Question

Please Help

Answer 1

Explanation:

The initial kinetic energy $KE_0$ is

$KE_0 = \frac{1}{2}mv_0^2 = \frac{1}{2}(400\:\text{kg})(10\:\text{m/s})^2$

$\:\:\:\:\:\;\:= 2×10^4\:\text{J} = 20\:\text{kJ}$

The final kinetic energy $KE$ is

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(400\:\text{kg})(30\:\text{m/s})^2$

$\:\:\:\:\:\:\:= 1.8×10^5\:\text{J} = 180\:\text{kJ}$

The work done W on the car is

$W = \Delta{KE} = KE - KE_0$

$\:\:\:\:\:\:\:= 180\:\text{kJ} - 20\:\text{kJ} = 1.6×10^5\:\text{J}$

The power expended P is

$P = \dfrac{W}{t} = \dfrac{1.6×10^5\:\text{J}}{15\:\text{s}} = 10667\:\text{Watts}$

$\:\:\:\:\:= 10.7\:\text{kW}$