into energy of the spring Ps.

Pe=Ke=Ps
Pe=mgh
Ke= 0.5 m V^2
Ps= 0.5k x^2

Lets find k (just to check )
mgh=0.5k x^2
k= 2mgh/x^2
k=2 x 9.3 x 9.81x8.0/(.5)^2
k=5840 N/m ( something is wrong since a spring constant 1086 N/m is given)

As the block compasses the spring
Pe=Ke+Ps
mgh=0.5 m V^2 + 0.5k x^2
V= sqrt[( 2mgh - k x^2)/m]

Which k would you like to use?
k= 1086 N/m or
k= 5840

Have fun