a 8.0 kg object is sliding down s frictionless inclined ramp at an acceleration of 2.6/s/s. what is the angle of the inclined plane

Answer:
[tex]15.4^{\circ}[/tex]
Explanation:
There is only one force acting along the direction parallel to the plane: the component of the weight parallel to the incline, which is given by
[tex]mg sin \theta[/tex]
where
m = 8.0 kg is the mass of the object
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]\theta[/tex] is the angle of the incline
Therefore, the equation of motion along this direction is
[tex]ma=mgsin \theta[/tex]
where
[tex]a=2.6 m/s^2[/tex] is the acceleration of the object
Re-arranging the equation, we can solve to find the angle of the incline:
[tex]sin \theta = \frac{a}{g}=\frac{2.6}{9.8}=0.265\\\theta = sin ^{-1} (0.265)=15.4^{\circ}[/tex]