A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of 130 N at an angle of 30.0° above the horizontal. The box has a mass of 24.0 kg, and the coefficient of kinetic friction between box and floor is 0.300.

(a) Find the acceleration of the box.

(b) The student now starts moving the box up a 10.0° incline, keeping her 130 N force directed at 30.0° above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?
m/s2, up the incline

(a) Well, well, well, looks like this student is getting her workout in! To find the acceleration of the box, we need to consider all the forces at play. We have the force the student is applying, the force of gravity, and the force of friction. Since the box is moving horizontally, we only consider the horizontal forces. Let's break it down.

The force the student pulls with has a horizontal component given by Fx = F * cos(theta), where F is the force of 130 N and theta is the angle of 30.0°. So, Fx = 130 N * cos(30.0°).

The force of friction can be determined using Ffriction = coefficient of kinetic friction * normal force. The normal force is simply the weight of the box, which is mass * gravity, so the normal force is 24.0 kg * 9.8 m/s^2.

Now, since the box is moving, we know that the sum of the forces equals mass times acceleration (Fnet = ma). The net force in the horizontal direction is the force applied by the student minus the force of friction. So, Fnet = Fx - Ffriction.

Now we can solve for acceleration by dividing both sides of the equation by mass: a = (Fx - Ffriction) / m.

Plug in the values and you'll find your answer for the acceleration. Give it a go!

(b) Ah, the student is making things interesting by moving the box up an incline! Let's calculate the new acceleration.

Since the force applied is directed at an angle of 30.0° above the line of the incline, we need to break it down into the components parallel and perpendicular to the incline.

The parallel component of the force applied is given by Fparallel = F * sin(theta), where F is the force of 130 N and theta is the angle of 30.0°.

Now, the force of friction will act opposite to the direction of motion, so it will be directed parallel to the incline. The force of friction is still determined by the coefficient of friction and the normal force, which, on an incline, is the weight of the box times the cosine of the angle of incline.

The net force in this case will be the force applied parallel to the incline minus the force of friction. We can now divide that by the mass to find the new acceleration.

I know it's a little more complicated, but give it a try! You've got this!

To solve this problem, we can break down the forces acting on the box and use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

(a) To find the acceleration of the box on a horizontal floor:
- Let's first resolve the force applied by the student into its horizontal and vertical components.
- The vertical component = F * sin(theta) = 130 N * sin(30.0°) = 65 N
- The horizontal component = F * cos(theta) = 130 N * cos(30.0°) = 112.87 N

- Now, let's calculate the force of kinetic friction acting on the box:
- Force of kinetic friction = coefficient of kinetic friction * normal force
- Normal force = mass * gravity (where gravity is approximately 9.8 m/s^2)
- Normal force = 24.0 kg * 9.8 m/s^2 = 235.2 N
- Force of kinetic friction = 0.300 * 235.2 N = 70.56 N

- The net force acting on the box horizontally is the difference between the applied horizontal force and the force of kinetic friction:
- Net force horizontally = Applied horizontal force - Force of kinetic friction
- Net force horizontally = 112.87 N - 70.56 N = 42.31 N

- Finally, we can use Newton's second law to calculate the acceleration:
- Net force horizontally = mass * acceleration
- 42.31 N = 24.0 kg * acceleration
- Acceleration = 42.31 N / 24.0 kg = 1.763 m/s^2

Therefore, the acceleration of the box on a horizontal floor is 1.763 m/s^2.

(b) To find the new acceleration of the box on a 10.0° incline:
- Let's first calculate the force of gravity acting on the box on the incline:
- Force of gravity = mass * gravity
- Force of gravity = 24.0 kg * 9.8 m/s^2 = 235.2 N

- Since the force applied by the student is still 130 N, the force parallel to the incline can be calculated as:
- Force parallel to incline = F_applied * cos(theta)
- Force parallel to incline = 130 N * cos(30.0°) = 112.87 N

- The force of kinetic friction on the incline can be calculated as before:
- Force of kinetic friction = coefficient of kinetic friction * normal force
- Normal force = mass * gravity * cos(theta)
- Normal force = 24.0 kg * 9.8 m/s^2 * cos(10.0°) = 233.27 N
- Force of kinetic friction = 0.300 * 233.27 N = 69.98 N

- The net force acting on the box along the incline is the difference between the applied force and the force of friction:
- Net force along incline = Force parallel to incline - Force of kinetic friction
- Net force along incline = 112.87 N - 69.98 N = 42.89 N

- Finally, we can calculate the new acceleration on the incline using Newton's second law:
- Net force along incline = mass * acceleration
- 42.89 N = 24.0 kg * acceleration
- Acceleration = 42.89 N / 24.0 kg = 1.787 m/s^2

Therefore, the new acceleration of the box on a 10.0° incline is 1.787 m/s^2.

To find the acceleration of the box, we need to consider the forces acting on it. Let's break down the problem step by step:

(a) Find the acceleration of the box:

1. The first step is to determine the force of friction acting on the box. We can do this by multiplying the coefficient of kinetic friction (0.300) by the normal force.

