UTexas Physics problem loop-the-loop, please help

So I haven't done this before but I honestly don't know where to start.

A bead slides without friction around a loop-the-loop. The bead is released from a height
14.1 m from the bottom of the loop-the-loop
which has a radius 5 m.
g = 9.8 m/s^2

What is its speed at the top of the loop-the-loop?

Please help, normally i would know how to do this but i don't know how to do it when mass is not given. Any response will help, thanks.

by conservation of energy:
KEi + PEi = KEf + PEf

KEi = 0
0 + mgh_1 = (1/2)mv^2 + mgh_2

Divide both sides by m
gh_1 = (1/2)v^2 + gh_2

isolating v...
2g(h_1 - h_2) = v^2

sqrt(2g(h_1 - h_2)) = v

h_1 = 14.1m
h_2 = diameter = 5m*2 = 10m
g = 9.8 m/s^2

plug in.

i think this is right xD

Always use conservation of energy in problems when you can't use kinematics,
first part,
as there is no non conservative forces acting,
we shall use conservation of energy
U1 + K1 =U2 + K2
where U=mgh
and K=1/2mv²
taking h=0 at bottom, and v1=0
mgh1=1/2mv²+mgh2
so, v=√[2g(h1-h2)]
=8.964 m/s

for your problem, second part,
from FBD,
mg+N=mv²/R
where V is speed at that moment,
answer=37.6 N