Lab help, can't get the same result as interactive physics

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SUMMARY

The discussion centers on a physics problem involving a 15 kg block on a 25-degree incline, interacting with a spring (k=150 N/m) and subject to kinetic friction (μk=0.4) and tension (20 N at 10 degrees). The user calculated the forces and attempted to apply the work-energy principle but arrived at a displacement of 0.447 m, which contradicted the interactive physics (IP) result of 0.792 m. The key error identified was the incorrect sign in the potential energy equation for the spring, which should be positive.

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Homework Statement


A 15 kg block is against, but not attached to, a spring w/ k=150 N/m and pushed back a distance of .75 m. All of this is on an incline of 25 degrees, w/ coefficient of kinetic friction equal to .4. To top it off, there's a string pulling w/ a Tension of 20 N at an angle of 10 degrees above the incline. How far does the block move when released?

Homework Equations


Wnc=Kf-Ki+GPf-GPi+Sf-Si
Wnc= nonconservative work
K=kinetic energy
GP=Universal Gravitational Potentiential Energy=mgh
S=Spring Work=-1/2(k)(x^2)
Fk=FN*mk
Fk=friction force
FN=normal force
mk=.4

The Attempt at a Solution


Well, first I found the x and y gravitational forces, w/ the positive x-axis pointing up the incline. That's (147sin(25),147cos(25))=(62.125,133.227)N
Then I found the same components for tension: (20cos(10),20sin(10))=(19.696,3.473)N

FN=Fwy-Ty (the problem says nothing about the block leaving the incline, and in face he told us to put a frictionless block above it in IP to prevent this). So FN=133.227-3.473=129.754 N
Fk=.4FN=51.9016 N

WT=20cos(10)d=19.696d
WFk=51.9016(cos180)(d)=-51.9016d

Now I start using energy equations (initial and final position have its velocity at 0, so K is ignored)
(15)(9.8)(dSin25)-15(9.8)(0)+0-(-1/2)(150)(.75^2)=19.696d-51.9016d
62.125d+42.1875=-32.2056d (I was already suspicious here since the answer would clearly be negative)
42.1875=-94.3306d

ignoring a negative answer, I'd get d=.447 m
IP (through Vector addition) gives me a displacement of .792m

Now I didn't do the IP myself, a partner did, but I need to know if I did anything wrong here.
 
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The potential energy stored in a spring is +(1/2) k x2 (no minus sign). The rest of your work seems okay.