A 10.0-kg block is at rest on a horizontal floor. If you push horizontally on the 10.0-kg block with a force of 20.0 N, it just starts to move.

(a) What is the coefficient of static friction?

b) A 12.0-kg block is stacked on top of the 10.0-kg block. What is the magnitude F of the force, acting horizontally on the 10.0-kg block as before, that is required to make the two blocks start to move?

Fb = 10kg * 9.8 = 98N = Force of block.

Fb = Fv = 98N,

a. Fh - u*Fv = 0,
20 - 98u = 0,
98u = 20,
u = 20 / 98 = 0.20.

b. Fb = (10 + 12)kg * 9.8 = 215.6N.
= Force of blocks.

Fv = Fb = 215.6N.

Fh - u*Fv = 0,
Fh = u*Fv = 0.2 * 215.6 = 43.1N = Force required to move the blocks.

To find the coefficient of static friction in part (a), we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. However, in this case, the block is at rest, so the net force is zero. Hence, we can set up the equation:

ΣF = 0

The forces acting on the block are the applied force and the force of static friction. Since the block is just starting to move, the maximum force of static friction is equal to the applied force. So we have:

F_applied - F_friction = 0

Substituting the values given, we get:

20.0 N - F_friction = 0

Then, solve for the force of static friction:

F_friction = 20.0 N

The coefficient of static friction can be found by dividing the force of static friction by the normal force. Since the block is on a horizontal floor and at rest, the normal force (N) is equal to the weight (W) of the block. So we have:

F_friction = μ_s * N

Substituting the known values, we get:

20.0 N = μ_s * (10.0 kg * 9.8 m/s^2)

Solving for μ_s:

μ_s = 20.0 N / (10.0 kg * 9.8 m/s^2)

Simplifying the equation:

μ_s = 0.204

Therefore, the coefficient of static friction is approximately 0.204 (rounded to three decimal places).

Now let's move on to part (b). In this case, there are two blocks stacked on top of each other. The force required to make both blocks start to move is the sum of the force needed for each block individually.

First, determine the force needed for the 12.0-kg block on top. The maximum force of static friction (F_friction1) acting on the 12.0-kg block is given by:

F_friction1 = μ_s * N

The normal force (N) acting on the 12.0-kg block is equal to its weight:

N = 12.0 kg * 9.8 m/s^2

Next, determine the force needed for the 10.0-kg block on the bottom. The maximum force of static friction (F_friction2) acting on the 10.0-kg block is also given by:

F_friction2 = μ_s * N

The normal force (N) acting on the 10.0-kg block is equal to the sum of the weight of the top block and the weight of the bottom block:

N = (12.0 kg + 10.0 kg) * 9.8 m/s^2

Finally, sum up the forces of static friction for the two blocks:

F = F_friction1 + F_friction2

Substituting the known values, you can calculate the magnitude (F) of the force required to make the two blocks start to move.