Question

A small block is released from rest at the top of a frictionless incline. The distance from the top of the incline to the bottom, measured along the incline, is 3.70 m. The vertical distance from the top of the incline to the bottom is 1.67 m. If
g = 9.80 m/s2, what is the acceleration of the block as it slides down the incline?

Answer 1

d = 3.70 m.
h = 1.67 m.

V^2 = Vo^2 + 2g*h = 0 + 19.6*1.67 = 32.73
V = 5.72 m/s. = Velocity at bottom of
incline.

d = 0.5g*t^2 = 1.67 m.
4.9t^2 = 1.67
t^2 = 0.341
t = 0.584 s. = Time to reach bottom.

V = Vo + a*t
a = (V-Vo)/t
V = 5.72 m/s
Vo = 0
Solve for a.

Answer 2

9.8