A box of books weighing 350 N is shoved across the floor of an apartment by a force of 460 N exerted downward at an angle of 35.0° below the horizontal. If the coefficient of kinetic friction between box and floor is 0.57, how long does it take to move the box 3.80 m, starting from rest?

To find the time it takes to move the box 3.80 m, we need to break down the problem into different steps:

Step 1: Find the horizontal and vertical components of the force exerted.

The force exerted downward at an angle of 35.0° can be broken down into its horizontal and vertical components using trigonometry:

F_horizontal = F * cos(angle)
F_vertical = F * sin(angle)

In this case:
F = 460 N (given)
angle = 35.0° (given)

F_horizontal = 460 N * cos(35.0°)
F_vertical = 460 N * sin(35.0°)

Please note that these values are constant throughout the motion since the force is being constantly applied.

Step 2: Calculate the gravitational force and the force of kinetic friction.

The gravitational force acting on the box can be calculated using the weight equation:

Weight = mass * gravity

Given that weight = 350 N, and the acceleration due to gravity is approximately 9.8 m/s^2, we can find the mass:

Weight = mass * gravity
350 N = mass * 9.8 m/s^2

Solving this equation gives us the mass of the box.

The force of kinetic friction acting on the box can be calculated using the equation:

Force of friction = coefficient of kinetic friction * Normal force

The normal force is equal in magnitude and opposite in direction to the vertical component of the force exerted (F_vertical).

Step 3: Calculate the net force acting on the box.

The net force is the sum of the horizontal component of the force exerted and the force of friction:

Net force = F_horizontal - Force of friction

Step 4: Calculate the acceleration of the box.

Using Newton's second law of motion, we can relate the net force to the acceleration:

Net force = mass * acceleration

Rearranging this equation, we can solve for acceleration:

acceleration = Net force / mass

Step 5: Calculate the time it takes to move 3.80 m.

We can use the kinematic equation to find the time it takes for the box to move 3.80 m:

distance = initial velocity * time + (1/2) * acceleration * time^2

The box starts from rest (initial velocity = 0) and moves a distance of 3.80 m. We know the acceleration from step 4. By substituting these values, we can rearrange the equation to solve for time.

Following the steps outlined above will help you find the time it takes to move the box 3.80 m.

To find the time it takes to move the box, we can break down the problem into steps:

Step 1: Resolve the force exerted downward into horizontal and vertical components.
Step 2: Calculate the normal force acting on the box.
Step 3: Calculate the frictional force acting on the box.
Step 4: Calculate the net force acting on the box.
Step 5: Determine the acceleration of the box.
Step 6: Use the acceleration to calculate the time taken to move the box.

Let's go through each step in detail:

Step 1: Resolve the force exerted downward into horizontal and vertical components.
The horizontal component of the force is given by:
F_horizontal = F * cos(θ) = 460 N * cos(35.0°)

Step 2: Calculate the normal force acting on the box.
The normal force acting on the box is equal to its weight, given by:
N = mg = 350 N

Step 3: Calculate the frictional force acting on the box.
The frictional force is given by:
f_friction = μ * N = 0.57 * 350 N

Step 4: Calculate the net force acting on the box.
The net force acting on the box is given by:
F_net = F_horizontal - f_friction

Step 5: Determine the acceleration of the box.
The acceleration of the box is given by Newton's second law:
F_net = ma, where m is the mass of the box.
a = F_net / m

Step 6: Use the acceleration to calculate the time taken to move the box.
Using the equation of motion: s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.

We are given the displacement s = 3.80 m and the acceleration a from step 5. Solving for t, we get:

3.80 = 0 + (1/2)(a)(t^2)

Rearranging the equation:

3.80 = (1/2)(a)(t^2)
7.60 = (a)(t^2)

Thus:

t^2 = 7.60 / a
t = sqrt(7.60 / a)

Now, let's calculate the values step by step:

Step 1:
F_horizontal = 460 N * cos(35.0°) = 377.92 N (rounded to three decimal places)

Step 3:
f_friction = 0.57 * 350 N = 199.5 N

Step 4:
F_net = F_horizontal - f_friction = 377.92 N - 199.5 N = 178.42 N (rounded to two decimal places)

Step 5:
a = F_net / m
= 178.42 N / (350 N / 9.8 m/s^2)
= 178.42 N / 35.714 kg
= 4.994 m/s^2 (rounded to three decimal places)

Step 6:
t = sqrt(7.60 / a)
= sqrt(7.60 / 4.994)
= sqrt(1.520)
≈ 1.23 seconds (rounded to two decimal places)

Therefore, it takes approximately 1.23 seconds to move the box a distance of 3.80 m starting from rest.