Question

The velocity of a 1.10 kg block sliding down a frictionless inclined plane is found to be 1.11 m/s. 1.90 s later, it has a velocity of 7.38 m/s. What is the angle of the plane with respect to the horizontal? Use "deg" as your units.

Answer 1

along the incline,

using Fnet = ma

mgsin@ = ma

a = gsin@

USing , v = u + at

7.38 = 1.11 + gsin@*1.90

9.81 x sin@ x 1.90 = 6.27

sin@ = 0.336

@ = sin^-1(0.336) =19.66 Degrees .