A baggage handler drops your 8kg suitcase onto a conveyor belt running at 1.8 m/s. The materials are such that the static coefficient of friction is .4 and the kinetic coefficient of friction is .6. How far is your suitcase dragged before it is riding smoothly on the belt?
v_f= v_i+a(delta t)
f_s: force of static friction
f_k: force of kinetic friction
The Attempt at a Solution
I've come up with two free body diagrams. The y direction seems to be negligible since mg=n and they cancel out. In the x direction, I have F_belt pointing in the positive x-direction. I am confused about how to place f_k. I think that it too points in the positive x-direction, and heres my logic: it is not moving faster than the conveyor belt, so its motion would appear opposite the conveyor belt until it too moves at 1.8 m/s.
Also, there would be a free body diagram for when the suitcase moves at the velocity of the conveyor belt. This diagram would have F_belt pointing in the positive x-direction and f_s pointing in the negative x-direction.
F_x=8kg(0m/s^2)=F_belt - f_s
F_belt= .4 (8kg) (9.8m/s^2)
F_x= 8kg*a = F_belt + f_k
= 31.4N + .6(8kg)(9.8m/s^2) = 78.44 N
78.44/8= 9.8 m/s^2 = acceleration of suitcase
1.8m/s=0m/s + 9.8 m/s^2(delta t)
delta t= .18 seconds
This isn't the answer, according to my book*.
*I changed the numbers, but this is the method I used with the numbers I had and I didn't get the right answer.