What Is the Correct Equation for a Box on a Frictionless Incline?

[Elo21]

Homework Statement

A 12 kg box is being held on a 23 degree frictionless incline. Find the box's equation once released.

Homework Equations

Net force=ma

Fpull-Fparallel=ma

The Attempt at a Solution

12kg x 9.8m/s = 117.6 N

(Fpull-mgsin23)= ma

sub. known values in

117.6 N-(117.6N (sin23))= 12kg x a

a$$\rightarrow$$equals about 6.0 m/s


I know I am doing something wrong. I am reviewing for a final, and am trying to remember things. Please help me figure out what I did wrong.
THANK YOU! :)

 

[dacruick]

i think its cos23 instead of sign.

 

[kerimek]

I would like to point out that the equations used in this attempt at a solution are correct, but the calculation for the net force is incorrect. The correct calculation for the net force would be 12kg x 9.8m/s^2 = 117.6 N. Also, the assumption that the acceleration is equal to 6.0 m/s is incorrect. The correct equation for the acceleration in this scenario would be a = gsin23, which would result in an acceleration of approximately 4.65 m/s^2. Additionally, in this scenario, the box will slide down the incline with a constant acceleration, so the equation for the position of the box would be x = x0 + v0t + (1/2)at^2, where x0 is the initial position, v0 is the initial velocity (which is 0 in this case), and t is time. I hope this helps you in your review for your final. Good luck!