- #1

AionTelos

- 3

- 0

## 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)

[tex]N + mg = \frac{mv^2}{R}[/tex]

## The Attempt at a Solution

[tex]N = \frac{mv^2}{R} - mg[/tex]

[tex]N = \frac{(1.8kg)(11m/s)^2}{2m} - (1.8kg)(9.8m/s^2) = 91.26N[/tex]

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?