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

`15.4^(\circ)`

Step-by-step 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 = (a)/(g)=(2.6)/(9.8)=0.265\\\theta = sin ^(-1) (0.265)=15.4^(\circ)`