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

AI Thread Summary
To find the acceleration of a 12 kg box on a 23-degree frictionless incline, the net force equation is applied: Fpull - Fparallel = ma. The gravitational force acting on the box is calculated as 117.6 N. The component of this force parallel to the incline is found using the sine function, leading to the equation 117.6 N - (117.6 N * sin(23)) = 12 kg * a. The poster suspects an error in using sine instead of cosine for the incline's angle. The correct approach should clarify the relationship between the forces acting on the box.
Elo21
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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\rightarrowequals 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! :)
 
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i think its cos23 instead of sign.
 

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