Two workers are sliding 310kg crate across the floor. One worker pushes forward on the crate with a force of 440N while the other pulls in the same direction with a force of 230N 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?
.221
M*g = 310kg * 9.8N/kg = 3,038 N. = Wt. of the crate = Normal force(Fn).
Fap-Fk = M*a
(440+230)-Fk = M*0 = 0
670 - Fk = 0
Fk = 670 N. = Force of kinetic friction
uk = Fk/Fn = 670/3038 = 0.221
To solve this problem, we need to consider the forces acting on the crate and use Newton's second law:
ΣF = ma
First, let's determine the net force acting on the crate. Since it is sliding at a constant speed, the net force is zero.
1. Calculate the net force:
Net force = 440N - 230N
Net force = 210N
Next, let's determine the force of friction acting on the crate. The force of friction can be calculated using the equation:
F_friction = µ * Fn
where µ is the coefficient of kinetic friction and Fn is the normal force acting on the crate.
2. Calculate the normal force:
The normal force is equal in magnitude but opposite in direction to the net force, as the crate is in equilibrium (constant speed).
Normal force = -210N
3. Calculate the force of friction:
F_friction = µ * Fn
F_friction = µ * (-210N)
Since F_friction is equal to the net force (zero) in this case, we can set them equal to each other and solve for µ:
F_friction = µ * (-210N)
0 = µ * (-210N)
Since the crate is sliding with a constant speed, we know that the frictional force is equal to the applied force. Therefore, the magnitude of the force of friction is equal to the magnitude of the applied force:
|F_friction| = |210N|
4. Determine the coefficient of kinetic friction:
µ = |F_friction| / |Fn|
µ = |210N| / |(-210N)|
µ = 1
Therefore, the crate's coefficient of kinetic friction on the floor is 1.
To find the crate's coefficient of kinetic friction on the floor, we need to use the concept of net force. When an object is moving at a constant speed, the net force acting on it must be zero.
In this case, we have two forces acting on the crate: the force applied by the worker pushing forward (440N) and the force applied by the worker pulling with the rope (230N).
Since the crate is moving at a constant speed, the force of friction between the crate and the floor must be equal in magnitude and opposite in direction to the sum of these two forces.
First, let's find the net force acting on the crate. To do this, we need to subtract the force of pulling (230N) from the force of pushing (440N):
Net force = Force of pushing - Force of pulling
Net force = 440N - 230N
Net force = 210N
Since the crate is moving with a constant speed, the net force acting on it must be zero. Therefore, we can conclude that the force of friction between the crate and the floor is 210N.
The force of friction can be calculated using the equation:
Force of friction = coefficient of kinetic friction * normal force
where the normal force is the vertical force pressing the crate against the floor.
In this case, since the crate is not moving vertically, the normal force is equal to the weight of the crate (mg), where m is the mass of the crate and g is the acceleration due to gravity.
Given that the mass of the crate is 310kg, the weight can be calculated as follows:
Weight = mass * gravitational acceleration
Weight = 310kg * 9.8m/s^2 (gravitational acceleration)
Weight = 3038N
Now we can substitute the values into the equation to find the coefficient of kinetic friction:
Force of friction = coefficient of kinetic friction * normal force
210N = coefficient of kinetic friction * 3038N
Solving for the coefficient of kinetic friction:
coefficient of kinetic friction = 210N / 3038N
coefficient of kinetic friction ≈ 0.069
Therefore, the crate's coefficient of kinetic friction on the floor is approximately 0.069.