## College Physics (4th Edition)

(a) The average power that must be delivered by the motor is $2620~Watts$ (b) The average power that must be delivered by the motor is $7850~Watts$
(a) Let $m_1 = 1202~kg$ and let $m_2 = 801~kg$. We can find the energy $E$ that the motor must provide: $E = m_1gh - m_2gh$ $E = (m_1 - m_2)~gh$ $E = (1202~kg - 801~kg)~(9.80~m/s^2)(40.0~m)$ $E = 1.57\times 10^5~J$ We can find the average power output: $Power = \frac{Energy}{Time} = \frac{1.57\times 10^5~J}{60~s} = 2620~Watts$ The average power that must be delivered by the motor is $2620~Watts$ (b) We can find the energy $E$ that the motor must provide: $E = m_1gh$ $E = (1202~kg)(9.80~m/s^2)(40.0~m)$ $E = 4.71\times 10^5~J$ We can find the average power output: $Power = \frac{Energy}{Time} = \frac{4.71\times 10^5~J}{60~s} = 7850~Watts$ The average power that must be delivered by the motor is $7850~Watts$.