Spring, Mass, Incline (Conservation of Mech-E)

[rmunoz]

Homework Statement

In the figure below, a block of mass m = 11 kg is released from rest on a frictionless incline angled of angle θ = 30°. Below the block is a spring that can be compressed 2.0 cm by a force of 270 N. The block momentarily stops when it compresses the spring by 5.4 cm.

http://www.webassign.net/halliday8e/pc/halliday8019c08/halliday8019c08-fig-0045.htm"

(a) How far has the block moved down the incline to this stopping point?
m

(b) What is the speed of the block just as it touches the spring?
m/s

The Attempt at a Solution

I have already attained the correct answer for part A. The problem I'm having is with part B. I'm pretty confused with the distances that i should be using to solve this problem. After setting up the relationship Ugrav(o)=Ugrav(1) + K(1) = E(spring) i decided that the way i would attain velocity at the instant before it hits the spring would be through this equation:

V^2=[K(d+x)^2]/m -2gd

for K i plugged in (270/.02)
for d+x i plugged in .3652m
and for d i plugged in .3652m-.054m.

I feel like this is on the right track but if this equation is missing something, some advice would be highly appreciated!

If this is in fact the right equation, then the problem I'm having is with the distances. I can't figure out is how i should go about attaining the values (d+x) and d. I understand that one of these values should be the length acquired in part a (.3652m), but I'm not sure which one. This i believe should be d+x, but what is d by itself? Basically i keep getting the same damn answer and its just wrong!

 

[rmunoz]

can i get some help with this one?

 

[LowlyPion]

and for d i plugged in .3652m-.054m.

Isn't this the distance it moved from the initial release?

So isn't d *sin30 then the height h it descended to that point?

So isn't

V² = 2*g*h

where h is .155588?

 

[midge]

I calculated 1.89m/s.

Just before the block stops and compresses the spring, all of its Emech is transferred to KE (since U=0 because of y=0)

KE= .5mv^2
19.68 = .5(11)^2
v= 1.89 m/s

 

[rmunoz]

Lowly Pion, thank you so much for the help. It was just that last little boost of brain power that i needed to close this one out, but that got me through it, really appreciate it! Geez this site rocks