A row of six 1 N wooden blocks is being pushed across a tabletop at a constant speed by a tow tractor that exerts a force of 1.3 N on the row. What is the coeffiecient of kinetic friction between the wooden blocks and the tabletop?
To calculate the coefficient of kinetic friction, we need to use the formula:
μk = Fk / N
Where:
μk is the coefficient of kinetic friction,
Fk is the force of kinetic friction, and
N is the normal force.
In this case, we are given the force applied by the tractor, which is 1.3 N. However, we need to find the force of kinetic friction and the normal force.
The force of kinetic friction can be obtained by subtracting the applied force from the net force:
Fk = Fa - F
Where:
Fa is the applied force, and
F is the net force.
Since the wooden blocks are being pushed at a constant speed, we know that the net force is zero because friction is equal to the applied force. Therefore:
Fk = Fa - F = 0
Now we can calculate the normal force. The normal force is the force exerted by a surface perpendicular to the object. In this case, it is equal to the weight of the object since it is on a horizontal surface.
The weight (W) can be calculated by multiplying the mass (m) of the object by the acceleration due to gravity (g):
W = m * g
Since we only have the force applied by the tractor of 1.3 N, we can assume that the normal force is equal to this value:
N = 1.3 N
Now we have all the values we need to calculate the coefficient of kinetic friction:
μk = Fk / N
μk = 0 / 1.3 N
μk = 0
Therefore, the coefficient of kinetic friction between the wooden blocks and the tabletop is 0.
To find the coefficient of kinetic friction between the wooden blocks and the tabletop, we can use the equation:
\(F_f = \mu_k \times F_N\)
Where:
\(F_f\) is the force of friction
\(\mu_k\) is the coefficient of kinetic friction
\(F_N\) is the normal force
In this case, the force pulling the row of blocks is equal to the force of friction:
\(F_f = 1.3 \, \text{N}\)
The normal force is the weight of the blocks, which can be calculated as the mass of one block multiplied by the acceleration due to gravity:
\(F_N = m \times g\)
Given that the weight of one block is 1 N and there are 6 blocks in the row, the total weight is:
\(F_N = 6 \, \text{N}\)
Substituting these values into the equation, we can solve for the coefficient of kinetic friction:
\(1.3 \, \text{N} = \mu_k \times 6 \, \text{N}\)
Simplifying the equation:
\(\mu_k = \frac{1.3}{6} = 0.2167\) (rounded to four decimal places)
Therefore, the coefficient of kinetic friction between the wooden blocks and the tabletop is approximately 0.2167.