Mass on incline plane sliding into spring, need to find max compression

[BryceHarper]

Homework Statement

A 11 kg box slides 4.0 m down the frictionless ramp shown in the figure , then collides with a spring whose spring constant is 190 N/m. The angle of the ramp is 30°. What is the maximum compression of the spring?

Homework Equations

Ei=mgh
Ef=1/2kx^2

The Attempt at a Solution

Ei=Ef
mgh=1/2kx^2
mgLsinθ=1/2kx^2
11(9.8)(4sin30)=1/2(190)x^2
1.5m = Max compression

I swear this is right but it's not, can anyone help me out?

 

[Villyer]

Is the spring laying on the ramp, or is there a flat part before the mass hits the spring, making the spring horizontal?

Also, does the 4m refer to the distance it slid along the ramp, or the vertical displacement?

 

[BryceHarper]

The spring is laying on the ramp and the 4m refers to the distance it slid along the ramp.

 

[Villyer]

You made the statement that the energy of the block 4 meters above the uncompressed spring is equal to the energy of the compressed spring, but that is leaving out a portion of energy.

From the point when the block is released to the time is stopped (at the maximum compression point of the spring) what is its change in height?

 

[BryceHarper]

Ohhhhhhhhh!

So i should do:
1/2(4+Xmax)sinθ=1/2k(Xmax)^2

?

 

[Villyer]

The left side should read 11(9.8) instead of (1/2), but besides that yes.

 

[BryceHarper]

Whoops! I guess I was so excited to have finally figured it out that I wrote the equation wrong lol.

To verify

mg(4+Xmax)sinθ=1/2k(Xmax)^2

So...

1/2k(Xmax)^2 - mgsinθ(Xmax) - mg4sinθ = 0

And then use quadratic equation to solve for Xmax?

 

[Villyer]

That looks right to me.

I wonder what the nature of the roots will be of that equation?

 

[BryceHarper]

Got it! Xmax is equal to 1.8m

Thanks for the help Villyer!

 

[Villyer]

Of course