Normal force = mass of the box x acceleration due to gravity
Normal force = 24.0 kg x 9.8 m/s^2 (acceleration due to gravity)
Normal force = 235.2 N

Force of friction = coefficient of kinetic friction x normal force
Force of friction = 0.300 x 235.2 N
Force of friction = 70.56 N

2. We can now analyze the forces acting on the box along both the x-axis (horizontal) and y-axis.

Along the x-axis:
- The force of friction acts in the opposite direction of motion, so it has a negative sign. Therefore, it is -70.56 N.
- The horizontal component of the applied force is given by:
F_applied_x = force applied x cos(angle)
F_applied_x = 130 N x cos(30.0°)
F_applied_x = 112.43 N

Along the y-axis:
- The weight (force due to gravity) acts downwards and is given by:
Weight = mass x acceleration due to gravity
Weight = 24.0 kg x 9.8 m/s^2
Weight = 235.2 N
- The vertical component of the applied force is given by:
F_applied_y = force applied x sin(angle)
F_applied_y = 130 N x sin(30.0°)
F_applied_y = 65 N

3. Now we can calculate the net force acting on the box along the x-axis and y-axis.

Net force along the x-axis:
Net force_x = F_applied_x + force of friction
Net force_x = 112.43 N - 70.56 N
Net force_x = 41.87 N

Net force along the y-axis:
Net force_y = Weight - F_applied_y
Net force_y = 235.2 N - 65 N
Net force_y = 170.2 N

4. Since the box is moving along the x-axis, we can use Newton's second law to calculate the acceleration.

Net force_x = mass x acceleration
41.87 N = 24.0 kg x acceleration
acceleration = 41.87 N / 24.0 kg
acceleration ≈ 1.746 m/s^2

Therefore, the acceleration of the box is approximately 1.746 m/s^2.

(b) Find the new acceleration of the box when moving up a 10.0° incline:

1. The applied force and the normal force are now no longer perpendicular to the direction of motion. We need to resolve these forces into components parallel and perpendicular to the incline.

Force parallel to the incline (F_parallel) = force applied x cos(30.0° - 10.0°)
F_parallel = 130 N x cos(20.0°)
F_parallel ≈ 119.84 N

Force perpendicular to the incline (F_perpendicular) = force applied x sin(30.0° - 10.0°)
F_perpendicular = 130 N x sin(20.0°)
F_perpendicular ≈ 44.57 N

2. We can now analyze the forces acting on the box along both the x-axis (parallel to the incline) and y-axis (perpendicular to the incline).

Along the x-axis:
- The force of friction acts in the opposite direction of motion, so it has a negative sign. Therefore, it is -70.56 N.

Net force_x = F_parallel + force of friction
Net force_x = 119.84 N - 70.56 N
Net force_x = 49.28 N

Along the y-axis:
- The weight (force due to gravity) acts downwards and is given by:
Weight = mass x acceleration due to gravity
Weight = 24.0 kg x 9.8 m/s^2
Weight = 235.2 N

Net force_y = Weight - F_perpendicular
Net force_y = 235.2 N - 44.57 N
Net force_y = 190.63 N

3. We can now calculate the net force acting on the box.

Net force = √((Net force_x)^2 + (Net force_y)^2)
Net force = √((49.28 N)^2 + (190.63 N)^2)
Net force ≈ 197.74 N

4. Finally, we can use Newton's second law to calculate the new acceleration.

Net force = mass x acceleration
197.74 N = 24.0 kg x acceleration
acceleration = 197.74 N / 24.0 kg
acceleration ≈ 8.24 m/s^2

Therefore, the new acceleration of the box when moving up a 10.0° incline is approximately 8.24 m/s^2.

Wb = mg = 24kg * 9.8N/kg = 235 N. = Wt.

of box.

Fb = 235N @ 0 Deg. = Force of box.
Fp = 235*sin(0) = 0 = Force parallel to
floor.
Fv = 235*cos(0) = 235 N. = Force perpendicular to floor.

Fv' = Fv - 130*sin30 = 235 - 65 = 170 N. = Normal.

Fk = u*Fv' = 0.3*170 = 51 N' = Force of kinetic friction.

a. Fn = Fap - Fp - Fk = ma.
130*cos30 - 0 - 51 = 24*a.
112.58 - 51 = 24a.
24a = 61.58.
a = 61.58 / 24 = 2.57 m/s^2.

b. Fp = 235*sin10 = 40.81 N.
Fv = 235*cos10 = 231.43 N.
Fv' = Fv - 130*sin30 = 231.43-65 = 166.43 N.

Fk = u*Fv' = 0.3*166.43.43 = 49.93 N.

Fn = Fap - Fp - Fk = ma.
130*cos30 - 40.81 - 49.93 = 24a.
112.58 - 40.81 - 49.93 = 24a.
24a = 112.58 - 90.74 = 21.84.
a = 21.84 / 24 = 0.91 m/s^2.