Find velocity and Kinetic Energy of a Springy Toboggan

[SelHype]

Two clever kids use a huge spring with 750 N/m to launch their toboggan at the top of a 9.5 m high hill (see the figure). The mass of the kids plus toboggan is 95.0 kg. If the kids manage to compress the spring 2.7 m, what will be their speed at the bottom of the hill? Neglect friction. What fraction of their final kinetic energy was initially stored in the spring?



Relevant equations:
K = .5mv$$^{2}$$
$$U_{spring}$$ = .5kx$$^{2}$$
$$U_{grav}$$ = mgy

My attempt at a solution:
Okay so $$\frac{N}{m}$$ is $$\frac{kg}{s^{2}}$$.

To solve for $$\frac{m}{s}$$ I have the following hint from my professor:

If we measure height from the bottom of the hill, initially the mass has potential energy
from gravity and the spring. At the end there is just kinetic energy. Write down the condition
for the conservation of energy and solve for the final speed.


Okay so the height from the bottom of the hill is 9.5m. I am actually kind of lost on this problem. I don't have my book, I can't find something online that explains this well enough for me to solve this problem, and it is 5am and this homework is due at 8am...even if I don't get it turned in by 8, I still want to understand how to find this. Once I find out how to solve it I can do the other problems!

Hints are my favorites so I can figure it out on my own! Thanks muchly!

SelHype



EDIT:

I tried this:

(9.8)(9.5m) + .5(2.7m)(2.7m)(750 N/m)

5.5 m/s


Clearly I am wrong, though...Any thoughts?

 

[Redbelly98]

I tried this:

(9.8)(9.5m) + .5(2.7m)(2.7m)(750 N/m)

That's a reasonable start. This is the (potential) energy at the top of the hill, with the spring compressed ... except that you left out:

The mass in the "mgy" term.
The units for g = 9.8 ____.

Once you have that number calculated, write an expression for the energy when the toboggan is at the bottom of the hill.