A car of mass m = 1400 kg moves at a constant v_1 = 40 km/h…

A car of mass $m = 1400\ \mathrm{kg}$ moves at a constant $v_1 = 40\ \mathrm{km/h}$. By how much must the engine's force increase for the car to double its speed over a distance $s = 280\ \mathrm{m}$? Neglect friction.

Explanation of Method

Extra engine force to double a car's speed (repeat).

Step 1: Work–energy theorem

The extra force $\Delta F$ does the work that increases the kinetic energy from $\tfrac12 mv_1^2$ to $\tfrac12 m(2v_1)^2$: $\Delta F\,s = \tfrac12 m\,(4v_1^2 - v_1^2) = \tfrac32 m v_1^2$.

\[ \Delta F = \frac{3mv_1^2}{2s} = \frac{3 \cdot 1400 \cdot 11.11^2}{560}. \]

\[\Delta F \approx 926\ \mathrm{N}\]

$v^2\ (\mathrm{m^2/s^2})$ $s\ (\mathrm{m})$ $0$ $100$ $200$ $280$ $123$ $494$ $\text{slope } \frac{2\,\Delta F}{m}$ $(2v_1)^2$ $\Delta F = \frac{3mv_1^2}{2s} \approx 926\ \mathrm{N}$

Square of the speed against distance. A constant extra force $\Delta F$ makes $v^2$ grow linearly; it has to quadruple $v^2$ (double $v$) within 280 m.

Result

\[\boxed{\,\Delta F \approx 926\ \mathrm{N}\,}\]

Remark

Note. This problem is the same as Translational Dynamics 126; the original result (≈ 924–926 N) is correct.