As the car is moving, it also gains height.

Therefore, it has both kinetic and potential energy due to height.

At the top, the car then moves at a constant speed by exerting some force creating energy Fd.

The total energy the car spends is given by:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{m}}{{\mathbf{v}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{m}}{\mathbf{g}}{\mathbf{h}}{\mathbf{+}}{\mathbf{F}}{\mathbf{d}}}$

At a constant speed, after climbing the hill, the car covers a distance given d = vt

d = (25.0)(1 × 3600) = 90000m

The energy released by the battery is E = qV

A battery-operated car utilizes a 12.0 V system. Find the charge the batteries must be able to move in order to accelerate the 750 kg car from rest to 25.0 m/s, make it climb a 2.00×10^{2} m high hill, and then cause it to travel at a constant 25.0 m/s by exerting a 5.00×10^{2} N force for an hour.

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