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? (force applies upper crate)

Answer 1

max force that can be applied to top crate is no more than friction holding it on.
friction- 4.62*.870*9.8 N and this is the force applied.

now fricton on the floor: (4.62+2.19).440*9.8

net force=mass*a
force applied-frictonFloor=m*a
solve for a.

d=1/2 a t^2
d is given, a is known now, solve for time t.