Final answer:
The angle of the incline down which a box can move at a constant velocity, given a coefficient of kinetic friction uk = 0.58, is most nearly calculated by taking the inverse tangent of uk. The result is approximately 30 degrees.
Step-by-step explanation:
To find the angle of the incline down which a box can move at a constant velocity given a coefficient of kinetic friction (uk), we can equate the force due to gravity along the incline to the force of friction. The component of the weight down the slope is mg sin θ, where θ is the angle of the incline and m is the mass of the object. The frictional force opposing this motion is μ_k mg cos θ, where μ_k is the coefficient of kinetic friction. For the object to slide at constant velocity, these two forces must be equal, giving us the equation μ_k mg cos θ = mg sin θ. By dividing both sides of the equation by mg cos θ, we can solve for the angle θ:
μ_k = tan θ
Substituting the given coefficient of kinetic friction, uk = 0.58, we get:
0.58 = tan θ
To find the angle θ in degrees, we use the inverse tangent function:
θ = atan(0.58)
By calculating the inverse tangent of 0.58, we can determine that the angle of the incline, in degrees, is most nearly 30 degrees.