What is the least amount of time in which this motion can be completed without the top block sliding on the lower block

The complete question is missing, so i have attached it.

t = 1.75 s

Explanation:

Looking at the image depicting the question with the block, we can say that there will be 2 frictional forces acting on the bottom of the block which we will name F1 and F2.

F1 is frictional force in the forward direction while F2 is frictional force in the backward direction.

Now at the top, F1 will still be acting on top but this time it will be in the backward direction while a second force F will be in the forward direction.

This means that for the top of the block, we have a net force of;

F - F1 = ma

a = (F - F1)/m

We have 4kg as mass of top block.

Thus;

a = (F - F1)/4

At the bottom, net force will be;

F1 - F2 = ma

a = (F1 - F2)/m

Bottom mass is 3kg.

a = (F1 - F2)/3

Since we are told the top block sliding on the lower block, it means they will move with the same acceleration.

Thus;

a = (F - F1)/4 = (F1 - F2)/3

F1 = μmg

0.6 is the coefficient of friction between top and bottom block. Thus;

F1 = 0.6 × 4 × 9.81

F1 = 23.544 N

Similarly, we are told coefficient of friction between bottom block and floor is 0.2

Thus;

F2 = 0.2(3 + 4) × 9.81

F2 = 0.2 × 7 × 9.81

F2 = 13.734 N

From earlier, we saw that;

a = (F1 - F2)/3

Thus;

a = (23.544 - 13.734)/3

a = 3.27 m/s²

Now, we are told the block F causes both blocks to slide a distance of 5m.

Thus, to find the time we will use Newton's equation of motion.

s = ut + ½at²

Since they began from rest, then u = 0.

Thus;

5 = ½ × 3.27 × t²

5 = 1.635t²

t² = 5/1.635

t = √(5/1.635)

t = 1.75 s

e) none of the above answers are correct

powerful computers now, not then