Problem calculating speed at the top of a loop-the-loop

[mr_miyagi]

This is the problem:
A 15.0 kg box is slid down a 6.00 meter high ramp to pick up speed to do a vertical loop-the-loop whose radius is 2.00 meters. All surfaces are conveniently frictionless.

Part A) asks to find the Speed at the top of the loop-the-loop.
Part B) asks to find the Normal force on the box at the top of it's motion.

I'm just not sure I've solved it correctly.
Here's what I did it:
A)
PE_i + KE_i = PE_f + KE_f
(1/2)*m*v_i² + m*h_i*g = (1/2)*m*v_f² + m*h_f*g
0 + 6mg = (1/2)*m*v_f² + 4*m*g
6g = v_f²/2 + 4g
v_f = sqrt(4g) = 6.26m/s

B) Normal = Centripedal Force - mg = m*v²/r - mg = 15*(6.26²/2) - 15*9.8 = 117.6 N

 

[gordon831]

The work in part (a) and (b) is good! Can you check your calculator result in part (b) though? I calculate a different number: \frac{(15kg)(6.26\frac{m}{s})^2}{(2m)}-(15kg)(9.8\frac{m}{s^2}) = 147N

 

[mr_miyagi]

The work in part (a) and (b) is good! Can you check your calculator result in part (b) though? I calculate a different number: \frac{(15kg)(6.26\frac{m}{s})^2}{(2m)}-(15kg)(9.8\frac{m}{s^2}) = 147N

I see my mistake. I've divided in the wrong place.

B) Normal = Centripedal Force - mg = m*v2/r - mg = 15*(6.262/2) - 15*9.8 = 117.6 N

Redoing it made get to 147N :D
Thanks for the help!