Conservation/Conversion of energy

[drkidd22]

Homework Statement

A mass of 4.5kg is dropped from a height of 3.3m above a spring. spring const is 25N/cm. How much is the spring compressed.

Homework Equations

F = 4.5*9.8 = 44.1

PE = mgh = 4.5*3.3*9.8 = 145.53

v = sqrt(2(9.8)(3.3)) = 8.04m/s

The Attempt at a Solution

The above are the few things I was able to grab out of the given information. The issue I'm having is plugging this into an equation to find the spring

 

[Delphi51]

Energy is conserved in this problem. If you write
initial energy = final energy
and then fill in the details of the kinds of energy involved you will be able to solve for the spring compression.

 

[drkidd22]

so I get that mgh = 1/2mv^2+mgh?

145.53 = .5(4.5)(8.04)^2+(4.5)(9.8)h

.7m?

I don't think I did it right

 

[drkidd22]

Well I looked into a few of the formulas on the book and I get to a quadratic equation, but I'm still missing something here and I'm not sure what it is.

.5(.25)Y^2-(4.5)(9.8)Y+4.5(9.8)(-3.3)
0.125Y^2-44.1Y+145.53.

356? That's not possible so any adive on this will be appreciated.

Thnsx

 

[drkidd22]

Got it done. My equation above is right, the only problem was that I had to convert N/cm to N/m.

Thanks

 

[Delphi51]

You didn't consider the spring!
Initially you have the mgh. When the mass hits the spring, it comes to a stop and you only have spring energy ½kx².