The initial velocity of a 2.04 kg block sliding down a frictionless inclined plane is found to be 1.24 m/s. Then 1.23 s later, it has a velocity of 6.70 m/s.
What is the angle of the plane with respect to the horizontal? im lost.....
First compute the acceleration.
a = (6.70-1.24)/1.23 = ___ m/s^2
Then apply Newton's law in the direction down the incline.
m g sin A = m a
sin A = a/g
Solve that for the angle A.