Two workers are sliding 330 kg crate across the floor. One worker pushes forward on the crate with a force of 410 N while the other pulls in the same direction with a force of 240 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor?

Wc = M*g = 330 * 9.8 = 3234 N. = Normal force, Fn.

Fap = 410 + 240 = 650 N. = Force applied.

Fap-Fk = M*a.
Fap-Fk = M*0 = 0.
Fk = Fap = 650 N. = Force of kinetic friction.

u = Fk/Fn.

I need answers of find the magnitude of normal force acting on the skateboard on 10 kg àt point A and 10 m at point B

Well, well, well, looks like we have a friction problem here! Don't worry, friction can be quite the "slippery" subject, but I'm here to give you a "slide" of information.

To determine the coefficient of kinetic friction, we need to compare the force applied to the crate with the frictional force acting against it.

Now, since the crate is sliding with a constant speed, we know that the net force acting on it is zero. This means that the force of friction must be equal in magnitude and opposite in direction to the applied force.

The applied force consists of the pushing force (410 N) and the pulling force (240 N), giving us a total applied force of 410 N + 240 N = 650 N.

Since the net force is zero, the frictional force must also be 650 N.

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

Now, the normal force is the force exerted perpendicular to the surface by the crate. Since the crate is on a horizontal surface, the normal force is equal in magnitude and opposite in direction to the weight of the crate, which is given by F_weight = m * g, where m is the mass of the crate and g is the acceleration due to gravity.

With the mass of the crate being 330 kg and the acceleration due to gravity being approximately 9.8 m/s^2, we have F_weight = 330 kg * 9.8 m/s^2 = 3234 N.

Since the normal force is equal in magnitude and opposite in direction to the weight, the normal force is 3234 N.

Now, substituting the values into the equation F_friction = μ * F_normal, we have 650 N = μ * 3234 N.

Solving for μ, we find μ = 650 N / 3234 N ≈ 0.20.

So, the crate's coefficient of kinetic friction on the floor is 0.20. Keep in mind that this coefficient is specific to this particular crate and the floor it is on, so don't go around applying it to all your frictional situations!

I hope that explanation didn't rub you the wrong way. If you need any more assistance, just give me a shout!

To find the coefficient of kinetic friction, we need to use Newton's second law of motion:

ΣF = m * a

Where ΣF is the net force acting on the crate, m is the mass of the crate, and a is the acceleration of the crate. Since the crate is sliding with a constant speed, we know that the acceleration is zero (a = 0).

The net force is the sum of the pushing force and the pulling force, minus the force of kinetic friction:

ΣF = F_push - F_pull - F_friction

F_push = 410 N (pushing force)
F_pull = 240 N (pulling force)

Now, we need to determine the force of kinetic friction. The force of kinetic 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 acting on the crate. The normal force is equal to the weight of the crate, which is the product of the mass and the acceleration due to gravity:

N = m * g

g = 9.8 m/s² (acceleration due to gravity)

Substituting the values, we get:

F_friction = µ * m * g

Now, we can rewrite the net force equation:

ΣF = F_push - F_pull - µ * m * g

Since the crate slides with a constant speed, there is no acceleration, therefore:

ΣF = 0

Now, we can solve for the coefficient of kinetic friction (µ):

0 = 410 N - 240 N - µ * 330 kg * 9.8 m/s²

Simplifying:

0 = 170 N - 3234 µ

Rearranging the equation:

3234 µ = 170 N

µ = 170 N / 3234

µ ≈ 0.053

Therefore, the crate's coefficient of kinetic friction on the floor is approximately 0.053.