What is the normal force on a block at the top of a loop?

[AionTelos]

Homework Statement

A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop the loop of radius R = 2.0 m. On the floor the speed of the block is observed to be 11 m/s.

What is the magnitude of the normal force exerted on the block at the top of the loop?

Homework Equations

(Flipping the y axis)
$$N + mg = \frac{mv^2}{R}$$

The Attempt at a Solution

$$N = \frac{mv^2}{R} - mg$$
$$N = \frac{(1.8kg)(11m/s)^2}{2m} - (1.8kg)(9.8m/s^2) = 91.26N$$

The solution says it's 20.7N, I'm not sure where I'm messing up. Though, I feel that maybe I should be using PE=KE?

 

[AionTelos]

I think I got it, I'm using the velocity at the bottom of the loop instead of at the top, will post the correct solution once I get it.

 

[AionTelos]

For anyone else wondering here was the solution I came too.
$$KE_i = PE_f + KE_f$$
$$\frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_f^2$$
h = -2R
$$v_i^2-4gR = v_f^2$$
$$N+mg=\frac{mv^2}{R}$$
$$N=\frac{m(v_i^2+4gR)}{R}-mg$$
m=1.8kg
v_i=11$\frac{m}{s}$
g=9.8$\frac{m}{s^2}$
R=2m
$$N=\frac{1.8((11)^2-4(9.8)(2))}{(2)}-(1.8)(9.8)$$

Phew, everything check out?

 

[haruspex]

Looks right now.