A 5 kg box is released on an incline and accelerates down the incline at 5.86 m/s^2. To the nearest tenth of a degree, what is the angle of the incline?
As shown in figure, the acceleration along the incline is gsin theta where theta is the angle of incline
since no value for coefficient of friction is given hence we will assume no frictional force
hence gsin theta = 5 .86
taking g = 9.8
sin theta = 0.6
hence theta = 36.9 hence rounding off to nearest 10th we get theta = 40 degree
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