A 25 kg box sits on an incline of 30 degrees. The coefficient of kinetic friction between the box and the incline is 0.29. Find the acceleration of the box down the incline.

Answer 1

Considering the + direction to be down the incline

The gravity force on the block in the + direction = mgsin(30) = 25(9.8)(.5) = 122.5 N

The normal force pushing up on the block = mgcos(30) = 25(9.8)(.866) = 212.18 N

The restraining friction force in the - direction = .29N = 61.53 N

So, Sum of the forces = 122.5 - 61.53 = ma = 25a so,

Acceleration a = (122.5-61.53)/25 = 2.44 m/sec²