Problem — Kinetic energy of a car at 80 km/h

Kinetic energy of a car. Calculate the kinetic energy of a $1200\;\mathrm{kg}$ car travelling at $80\;\mathrm{km/h}$. What fall height corresponds to the same energy?

Solution

Data: $m = 1200\;\mathrm{kg}$, $v = 80\;\mathrm{km/h}$. Convert the speed to SI units:

$v = \frac{80}{3{,}6} = 22{,}2\;\mathrm{m/s}$

Kinetic energy:

$E_k = \tfrac{1}{2}mv^2 = \tfrac{1}{2}(1200)(22{,}2)^2 = 600\cdot 493{,}8 \approx 2{,}96\cdot 10^5\;\mathrm{J}$

$\ev{E_k \approx 2{,}96\cdot 10^5\;\mathrm{J}}$

Equivalent height. Find the height $h$ from which the car, falling freely, would gain the same energy: $mgh = E_k$, i.e. $h = E_k/(mg) = v^2/(2g)$:

$h = \frac{v^2}{2g} = \frac{(22{,}2)^2}{2\cdot 9{,}81} = \frac{493{,}8}{19{,}62} \approx 25{,}2\;\mathrm{m}$

$\ev{h \approx 25\;\mathrm{m}}$

Physical reading. Crashing at $80\;\mathrm{km/h}$ is equivalent to falling from about $25\;\mathrm{m}$, i.e. from an 8-9 storey building. The conversion makes tangible how much energy is involved in a road impact.

Collegamenti

Argomenti: Work and energy Concetti: Kinetic energy · Gravitational potential energy Competenze: Symbolic set-up