Question

Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton’s Laws, and (3) solve the equations of motion. A block starts from rest at a height of 3.9 m on a fixed inclined plane. The acceleration of gravity is 9.8 m/s 2 . 3.8 kg µ = 0.12 31◦ What is the speed of the block at the bottom of the ramp? Answer in units of m/s.

Answer 1

8 . 55599 m / s.  
Let : h = 8 . 2 m , m = 5 . 1 kg , = 0 . 22 , = 22 , and v f = final speed . The normal force to the inclined plane is N = mg cos . The sum of the forces par- allel to the inclined plane is F net = ma = mg sin - mg cos a = g sin - g cos Since v 2 f = v 2 + 2 ax = 2 ad (1) along the plane, and neglecting the dimension of the block, the distance, d, to the end of the ramp is d sin = h d = h sin (2) therefore v f = r 2 ah sin = r 2 g h (sin - cos ) sin = p 2 g h (1- cot ) (3) = q 2(9 . 8 m / s 2 )(8 . 2 m)[1- (0 . 22)cot22 ] = 8 . 55599 m / s .