Final answer:
The suitcase is dragged 1.125 meters before it is riding smoothly on the conveyor belt.
Step-by-step explanation:
To determine how far the suitcase is dragged before it is riding smoothly on the conveyor belt, we need to consider the frictional forces acting on the suitcase. The initial force between the suitcase and the conveyor belt is the force of gravity, which is equal to the weight of the suitcase, given by the equation Fg = m * g, where m is the mass of the suitcase (10 kg) and g is the acceleration due to gravity (9.8 m/s²).
The maximum static friction force can be found using the equation Ff = µs * N, where µs is the coefficient of static friction (0.50) and N is the normal force. In this case, the normal force is equal to the weight of the suitcase, so N = Fg. If the force required to overcome static friction is greater than the force of gravity, the suitcase will start moving.
Once the suitcase is moving, the force of friction becomes the kinetic friction force, which can be found using the equation Ff = µk * N, where µk is the coefficient of kinetic friction (0.20) and N is the normal force. The normal force is still equal to the weight of the suitcase, so N = Fg. The distance the suitcase is dragged before it is riding smoothly on the belt can be calculated using the equation d = (v² - u²) / (2 * a), where v is the final velocity (equal to the velocity of the conveyor belt, 1.5 m/s), u is the initial velocity (equal to 0 m/s), and a is the acceleration (equal to the acceleration due to friction).
Using the given values, we can solve for the distance: d = (1.5² - 0²) / (2 * 0.20 * 10) = 1.125 m. Therefore, the suitcase is dragged 1.125 meters before it is riding smoothly on the conveyor belt.