A box of books wighing 339 N is shoved across the floor by a force of 506 N exerted downward at an angle of 33° below the horzontal. The acceleration of gravity is 9.81m/s^2. If µk between the box and the floor is 0.59, how long does it take to move the box 5.27 m, starting from rest? Answer in units of s.

break the pushing force into horiztonal and vertical components. The vertical force adds to weight to determine force normal,and friction. You have the horizontal force, the opposingfriction, determine acceleration. Thence distance is known, find time.

sry but im a blonde so not even half of that just made sense! :(

wat is the horizontal force??

To solve this problem, we'll need to break it down into several steps:

Step 1: Find the normal force (Fn)
The normal force (Fn) is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force counters the weight of the box, which is given as 339 N. Since the box is on a flat surface, the normal force is equal in magnitude but opposite in direction to the weight.

Fn = -339 N

Step 2: Resolve the applied force into horizontal and vertical components
The applied force of 506 N is exerted downward at an angle of 33° below the horizontal. To determine the horizontal and vertical components of this force, we use trigonometric functions.

F(horizontal) = F * cos(angle)
F(horizontal) = 506 N * cos(33°)

F(vertical) = F * sin(angle)
F(vertical) = 506 N * sin(33°)

F(horizontal) ≈ 423.91 N
F(vertical) ≈ -269.93 N (Note: Negative because it is acting in the opposite direction of the weight)

Step 3: Calculate the net force in the horizontal direction
The net force (Fnet) in the horizontal direction is the sum of the applied force's horizontal component and the force of friction (Ffriction).

Fnet = F(horizontal) + Ffriction

Step 4: Determine the force of friction (Ffriction)
The force of friction (Ffriction) can be calculated using the equation:

Ffriction = µk * Fn

Given that µk is 0.59 and we calculated the normal force (Fn) as -339 N, we can substitute these values in:

Ffriction = 0.59 * (-339 N)

Ffriction ≈ -199.01 N

(Note: The negative sign indicates that the force of friction is acting in the opposite direction of the applied force)

Step 5: Calculate the net force in the horizontal direction
Now we can substitute the values we calculated into the equation for net force:

Fnet = F(horizontal) + Ffriction
Fnet = (423.91 N) + (-199.01 N)
Fnet = 224.90 N

Step 6: Apply Newton's second law of motion to find acceleration
Newton's second law of motion states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:

Fnet = m * a

In this case, we're given the weight of the box, which is equal to the force of gravity acting on it: weight = m * g

339 N = m * 9.81 m/s^2

Solving for the mass (m):

m = 339 N / 9.81 m/s^2
m ≈ 34.57 kg

Now we can substitute the mass into the equation Fnet = m * a:

224.90 N = 34.57 kg * a

Solving for the acceleration (a):

a = 224.90 N / 34.57 kg
a ≈ 6.50 m/s^2

Step 7: Use kinematic equation to find the time (t)
We know the initial velocity (u) is 0 m/s, the displacement (s) is 5.27 m, and the acceleration (a) is approximately 6.50 m/s^2. We can use the kinematic equation:

s = u * t + (1/2) * a * t^2

Since the initial velocity (u) is 0 m/s, this equation simplifies to:

s = (1/2) * a * t^2

Rearranging the equation to solve for time (t):

t^2 = (2 * s) / a
t = sqrt((2 * 5.27 m) / 6.50 m/s^2)
t ≈ 1.08 s

Therefore, it takes approximately 1.08 seconds to move the box 5.27 m, starting from rest.