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. A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s.
. A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s. (a) What is the average power of the elevator motor during this time interval?
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Final answer:To find theaverage powerof the elevator motor during its acceleration phase, use the formula P = F * v. When the elevator is at its cruising speed, the motor power required to maintain this speed is zero.Explanation:To find the average power of theelevator motor, we can use the formula P = F * v, where P is power, F is force, and v is velocity. The force can be calculated using Newton's second law, F = m * a, where m is mass and a is acceleration. Since the elevator starts from rest and reaches its cruising speed, we know that the final velocity is 1.75 m/s and the time interval is 3.00 s. Using these values, we can calculate the acceleration and then the force. Finally, we can find the average power by multiplying the force by the velocity.When the elevator is moving at its cruising speed of 1.75 m/s, it has reached a state ofconstant velocity, meaning there is no net external force acting on it. Therefore, the motor power required to maintain this speed would be zero.Learn more about power of elevator motor here:brainly.com/question/3107377#SPJ11...