The velocity of a 1.5 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. 1.2 s later, it has a velocity of 4.37 m/s. What is the angle of the plane with respect to the horizontal (in degrees)?

Answer 1

It is accelerating at a rate (4.37-1.13)/1.2 = 2.7 m/s^2

The Force pushing it down the ramp is
F = a/M = 1.8 N

With no friction, 1.8 N must be the component of the weight down the ramp, which is M g sin A. Therefore
1.8 N = (1.5)(9.8) sinA
sin A = 0.1224
A = 7.0 degrees