Kinematics ramp and crate problem

[Fera09]

Homework Statement

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1480 N will move with speed 2.1 m/s at the top of a ramp that slopes downward at an angle 23.0 degrees. The ramp will exert a 578 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp.
Calculate the maximum force constant of the spring k_max that can be used in order to meet the design criteria.

Homework Equations

h_inital= (sin23=h/8) = 3.126
h_final= 0
(spring)x_initial= 0
(spring)x_final= ?
v_final= 0
v_initial= 2.1

W_friction = F_friction * Distance

Change in gravitational potential energy = mgh_final-mgh_initial

Change in elastic potential energy = (1/2)kx_final² - (1/2)kx_initial²

Change in Kinetic Energy = (1/2)mv_final²-(1/2)mv_initial

The Attempt at a Solution

Since the final velocity, final height, and initial x for the spring are all equal to zero I got the equation..

W_friction + mgh_inital + (1/2)mv_inital² = (1/2)kx_final²

And then I solved for k..

k= (2(W_friction+mgh_inital+(1/2)mv_initial²))/x²I don't know if what I did it's right, but if it is.. I can't solve for k since I don't know x, the distance that the spring compressed, and I don't know how to find it =[[
Help pleaseee

 

[Chewy0087]

W_friction + mgh_inital + (1/2)mv_inital² = (1/2)kx_final²

this is perfect, although bear in mind that the work against friction will be negative (because it's energy that it has lost).

what I would do from here is put in the values that you know in order to get a nice equation relating k & x.

next you can set up a simultaneous equation, because you also know hooke's law don't you? that the force is proportional to the spring constant multiplied by the extension

F = kx, you know what K is (bear in mind F is a vector - if it's still on the ramp the only bit the spring is supporting is the horizontal component of the force)

do you see why this works?

 

[Fera09]

I think so.. but what is the force of the spring in hooke's law?

 

[Chewy0087]

Hooke's law is as follows

the force impressed on a spring is equal to the extension of the spring caused by that force multiplied by the spring constant.

so F = kx

however like I said bear in mind that the spring is on the angle and is only supporting the box's horizontal motion (the floor is supporting the box vertically) so perhaps it will look something like

mg cos(theta) = kx

have you set up the simultaneous equation yet? you had the right formula you just needed to put in the numbers.

 

[Fera09]

Okay, so I set..

mgcos(theta)=k*x

my equation for K was..
(2(W_friction+mgh_inital+(1/2)mv_initial²))/x²

And i got x= .49 m
which I then substituted into my equation for k, but I got the wrong answer
is my equation for k wrong?