## College Physics (4th Edition)

The engine must supply an average mechanical power of $60,000~Watts$ during this time interval.
We can find the change in the car's kinetic energy: $\Delta KE = \frac{1}{2}mv_2^2-\frac{1}{2}mv_1^2$ $\Delta KE = \frac{1}{2}m~(v_2^2-v_1^2)$ $\Delta KE = \frac{1}{2}(1200~kg)~[(30.0~m/s)^2-(20.0~m/s)^2]$ $\Delta KE = 3.00\times 10^5~J$ The engine does this amount of work in 5.0 seconds. We can find the average power output: $Power = \frac{Energy}{Time} = \frac{3.00\times 10^5~J}{5.0~s} = 60,000~Watts$ The engine must supply an average mechanical power of $60,000~Watts$ during this time interval.