Final answer:
The force needed to keep the conveyor belt moving at a constant velocity, given the specifics of the grains and belt speed, is 20 N and the minimum power of the motor driving the conveyor belt is 40 W.
Step-by-step explanation:
The question involves a physics concept related to forces, work, and power. Specifically, we are examining a scenario where grains fall onto a conveyor belt, and we must determine the force needed to maintain the belt's velocity and the power of the motor driving the conveyor belt. To find the force necessary to maintain constant velocity, consider that, due to Newton's third law, the momentum of the grains falling onto the conveyor belt will exert an equal and opposite force on the conveyor, necessitating a force from the motor to keep the belt moving at a constant speed. Now, if we consider the grains falling at a rate of 10 kg/s and the belt moving horizontally at 2 m/s, we can determine this force using the formula F = Δp/Δt, which relates force to the change in momentum over time. The minimum power needed by the motor can be computed with the formula P = Fv, where P is the power, F is the force, and v is the velocity of the belt.
In this scenario, the force must equal the rate of momentum transfer, so F = (10 kg/s)(2 m/s) = 20 N, which matches option (a). The minimum power for the motor is given by P = (20 N)(2 m/s) = 40 W, aligning with option (b). Therefore, the force required to maintain constant belt velocity is 20 N, and the minimum power for the motor driving the conveyor belt is 40 W.