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?
Wnc= nonconservative work
GP=Universal Gravitational Potentiential Energy=mgh
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
Now I start using energy equations (initial and final position have its velocity at 0, so K is ignored)
62.125d+42.1875=-32.2056d (I was already suspicious here since the answer would clearly be negative)
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.