Two wooden crates rest on top of one another. the smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. there is no friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. a massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).a)the rope is pulled with a tension t = 234 n (which is small enough that the top crate will not slide). what is the acceleration of the small crate? b)in the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate? c)what is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide? d)the tension is increased in the rope to 1187 n causing the boxes to accelerate faster and the top box to begin sliding. what is the acceleration of the upper crate? m/s2e)as the upper crate slides, what is the acceleration of the lower crate?

a)  a = 2 m/s² , b) fr = -52 N , c)  fr = 193.55 N  and d)   a = 6.1 m/s²

Explanation:

For this long problem let's use Newton's second law

         F = ma

a) Consider the boxes as a whole, because they do not move

    F = (m1 + m2) a

    a = F / (m1 + m2)

    a = 234 / (25 +91)

    a = 2 m / s²

b) The frictional force between the two boxes is the same, we calculate for the lower box

    F -fr = m2 a

     fr = m2 a -F

    fr = 91 2 - 234

    fr = -52 N

c) Let's calculate the maximum friction force of the upper case

    fr = μ N

    fr = 0.79 25  9.8

    fr = 193.55 N

So that the boxes do not slip, it is necessary that the force applied does not exceed the friction

     F = m2 a + fr

the acceleration can be zero and the maximum friction force

     F = fr = 193.55 N

d) The acceleration of the upper case is created by the friction force,

     fr = m1 a1

     a1 = fr / m1

But with the box they are sliding the friction force is reduced to the kinetic friction force

     fr = μ N

     frc = 0.62 25 9.8

     frc = 151.9 N

     a = 151.9 / 25

     a = 6.1 m / s²

e) The acceleration of the lower case

     F -fr = m2 a

     a = (F-frc) / m2

     a = (1187 - 151.9) / 91

     a = 11.4  m/s²