# Determine the energy required to accelerate a 1300-kg car from

Determine the energy required to accelerate a 1300-kg car from 10 to 60 km/h on an uphill road with a vertical rise of 40 m.
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peterpan7117i
Givens:
$m=1300kg$
${V}_{1}=10k\frac{m}{h}$
${V}_{2}=60k\frac{m}{h}V$
$\mathrm{\Delta }z=40m$
$g=9.81\frac{m}{{s}^{2}}$
Solution:
Note : $1k\frac{m}{h}=\frac{1000m}{60×60s}=\frac{5}{18};\frac{m}{s}$
The required energy is the sum of the change in potential energy and the change in kinetic energy.
$\mathrm{\Delta }E=mg\mathrm{\Delta }z+m\frac{\mathrm{\Delta }{V}^{2}}{2}=mg\mathrm{\Delta }z+m\frac{{V}_{2}^{2}-{V}_{1}^{2}}{2}=1300\left[9.81×40+\frac{{60}^{2}-{10}^{2}}{2}×\frac{{5}^{2}}{{18}^{2}}\right]$
$=685660.1235J=685.66kJ$