A 95kg box initially at rest on a horizonal floor requires a 650N horizontal force to set it in motion. After the box is in motion, a horizontal force of 560N keeps it moving at a constant velocity. Determine uS and uK the clock and the floor.
uS = 650/(M*g)
uK = 560/(M*g)
M is the mass in kg and g is the acceleration of gravity in m/s^2.
Do the calculations
10.82
To solve this problem, we can use the concept of friction and the equations of motion. Let's break down the steps to find the coefficients of static friction (uS) and kinetic friction (uK).
Step 1: Find the coefficient of static friction (uS)
We know that the applied force required to set the box in motion is 650N, and the weight of the box is 95kg * 9.8m/s^2 = 931N. The maximum static friction force (fs_max) is given by:
fs_max = uS * N
where N is the normal force, which is equal to the weight of the box.
fs_max = uS * 931N
Since the applied force is equal to the maximum static friction force, we have:
650N = uS * 931N
Solving for uS:
uS = 650N / 931N
Step 2: Find the coefficient of kinetic friction (uK)
Once the box is in motion, the horizontal force required to keep it moving at a constant velocity is 560N. The kinetic friction force (fk) is given by:
fk = uK * N
Again, N is the normal force.
fk = uK * 931N
Since the applied force is equal to the kinetic friction force, we have:
560N = uK * 931N
Solving for uK:
uK = 560N / 931N
Step 3: Calculate the values of uS and uK
Now we can calculate the values of uS and uK.
uS = 650N / 931N ā 0.698
uK = 560N / 931N ā 0.601
Therefore, the coefficient of static friction (uS) is approximately 0.698, and the coefficient of kinetic friction (uK) is approximately 0.601.
To determine the coefficient of static friction (uS) and the coefficient of kinetic friction (uK) between the box and the floor, we can use the given information about the forces required to set the box in motion and to keep it moving at a constant velocity.
1. Finding uS (coefficient of static friction):
The coefficient of static friction represents the friction force between two surfaces when they are not slipping relative to each other.
The force required to set the box in motion is given as 650N. Since the box is initially at rest, this is the maximum static friction force (uS * normal force). Therefore, we have:
650N = uS * normal force
The normal force is equal to the weight of the box, which is m * g (mass * acceleration due to gravity). So, we have:
650N = uS * (95kg * 9.8 m/s^2)
Solving for uS:
uS = 650N / (95kg * 9.8 m/s^2)
2. Finding uK (coefficient of kinetic friction):
The coefficient of kinetic friction represents the friction force between two surfaces when they are sliding relative to each other.
The force required to keep the box moving at a constant velocity is given as 560N. This force is equal to the kinetic friction force (uK * normal force). Therefore, we have:
560N = uK * normal force
Again, the normal force is equal to the weight of the box. So, we have:
560N = uK * (95kg * 9.8 m/s^2)
Solving for uK:
uK = 560N / (95kg * 9.8 m/s^2)
Calculating these values, uS is approximately 0.696 and uK is approximately 0.606.
So, the coefficient of static friction (uS) between the box and the floor is 0.696, and the coefficient of kinetic friction (uK) is 0.606.