Question

Two crates rest on top of one another.There is a crate of mass 4.62 kg on top of another crate of mass 2.19 kg. The coefficient of friction between the lower crate and the floor is μk = 0.440 and the coefficient of static friction between the two crates is μs = 0.870. If the crates start at rest and a force is applied to the right so that both move a distance of 8.70 m, what is the minimal amount of time required in which this can be accomplished without the top crate sliding on the lower crate? (THANK YOU)

Answer 1

max force on top crate= 4.62*.870*9.8=39.39012N

so the time to accelerate
d=1/2 a t=1/2 (force/masstop)*t
8.70=1/2 (9.8*.870) t
solve for time
All this assumes the Force is applied to the bottom crate.

If the force applied is to the upper crate, please repost.