To determine the coefficient of kinetic friction between the surface and the box, we can use the given information about the forces acting on the rope.
The force exerted on the rope (Frope) is equal to the sum of the force of tension in the rope (Ftension) and the force of friction (Ffriction). In this case, the box is being pulled at a constant speed, so the net force on the box is zero. Thus, the force of tension in the rope is equal in magnitude and opposite in direction to the force of kinetic friction:
Ftension = Ffriction
The force exerted on the rope is given as 260 N, which is equal to the force of tension (Ftension). Therefore, we have:
Ftension = Ffriction = 260 N
The weight of the box (W) is given as 450 N.
Now, let's analyze the forces acting on the box:
- The vertical component of the weight (W) is balanced by the normal force (N) from the floor, as the box is on a level floor:
N = W = 450 N
- The horizontal component of the weight (W) is counteracted by the force of friction (Ffriction) and the force of tension (Ftension):
Ffriction + Ftension = W * sin(30°)
Now, let's substitute the known values and solve for the coefficient of kinetic friction (μ):
260 N + 260 N = 450 N * sin(30°)
520 N = 450 N * 0.5
520 N = 225 N
To solve for the coefficient of kinetic friction (μ), we need the ratio of the force of friction (Ffriction) to the normal force (N):
μ = Ffriction / N
Substituting the values:
μ = 225 N / 450 N
μ = 0.5
Therefore, the coefficient of kinetic friction between the surface and the box is 0.5.