A 5.0-kg block is at rest on a horizontal floor. If you push horizontally on the 5.0-kg block with a force of 10.0 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 5.0-kg block. What is the magnitude F of the force, acting horizontally on the 5.0-kg block as before, that is required to make the two blocks start to move?

Duplicate post. Look for the answer to one of your other posts of the same questijon

To find the coefficient of static friction in part (a), we can use the formula:

Fs = μs * N

where Fs is the force of static friction, μs is the coefficient of static friction, and N is the normal force.

Since the block is at rest on a horizontal floor, the normal force is equal to the weight of the block, which is given by:

N = m * g

where m is the mass of the block and g is the acceleration due to gravity.

Given that the mass of the block is 5.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the normal force:

N = 5.0 kg * 9.8 m/s^2
N = 49 N

Since the block just starts to move when a horizontal force of 10.0 N is applied, we know that the force of static friction is equal to 10.0 N. Using the formula mentioned earlier, we can find the coefficient of static friction:

10.0 N = μs * 49 N

Solving for μs:

μs = 10.0 N / 49 N
μs ≈ 0.204

Therefore, the coefficient of static friction is approximately 0.204.

Moving on to part (b), to find the force F required to make the two blocks start to move, we need to consider the additional weight of the second block. The total normal force acting on the 5.0-kg block is now the sum of the weight of the 5.0-kg block and the weight of the 12.0-kg block:

N = (5.0 kg + 12.0 kg) * 9.8 m/s^2
N = 166.8 N

Using the same formula as before, we can calculate the force of static friction required for the blocks to start moving:

Fs = μs * N
10.0 N = μs * 166.8 N

Solving for μs:

μs = 10.0 N / 166.8 N
μs ≈ 0.06

Therefore, the coefficient of static friction between the blocks is approximately 0.06.

Since the force F required to make the blocks start moving is the force of static friction, we have:

F = 10.0 N

So, the magnitude of the force acting horizontally on the 5.0-kg block, as before, that is required to make the two blocks start to move is 10.0 N.