What is the Spring Constant for a Compressed Mass on a Vertical Spring?

[shahar_rod]

Homework Statement

The left side of the figure shows a light (`massless') spring of length 0.330 m in its relaxed position. It is compressed to 71.0 percent of its relaxed length, and a mass M= 0.210 kg is placed on top and released from rest (shown on the right).

The mass then travels vertically and it takes 1.50 s for the mass to reach the top of its trajectory. Calculate the spring constant, in N/m. (Use g=9.81 m/s2). Assume that the time required for the spring to reach its full extension is negligible.

Homework Equations

F=-kx

The Attempt at a Solution

I'm not too sure about how I consider the time it takes for the mass to go back up.
I know that the energy while the mass is on the spring is mg (downwards) and -kx upwards.

Then when the ball goes back up - its KE + mgh is equal to mg-kx?

thanks!

 

[jegues]

Calculate the maximum height of the object above the compressed spring, then you'll have a value of gravitational potential energy that is equal to the potential energy initially stored in the compressed spring.

Then you simply solve for k.

 

[shahar_rod]

if so then why do i need the t=1.5 sec? it doesn't make any sense...

is this what you suggest:
mgh = 0.21*0.71*0.33*9.81= -k(0.29*0.33)

 

[denverdoc]

The mass flies free of the spring and reaches an unknown height. use t to determine that height and the relation

mgh = 1/2 K*x^2

where x is the displacement of the spring which we are not given directly, but can be figured fout rom the data given.