Displacement of spring, block down an incline

[booksrmylife]

Homework Statement

An inclined plane of angle θ = 20.0° has a spring of force constant k = 460 N/m fastened securely at the bottom so that the spring is parallel to the surface as shown in the figure below. A block of mass m = 2.55 kg is placed on the plane at a distance d = 0.288 m from the spring. From this position, the block is projected downward toward the spring with speed v = 0.750 m/s. By what distance is the spring compressed when the block momentarily comes to rest?

http://www.webassign.net/serpse8/7-p-063.gif (this shows an image if you can get there)

Homework Equations

dE=dk+du=0

Kf-Ki+Dfsp-Disp+Dfg-Dig=0
Ki=1/2mv^2
Dfsp=1/2kx^2
Dig= mg(h+x)

The Attempt at a Solution

Kf,Disp, and Dfg all Equal zero
so by substitution:

-1/2mv^2+1/2kx^2-mg(h+x)=0
-2.55(.75^2)/2+230x^2-24.99sin20(.288)-24.99x=0

230x^2-24.99x-3.18=0

so x should be .183, but that's wrong

 

[Mandeep Deka]

Dig= mg(h+x)

How do you get that?
As far as i can see, i think it should have been mg[h + x(sin20.0)]
Don't you think so??



Hope you understood the mistake!