A box of books weighing 229 N is shoved across the floor by a force of 420 N exerted downward at an angle of 35° below the horizontal.If µk between the box and the floor is 0.57, how long does it take to move the box 9 m, starting from rest? (If the box will not move, enter 0.)

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

First, let's break down the forces acting on the box:

1. The force of gravity (weight):
The weight of the box is given as 229 N, directed vertically downward.

2. The applied force:
The force exerted downward at an angle of 35° below the horizontal is 420 N. We can break it down into its horizontal and vertical components.

The horizontal component of the force (Fa) can be found using trigonometry:
Fa = 420 N * cos(35°)

The vertical component (Fv) can also be found using trigonometry:
Fv = 420 N * sin(35°)

3. The force of friction:
The force of friction (Ff) is opposing the motion and can be calculated using the equation:
Ff = µk * Fn

Where:
- µk is the coefficient of kinetic friction (given as 0.57)
- Fn is the normal force, which is equal to the weight of the box in this case (229 N)

Now, let's calculate the force of friction and check if the box will move:
Ff = µk * Fn
= 0.57 * 229 N

If the force of friction (Ff) is greater than the horizontal component of the applied force (Fa), the box will not move, and the time taken will be 0. Therefore, we need to check if Ff ≤ Fa.

If Ff ≤ Fa, the box will start moving. We can then calculate the acceleration (a) using Newton's second law:

Fnet = ma

Since the applied force (Fa) is greater than the force of friction (Ff), the net force (Fnet) can be calculated as:

Fnet = Fa - Ff

Using the known relationship Fnet = ma and rearranging the equation, we can calculate the acceleration (a):

a = Fnet / m

The mass (m) is equal to the weight (given as 229 N) divided by the acceleration due to gravity (approximated as 9.8 m/s^2).

Now that we know the acceleration, we can use one of the kinematic equations to solve for time (t) when starting from rest:

d = vit + (1/2)at^2

Where:
- d is the distance traveled (given as 9 m)
- vi is the initial velocity (0 m/s since it starts from rest)
- a is the acceleration (calculated)
- t is the time we need to find.

By rearranging the equation, we can solve for time (t):

t = sqrt((2 * d) / a)

Putting all the values together, we can now calculate the time it takes to move the box 9 m, starting from rest.