4 kg block rests on a smooth horizontal surface,upper surface of 4 kg is rough.A block of mass 2 kg is placed on its upper surface.The acceleration of upper block with respect to earth when 4 kg mass is pulled by a horizontal force of 30N?coefficient of static friction=0.8 and coefficient of kinetic friction=0.6 I have been working to solve this but not getting it as i do not know what is force of friction on 2 kg block .I know it is coefficient of friction divided by normal reaction but what would be normal reaction?

We do not know yet if the top block slips.

First, assume it does not slip.
F = m a
30 = (4+2) a
a = 5 m/s^2

Now
How much force is required to accelerate the top block at 5 m/s^2?
F = 2 (5) = 10 N required
How much do we have?
Fnormal = m g = 2*9.81 = 19.62 N
so friction max = .8 * 19.62 = 15.7 N

But we only need 10 N to keep that top block stuck on the bottom one, so it does not slide and we are finished.

To find the force of friction on the 2 kg block, we first need to determine the normal reaction acting on it. The normal reaction is the force exerted by a surface perpendicular to the surface. In this case, the 2 kg block is resting on the upper surface of the 4 kg block.

The normal reaction depends on the weight of the block and the contact between the two blocks. As both blocks are at rest, the normal reaction is equal to the weight of the 2 kg block.

The weight of an object can be calculated using the equation:

Weight = mass × acceleration due to gravity (g)

The acceleration due to gravity on Earth is approximately 9.8 m/s².

Therefore, the weight of the 2 kg block would be:

Weight = 2 kg × 9.8 m/s² = 19.6 N

Since the normal reaction is equal to weight, the normal reaction on the 2 kg block is 19.6 N.

Now we can calculate the force of friction on the 2 kg block using the coefficient of friction. In this case, we have both the coefficient of static friction (μs = 0.8) and the coefficient of kinetic friction (μk = 0.6).

If the upper block is not moving, we need to consider static friction. The force of static friction (Fs) can be calculated using the equation:

Fs ≤ μs × Normal Reaction

In this case:

Fs ≤ 0.8 × 19.6 N

Here, "≤" means "less than or equal to."

The force of friction on the 2 kg block will depend on the magnitude of the force of static friction acting on the 4 kg block. If the force of static friction is less than the force applied, the blocks will start moving, and the force of kinetic friction will take effect.

On the other hand, if the force of static friction is equal to the applied force, the blocks will remain at rest. We need to find the maximum value of the force of static friction because it determines whether the blocks will move or not.

Therefore, we can equate the maximum force of static friction to the applied force:

Fs = 30 N

Now we have Fs = 0.8 × 19.6 N.

Fs = 15.68 N

So, the force of friction on the 2 kg block is 15.68 N when the 4 kg block is pulled by a horizontal force of 30 N.