physics
posted by Ramon on .
A 0.150 kg block moving vertically upward collides with a light vertical spring and compresses it 4.50cm before coming to rest.
If the spring constant is 54.0 N/m, what was the initial speed of the block? (Ignore energy losses to sound and other factors during the collision.)

Thanks for including the dimensions (kg, etc) of the physical quantities this time.
When the spring is fully compressed, the initial kinetic energy (1/2)MV^2 is converted to gravitational energy MgY and spring potential energy (1/2)kY
Y is the vertical distance the spring is compressed (0.045 m)
g is the acceleration of gravity (9.81 m/s^2)
M is the 0.150 kg mass
k is the 54.0 N/m spring constant
(1/2)MV^2 = MgY + (1/2)kY^2
V^2 = 2gY + (k/M)Y^2
Solve for V by plugging in the numbers. The 2gY gravity term may turn out to be negligible compared to (k/M)Y^2 
I just want to thank DRWLS. I sent to Cramster and the answer was wrong several times by them. Again thanks for help.

1.611m/s