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.
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)
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