Q.A car of mass $1000\ \text{kg}$, moving at $72\ \text{km/h}$, is brought to rest entirely by friction over a distance of $40\ \text{m}$ on a level road. Find the coefficient of kinetic friction between the tyres and the road, and the total heat generated in bringing the car to rest (take $g = 9.8\ \text{m/s}^2$).

Given: mass $m=1000\ \text{kg}$, speed $u=72\ \text{km/h} = 72\times\dfrac{1000}{3600} = 20\ \text{m/s}$, stopping distance $d=40\ \text{m}$, $g=9.8\ \text{m/s}^2$, friction alone brings the car to rest.

Initial kinetic energy: $K_i = \frac{1}{2}mu^2 = \frac{1}{2}(1000)(20)^2 = \frac{1}{2}(1000)(400) = 2\times10^5\ \text{J}$

Coefficient of kinetic friction. Since friction alone does all the (negative) work that brings the car to rest, the work done against friction, $\mu mg d$, must equal this entire initial kinetic energy: $\mu mgd = K_i \quad\Longrightarrow\quad \mu = \frac{K_i}{mgd} = \frac{2\times10^5}{1000\times9.8\times40} = \frac{2\times10^5}{392{,}000} \approx 0.51$ …

Convert the speed to m/s, find the initial KE, set it equal to the work done against friction $\mu mgd$ to solve for $\mu$, and note that this same lost …

• Forgetting to convert $72\ \text{km/h}$ to $20\ \text{m/s}$ before substituting.
• Assuming some of the kinetic energy simply "disappears" rather than recognising it must reappear as heat, since …