A worker pulls a 200-N packing crate at constant velocity across a rough floor by exerting a force F = 55.0 N at an angle of 35.0 degrees above the horizontal. What is the coefficient of kinetic friction of the floor?

Wc = m * g = 200 N.

m * 9.8 = 200
m = 20.41 kg. = Mass of crate.

F = 55N.[35o.]

55*cos35 - Fk = m*a
45.05 - Fk = m*0 = 0
Fk = 45.05 = Force of kinetic friction.

u = Fk/Wc = 45.05/200 = 0.225

Well, well, well, looks like we have ourselves a physics problem! Let's clown around with it and find the coefficient of kinetic friction.

First things first, let's break down the forces at play here. We have the weight of the crate, which is 200 N, pulling it downwards. Then we have the force applied by the worker, which we'll call F, pulling it at an angle of 35.0 degrees above the horizontal.

Since the crate is moving at a constant velocity, that means that the force of friction must be equal in magnitude and opposite in direction to the force F applied by the worker. If only it was as opposed to exercise as the crate is to motion!

Now, let's find the horizontal component of the force F. This can be found by multiplying the magnitude of the force by the cosine of the angle. So, F horizontal = F * cos(35.0).

Since the force of friction is equal in magnitude and opposite in direction to the horizontal component of F, we can say that the force of friction is also equal to F horizontal. Newton would be proud!

Now, let's use this force of friction to find the coefficient of kinetic friction. We know that the force of friction is equal to the coefficient of kinetic friction multiplied by the normal force, which in this case is just the weight of the crate. So, force of friction = coefficient of kinetic friction * weight.

Substituting in our values, we get F horizontal = coefficient of kinetic friction * weight. Let's rearrange that to find our answer.

Coefficient of kinetic friction = F horizontal / weight.

Plugging in our given values, we get:

Coefficient of kinetic friction = (55.0 N * cos(35.0)) / 200 N.

Now it's just a matter of crunching the numbers to find our solution. I'll leave that part to you, my friend! Go on and calculate that coefficient of kinetic friction, and remember to have some fun along the way!

To find the coefficient of kinetic friction, we will use the formula:

\(F_{friction} = \mu_k \cdot F_{normal}\)

Where:
\(F_{friction}\) is the force of friction,
\(\mu_k\) is the coefficient of kinetic friction, and
\(F_{normal}\) is the normal force.

Given that the worker pulls the crate at a constant velocity, we know that the applied force is equal to the force of friction.

The force exerted by the worker can be resolved into its horizontal and vertical components as follows:

\(F_{x} = F \cdot \cos(\theta)\)
\(F_{y} = F \cdot \sin(\theta)\)

Given:
\(F = 55.0 \, \text{N}\)
\(\theta = 35.0 \, \text{degrees}\)

Using trigonometric functions, we can find the horizontal and vertical components of the applied force:

\(F_{x} = 55.0 \, \text{N} \cdot \cos(35.0^\circ)\)
\(F_{y} = 55.0 \, \text{N} \cdot \sin(35.0^\circ)\)

\(F_{x} = 55.0 \, \text{N} \cdot 0.819 \)
\(F_{y} = 55.0 \, \text{N} \cdot 0.574 \)

\(F_{x} \approx 45.0 \, \text{N} \)
\(F_{y} \approx 31.6 \, \text{N} \)

The normal force \(F_{normal}\) is equal to the weight of the crate, which is its mass multiplied by the acceleration due to gravity:

\(F_{normal} = m \cdot g\)

Given that the weight of the crate is equal to its mass times the acceleration due to gravity, and that the mass of the crate is:

\(m = \dfrac{F_{crate}}{g}\)

Given:
\(F_{crate} = 200 \, \text{N} \)
\(g \approx 9.8 \, \text{m/s}^2 \)

\(m = \dfrac{200 \, \text{N}}{9.8 \, \text{m/s}^2}\)
\(m \approx 20.41 \, \text{kg}\)

\(F_{normal} = 20.41 \, \text{kg} \cdot 9.8 \, \text{m/s}^2\)
\(F_{normal} \approx 200.0 \, \text{N}\)

Now that we have the applied horizontal force (\(F_x\)) and the normal force (\(F_{normal}\)), we can find the force of friction (\(F_{friction}\)) using the formula mentioned earlier:

\(F_{friction} = \mu_k \cdot F_{normal}\)

Since the crate is pulled at constant velocity, the force of friction is equal to the applied horizontal force:

\(F_{friction} = F_x\)

Therefore:

\(F_x = \mu_k \cdot F_{normal}\)

Solving for \(\mu_k\):

\(\mu_k = \dfrac{F_x}{F_{normal}}\)

Plugging in the values:

\(\mu_k = \dfrac{45.0 \, \text{N}}{200.0 \, \text{N}}\)

\(\mu_k \approx 0.225\) (rounded to three significant figures)

Therefore, the coefficient of kinetic friction of the floor is approximately \(0.225\).

To find the coefficient of kinetic friction, we first need to understand the forces acting on the packing crate.

The worker exerts a force F = 55.0 N at an angle of 35.0 degrees above the horizontal. This force can be split into two components:

1. The vertical component (F_vertical) of the force is F_vertical = F * sin(θ), where θ is the angle of 35.0 degrees.
2. The horizontal component (F_horizontal) of the force is F_horizontal = F * cos(θ).

Since the crate is pulled at a constant velocity, we know that the net force acting on the crate is zero. This means that the force of friction opposing the crate's motion must be equal in magnitude and opposite in direction to the horizontal component of the worker's force.

The force of friction (F_friction) can be calculated using the equation F_friction = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.

The normal force (N) is the force exerted by the floor on the crate and is equal in magnitude and opposite in direction to the vertical component of the worker's force (F_vertical).

Since the crate is at constant velocity, the normal force is equal to the weight of the crate, which is the force due to gravity acting on it. In this case, the weight of the crate is given as 200 N.

So, N = 200 N.

Now, we can equate the magnitudes of the horizontal component of the worker's force (F_horizontal) and the force of friction (F_friction).

F_horizontal = F_friction

F_horizontal = μ * N

F_horizonatal = μ * 200 N

Now, we can substitute the values and calculate the coefficient of kinetic friction (μ).

55.0 N * cos(35.0 degrees) = μ * 200 N

μ = (55.0 N * cos(35.0 degrees)) / 200 N