Answer:
a) -4 N
b) +4 N
Explanation:
Draw a free body diagram for each block.
For the large block, there are 2 forces: 12 N pushing to the right, and F pushing to the left.
For the small block, there is 1 force, F pushing to the right.
There are also weight and normal forces in the vertical direction, but we can ignore those.
Sum of forces on the large block in the x direction:
∑F = ma
12 − F = 4a
Sum of forces on the small block in the x direction:
∑F = ma
F = 2a
2F = 4a
Substitute:
12 − F = 2F
12 = 3F
F = 4
The small block pushes on the large block 4 N to the left (-4 N).
The large block pushes on the small block 4 N to the right (+4 N).
To develop this problem we will apply the concepts related to the kinematic equations of motion, specifically that of acceleration. Acceleration can be defined as the change of speed in an instant of time, mathematically this is

If a mobile is decreasing its speed (it is slowing down), then its acceleration is in the opposite direction to the movement. This would imply that the acceleration vector is opposite to the velocity vector.
Therefore the correct answer is B.
Answer:
0.000025s
Explanation:
Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s
The center of mass is given with this formula:

Velocity is:

So, for the velocity of the center of mass we have:

In our case it is: