Assume cathy has to drag her suitcase along the floor of the airport terminal by a rope. The rope makes a 40.0 degree angle with the horizantal. The suitcase has a mass of 36.0 kg and Cathy pulls on the rope with a force of 65.0 N.

a) What is the magnitude of the normal force acting on the suitcase due to the floor?
b) If the kinetic friction between the suitcase and the marble floor is 40.0 N, find the acceleration of the suitcase while Cathy pulls with 65.0 N force at 40 degrees.
c) Starting from rest, for how long a time must she pull with this force until the suitcase reaches a comfortable walking speed of 0.5 m/s^2.

To find the answer to these questions, we need to use several concepts from physics, including Newton's laws of motion and components of forces. Let's break down each part of the question step by step:

a) To calculate the magnitude of the normal force acting on the suitcase, we need to consider the force components. The normal force is the force exerted by a surface perpendicular to it. In this case, the normal force acts upward to counterbalance the weight of the suitcase.

Since the suitcase is on the floor and not moving vertically, the vertical forces must be balanced:

Normal force + Weight = 0

The weight of the suitcase can be calculated using the mass and the acceleration due to gravity (9.8 m/s^2):

Weight = mass * acceleration due to gravity
= 36.0 kg * 9.8 m/s^2

Solving for the normal force:

Normal force = -Weight
= -36.0 kg * 9.8 m/s^2

Since the normal force acts upward, it will be positive:

Normal force = 36.0 kg * 9.8 m/s^2

b) To find the acceleration of the suitcase, we need to consider the horizontal forces acting on it. The force Cathy applies is at an angle of 40.0 degrees with the horizontal.

We need to determine the component of Cathy's force that acts parallel to the ground, which will contribute to the acceleration. The component of force in the horizontal direction can be calculated using trigonometry:

Force parallel to the ground = Force * cos(angle)

Force parallel to the ground = 65.0 N * cos(40.0 degrees)

Now we can calculate the net force acting on the suitcase:

Net force = Force parallel to the ground - Force of friction

Since the suitcase is moving, the frictional force will be kinetic friction:

Net force = Force parallel to the ground - Kinetic friction

Net force = 65.0 N * cos(40.0 degrees) - 40.0 N

Finally, we can use Newton's second law of motion:

Net force = mass * acceleration

Rearranging the equation to solve for acceleration:

Acceleration = Net force / mass

Substituting the values into the equation:

Acceleration = (65.0 N * cos(40.0 degrees) - 40.0 N) / 36.0 kg

c) To calculate the time required for the suitcase to reach a speed of 0.5 m/s, we need to use the equation of motion:

final velocity = initial velocity + (acceleration * time)

Since the suitcase starts from rest, the initial velocity can be assumed as zero. Rearranging the equation:

time = (final velocity - initial velocity) / acceleration

Substituting the values into the equation:

time = (0.5 m/s - 0 m/s) / acceleration

To solve these problems, we will need to use Newton's second law (F = ma) and resolve forces into their components.

a) The normal force (Fn) acts perpendicular to the floor, so it counteracts the gravitational force (mg) pulling the suitcase downward. Since the suitcase is not accelerating vertically, the magnitudes of these two forces are equal. Therefore, the magnitude of the normal force is equal to the suitcase's weight (mass x gravity).

Fn = mg
Fn = 36.0 kg x 9.8 m/s^2
Fn = 352.8 N

So, the magnitude of the normal force acting on the suitcase is 352.8 N.

b) To find the acceleration of the suitcase, we need to consider the net horizontal force acting on it. This force is the horizontal component of the applied force minus the force of kinetic friction.

Fnet = Fapplied - Ffriction

The horizontal component of the applied force is given by:

Fx = Fapplied * cos(theta)

Substituting the given values:

Fx = (65.0 N) * cos(40.0 degrees)
Fx = 65.0 N * 0.766
Fx = 49.79 N

Now we can calculate the net force:

Fnet = 49.79 N - 40.0 N
Fnet = 9.79 N

The net force is equal to the mass of the suitcase multiplied by its acceleration:

Fnet = ma

Substituting the known values:

9.79 N = (36.0 kg) * a

Now solve for acceleration:

a = 9.79 N / 36.0 kg
a ≈ 0.272 m/s^2

Therefore, the acceleration of the suitcase is approximately 0.272 m/s^2.

c) To find the time it takes for the suitcase to reach a comfortable walking speed of 0.5 m/s^2, we can use the equation:

v = u + at

Where:
v = final velocity (0.5 m/s)
u = initial velocity (0 m/s)
a = acceleration (0.272 m/s^2)
t = time

Rearranging the equation:

t = (v - u) / a

Substituting the known values:

t = (0.5 m/s - 0 m/s) / 0.272 m/s^2
t ≈ 1.838 s

Therefore, Cathy must pull with the 65.0 N force for approximately 1.838 seconds to reach a comfortable walking speed of 0.5 m/s.