A 4.0-kg block is stacked on top of a 12.0-kg block, which is accelerating along a

horizontal table at a = 5.2m/s2 . Let u=uk=us.
(a) What minimum coefficient of friction u between the two blocks will prevent the 4.0-kg block
from sliding off?
(b) If u is only half this minimum value, what is the acceleration of the 4.0-kg block with respect to
the table?
(c) If u is only half this minimum value, what is the acceleration of the 4.0-kg block with respect to
the 12.0-kg block?
(d) What is the force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the
table is frictionless?

a) The minimum coefficient of friction u between the two blocks to prevent the 4.0-kg block from sliding off is 0.6.

b) If u is only half this minimum value, the acceleration of the 4.0-kg block with respect to the table is 2.6 m/s2.

c) If u is only half this minimum value, the acceleration of the 4.0-kg block with respect to the 12.0-kg block is 7.8 m/s2.

d) The force that must be applied to the 12.0-kg block in (a) and in (b), assuming that the table is frictionless, is 60 N.

To solve this problem, we need to analyze the forces acting on the blocks.

Let's start by analyzing the forces on the 12.0-kg block:
- The weight of the block (mg) acts vertically downward.
- The normal force (N) of the table acts vertically upward.
- The friction force (f) opposes the motion and acts horizontally.

The friction force can be expressed as f = u*N, where u is the coefficient of friction.

Now, let's analyze the forces on the 4.0-kg block:
- The weight of the block (mg) acts vertically downward.
- The normal force (N) of the 12.0-kg block acts vertically upward.
- The friction force (f) between the two blocks opposes the motion and acts horizontally.
- The force of tension (T) between the two blocks acts horizontally.

(a) To prevent the 4.0-kg block from sliding off, the maximum friction force must equal the force of tension. Therefore, f = T. Since f = u*N, we can set this equal to T and solve for u.

u * N = T

To find N, we need to find the normal force acting on the 12.0-kg block. Since the block is accelerating, the net force in the vertical direction must be non-zero. Therefore, N - mg = ma, where a is the acceleration. Rearranging this equation, we have N = mg + ma.

Substituting N into the equation for u * N = T, we get:
u * (mg + ma) = T

(b) If u is only half the minimum value, the equation for the friction force becomes:
0.5u * (mg + ma) = T

To find the acceleration of the 4.0-kg block with respect to the table, we need to determine the net force on the block. The net force is given by:
Net force = applied force - friction force

So, Net force = T - f

(c) To find the acceleration of the 4.0-kg block with respect to the 12.0-kg block, we need to consider the forces acting on the 4.0-kg block. The net force on the 4.0-kg block is given by:
Net force = T - f

(d) If the table is assumed to be frictionless, the force applied to the 12.0-kg block in both (a) and (b) is the force required to overcome its weight and accelerate it. The force required is given by:
Force = mass * acceleration

Now you have the general steps to solve each part of the problem. Apply these steps to the given masses, accelerations, and equations to find the specific answers for each part.

To solve this problem, we can use Newton's second law of motion and the concept of friction.

(a) To prevent the 4.0-kg block from sliding off, we need to find the minimum coefficient of friction, μ, between the two blocks.
The force of friction between the two blocks is equal to the product of the coefficient of friction and the normal force.

The equation for the friction force is given by:
F_friction = μ * m * g

Where:
F_friction is the force of friction
μ is the coefficient of friction
m is the mass of the 4.0-kg block
g is the acceleration due to gravity (approximately 9.8 m/s²)

The normal force acting on the 4.0-kg block is equal to its weight:
F_normal = m * g

Substituting the value of F_normal into the friction force equation:
F_friction = μ * (m * g)

Since the blocks are accelerating together, the net force acting on the 4.0-kg block is equal to its mass times its acceleration:
F_net = m * a

Setting the net force equal to the force of friction:
m * a = μ * (m * g)

We can rearrange the equation to solve for μ:
μ = (m * a) / (m * g)
μ = a / g

Substituting the given values:
μ = 5.2 m/s² / 9.8 m/s²
μ = 0.53

Therefore, the minimum coefficient of friction, μ, necessary to prevent the 4.0-kg block from sliding off is 0.53.

(b) If μ is only half the minimum value, the new coefficient of friction, μ_new, will be:

μ_new = 0.5 * 0.53
μ_new = 0.27

The acceleration of the 4.0-kg block with respect to the table, a_new, can be found using the same equation as before:

a_new = μ_new * g
a_new = 0.27 * 9.8 m/s²
a_new ≈ 2.65 m/s²

Therefore, the acceleration of the 4.0-kg block with respect to the table when μ is only half the minimum value is approximately 2.65 m/s².

(c) The acceleration of the 4.0-kg block with respect to the 12.0-kg block is the same as its acceleration with respect to the table. Therefore, the acceleration of the 4.0-kg block with respect to the 12.0-kg block is approximately 2.65 m/s².

(d) If the table is frictionless, there will be no force of friction between the two blocks. Therefore, the force that must be applied to the 12.0-kg block in both cases (a and b) to produce the given acceleration is equal to the net force acting on the 12.0-kg block, which is equal to its mass times its acceleration:

F_applied = m_12 * a

Substituting the given values:
F_applied = 12.0 kg * 5.2 m/s²
F_applied = 62.4 N

Therefore, the force that must be applied to the 12.0-kg block in both cases (a and b), assuming a frictionless table, is 62.4 N.