## College Physics (4th Edition)

The correct answer is: D. $1.5\times 10^5~\times sin~10^{\circ}$
The extra power would be required to increase the gravitational potential energy of the car. We can find the increase in gravitational potential energy each second: $\Delta U_g = mgh$ $\Delta U_g = mgd~sin~\theta$ $\Delta U_g = (1000~kg)(10~m/s^2)(15~m)~sin~10^{\circ}$ $\Delta U_g = 1.5\times 10^5~sin~10^{\circ}~J$ Since this amount of additional energy is required each second, the extra power that is required is $1.5\times 10^5~sin~10^{\circ}~W$ The correct answer is: D. $1.5\times 10^5~\times sin~10^{\circ}$