normal force = m g cos 45
so friction down slope = .3 m g cos 45
at 45 sin 45 = cos 45 = sqrt 2/2
weight down slope = m g sin 45
15 N = m g sin 45 + .3 m g cos 45
15 N = (m g sqrt 2/2)( 1 + .3)
30/1.3 sqrt 2 = m g = weight
if friction he other way then
F up = (m g sqrt 2/2)( 1 - .3)
= (m g sqrt 2/2)( .7)
a block moves up 45degree incline w/ constant speed under the action of a force of 15n applied parallel to the incline.If the coefficient of kinetic friction is 0.3, determine the weight of the block and minimum force required to allow the block to move down the incline at constant speed
2 answers
A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?