A block of mass 35kg is on a flat floor with a static friction coefficient of 0.36 and a kinetic friction coefficient of 0.29. A person attempts to slide the box from rest in a forward direction . Find the maximum frictional force experienced and the acceleration of the box once it is moving if:

a) the person pushes horizontally with 120N
b) the person pushes at an angle with a 25N vertical component directed down and a 200N horizontal component of the push directed forward
c)the person pulls with 120 N on a rope angle 30 degrees up from the horizontal.

To find the maximum frictional force experienced and the acceleration of the box, we will use the concept of friction and Newton's second law of motion. We will also take into account the different scenarios mentioned.

Before we proceed, let's define the equations for static and kinetic friction:

Static friction: F_s = μ_s * N
Kinetic friction: F_k = μ_k * N

where:
F_s = static friction force
F_k = kinetic friction force
μ_s = static friction coefficient
μ_k = kinetic friction coefficient
N = normal force (equal to the weight of the object, in this case)

Now, let's analyze each scenario:

a) The person pushes horizontally with 120N:

In this case, the force applied by the person (120N) is greater than the maximum static friction force. Thus, the static friction force will equal the force applied by the person.

F_s = 120N

To calculate the acceleration, we can use Newton's second law:

F_net = m * a

In this case, the net force is equal to the force applied by the person minus the kinetic friction force:

F_net = F_applied - F_k

Since the box is initially at rest, the net force is equal to the force applied:

F_net = F_applied

Now, we can rearrange the equation to solve for acceleration:

F_net = m * a
F_applied = m * a
120N = 35kg * a

Solving for a, we find:

a = 120N / 35kg
a ≈ 3.43 m/s² (approximately)

b) The person pushes at an angle with a 25N vertical component down and a 200N horizontal component forward:

In this case, we need to calculate the horizontal and vertical components of the friction force separately.

Vertical component of friction: F_v = μ_s * N

Since the box is not accelerating vertically, the vertical component of the applied force must be equal to the vertical component of friction:

25N = μ_s * N
25N = 0.36 * N

Solving for N, we find:

N = 25N / 0.36
N ≈ 69.44 N (approximately)

Now, we can calculate the horizontal component of friction:

F_h = μ_k * N
F_h = 0.29 * 69.44N
F_h ≈ 20.11 N (approximately)

To find the acceleration, we use the same equation as before:

F_net = m * a

Substituting the horizontal component of the friction force as the net force:

F_net = F_h

a = F_h / m
a ≈ 20.11 N / 35kg
a ≈ 0.57 m/s² (approximately)

c) The person pulls with 120 N on a rope angle 30 degrees up from the horizontal:

To determine the acceleration in this scenario, we need to find the horizontal and vertical components of the applied force. The horizontal component will contribute to the acceleration, and the vertical component will affect the normal force.

Horizontal component of the applied force: F_h = F_applied * cos(angle)

F_h = 120N * cos(30°)
F_h ≈ 104.04 N (approximately)

Vertical component of the applied force: F_v = F_applied * sin(angle)

F_v = 120N * sin(30°)
F_v ≈ 60 N (approximately)

The normal force will be affected by the vertical component of the applied force:

N = weight - F_v

Since weight = m * g, where g is the acceleration due to gravity (approximately 9.8 m/s²):

N = m * g - F_v
N = 35 kg * 9.8 m/s² - 60 N
N ≈ 329.5 N (approximately)

Now, we can calculate the horizontal component of friction:

F_h = μ_k * N
F_h = 0.29 * 329.5 N
F_h ≈ 95.47 N (approximately)

Using the same equation as before to find the acceleration:

F_net = m * a

Substituting the horizontal component of the friction force as the net force:

F_net = F_h

a = F_h / m
a ≈ 95.47 N / 35kg
a ≈ 2.73 m/s² (approximately)

So, in summary:

a) The maximum frictional force experienced is 120N, and the acceleration of the box is approximately 3.43 m/s².
b) The maximum frictional force experienced is approximately 20.11 N, and the acceleration of the box is approximately 0.57 m/s².
c) The maximum frictional force experienced is approximately 95.47 N, and the acceleration of the box is approximately 2.73 m/s².