Gravitational/Spring Potential Energy problem

[pelmel92]

Homework Statement

A block of mass m starts from rest at a height h and slides down a frictionless plane inclined at angle θ with the horizontal, as shown below. The block strikes a spring of force constant k. Find the distance the spring is compressed when the block momentarily stops. (Let the distance the block slides before striking the spring be . Use the following as necessary: m, θ, k, , and g.)

Homework Equations

PE_spring = .5kx^2
PE_grav = mgh

The Attempt at a Solution

Setting the two above equations equal to each other and solving for x doesn't work... tries so far: √(2mg/k), √(-2mg/k), and one where i tried to account for the extra change in height past the natural length of the spring where i got
x=(-mgsin(ø)+√((mgsin(ø))^2 -2kmgh))/k

 

[tiny-tim]

hi pelmel92!

... tries so far: √(2mg/k), √(-2mg/k), and …

erm … what happened to h ?

 

[pelmel92]

haha whoops, i copied that over without the h's, but the answer comes up wrong even as √(2mgh/h) or √(-2mgh/k)

 

[tiny-tim]

ah … without seeing the diagram, i thought the the spring was horizontal, but now i realize it's on the slope!

now i understand your x=(-mgsin(ø)+√((mgsin(ø))^2 -2kmgh))/k !

i think you have a sign wrong ​

 

[pelmel92]

thanks for giving me a hand :) do you mean i should subtract the square root instead of add? or should i have originally set .5kx^2 equal to -mg∆h? I'm a little confused as to how to handle this problem :/

 

[tiny-tim]

perhaps I've got the equation wrong …

can you show us the quadratic equation you have? ​

 

[pelmel92]

well i started off with .5kx^2=mg∆H
but since the question defined h as the height the object fell when it first touches the uncompressed spring, i thought i would probably need to consider that in my ∆H.
so from there P.E.=mg(h+xsin(ø))=.5kx^2
then in quad format:
0=.5kx^2 - mgsin(ø)x -mgh

 

[pelmel92]

if it helps, here's the diagram they give
http://www.webassign.net/tipler6/7-p-094.gif