Question
The initial velocity of a 2.56 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. Then 1.08 s later, it has a velocity of 3.99 m/s.What is the angle of the plane with respect to the horizontal?
Answers
drwls
Acceleration = a = (3.99-1.13)/1.08
= 2.648 m/s^2
Force along the plane = M*g sinA = M*a
sinA = a/g = 0.270
A = sin^-1(0.270)= 15.7 degrees
= 2.648 m/s^2
Force along the plane = M*g sinA = M*a
sinA = a/g = 0.270
A = sin^-1(0.270)= 15.7 degrees