Question

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 1

Answer:

$15.4^{\circ}$

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

$mg sin \theta$

where

m = 8.0 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration of gravity

$\theta$ is the angle of the incline

Therefore, the equation of motion along this direction is

$ma=mgsin \theta$

where

$a=2.6 m/s^2$ is the acceleration of the object

Re-arranging the equation, we can solve to find the angle of the incline:

$sin \theta = \frac{a}{g}=\frac{2.6}{9.8}=0.265\\\theta = sin ^{-1} (0.265)=15.4^{\circ}$