Answer:
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