# Max value of spring constant

## Homework Statement

You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1500 N will move with speed 2.2 m/s at the top of a ramp that slopes downward at an angle 23.0 degrees. The ramp will exert a 586 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 7.6 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.

F = -kx
PEs = 1/2 kx^2
PEg = mgh
KE = 1/2 mv^2

## The Attempt at a Solution

1/2mv^2 + mgh = Ff*d + 1/2 kx^2
kx = 586 N

k = 586/x

1/2(1500/9.8)(2.2)^2 + (1500/9.8)(7.6sin(23)) = (586*7.6) + 1/2 kx^2

371.143 = 1/2(586/x)(x^2)
x = 1.27 m
k(1.27) = 586

k = 461.4 N/m -- this apparently is not the right answer; what am i doing wrong?