Help with physics problem?

Question

What is the speed of the bead at point C?
Here are the details...Point A is 100cm above the potential energy line, B is on the potential energy line and C is 80cm is above the potential energy line. (It's on a graph) Thats all the information it gave me. Can you please show me how to do it.
Thank you.

Answer 1

if the bead is at rest at point A then at point C it will have kinetic energy equal to the loss in potential energy:

.5*m*v^2=m*g*h
h=.20m
g=9.81

v=sqrt(.40*9.81)m/s
j

Answer 2

If point A is 100cm above the potential energy line, the potential energy is mass*acceleration of gravity*height or M(9.8m/s/s)(1m) or 9.8(N/Kg)*M. If point B is at 80cm, the potential energy is M(9.8m/s/s)(.8m) or 7.84(N/Kg)*M. The kinetic energy is .5mv^2 and that is equal to 9.8(N/Kg)*M-7.84(N/Kg)*M:
.5mv^2 = 9.8(N/Kg)*m-7.84(N/Kg)*m //devide both sides by m
.5v^2 = 1.96(N/Kg)
(v^2)^.5 = (3.92(N/Kg))^.5 //square root both sides, also N/Kg is the same as (m/s)^2.
v = 1.98m/s
I hope I didn't make it seem complicated.