Question

A small block with mass 0.200 kg is released from rest at the top of a frictionless incline. The block travels a distance 0.440 m down the incline in 2.00 s. The 0.200 kg block is replaced by a 0.400 kg block. If the 0.400 kg block is released from rest at the top of the incline, how far down the incline does it travel in 2.00 s?

Answer 1

0.200 kg block:
d = 0.5a*t^2 = 0.44 m.
0.5a*2^2 = 0.44
2a = 0.44
a = 0.22 m/s^2.

0.400 kg block:
d = 0.5a*t^2 = 0.5 * 0.22*2^2 = 0.44 m