Final answer:
A box of mass m is initially at rest at the top of an inclined plane, which has a height of 6.7 m and makes an angle of θ = 23° with respect to the horizontal, the distance the box travels after the end of the inclined plane is approximately 2.55 meters.
Step-by-step explanation:
To find the distance d that the box travels after the end of the inclined plane, we need to consider the forces acting on the box.
Initially, the box is at rest at the top of the inclined plane, so the gravitational force acting down the plane can be broken down into two components: mg sin(θ) down the plane and mg cos(θ) perpendicular to the plane.
The net force acting down the plane is given by Fnet = mg sin(θ) - μp mg cos(θ), where μp is the coefficient of kinetic friction between the box and the plane.
Since the box is moving at a constant velocity, the net force is zero, so mg sin(θ) - μp mg cos(θ) = 0.
Solving this equation for the distance d, we get d = (μp cos(θ) / sin(θ)) * h, where h is the height of the inclined plane.
Plugging in the values given, we have:
d = (0.1 * cos(23°) / sin(23°)) * 6.7.
d ≈ 2.55 m.
So therefore the distance the box travels after the end of the inclined plane is approximately 2.55 meters.