(I) How much work must be done to stop a 925-kg car traveling at 95 km/h?

The mass of the car is\(m = 925\;{\rm{kg}}\).

The initial speed of the car is\({v_{\rm{o}}} = 95\;{\rm{km/h}} = 26.39\;{\rm{m/s}}\).

The final speed of the car is \(v = 0\).

Work done is equal to the change in the kinetic energy of an object.It is negative when the object slows down because the work is done against the motion of the object.

The initial kinetic energy of the car is \(K = \frac{1}{2}mv_{\rm{o}}^2\) .

The final kinetic energy of the car is zero as the final speed of the car is zero.

Then, the change in kinetic energy is:

\(\begin{array}{c}\Delta K = 0 - K\\ = - K\end{array}\)

Thus, the work done is:

\(\begin{array}{c}W = - K\\ = - \frac{1}{2}mv_{\rm{o}}^2\\ = - \frac{1}{2} \times \left( {925\;{\rm{kg}}} \right) \times {\left( {26.39\;{\rm{m/s}}} \right)^2}\\ = - 3.22 \times {10^5}\;{\rm{J}}\end{array}\)

Hence, the work done on the car is \( - 3.22 \times {10^5}\;{\rm{J}}\).