Answer:
[tex]30.7^{\circ}[/tex]
Explanation:
Assuming there is no friction, the only force acting on the ice cube along the incline is the component of the weight of the cube parallel to the incline, that is:
[tex]mg sin \theta[/tex]
where
m is the mass of the cube
g is the acceleration of gravity
[tex]\theta[/tex] is the angle of the incline
The equation of the forces along the incline therefore is
[tex]mg sin \theta = ma[/tex]
where a is the acceleration.
Here we have the following data:
[tex]m=0.2843276 kg[/tex]
[tex]g=9.8 m/s^2[/tex]
[tex]a=5 m/s^2[/tex]
Solving for [tex]\theta[/tex], we find the angle of the incline:
[tex]\theta = sin^{-1} (\frac{a}{g})=sin^{-1}(\frac{5}{9.8})=30.7^{\circ}[/tex]
