Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity [tex]V_{i}[/tex] = 60 mph = 26.8224 m/s
Final velocity [tex]V_{f}[/tex] = 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk = [tex]\frac{1}{2}[/tex]m( [tex]V_{i}[/tex]² - [tex]V_{f}[/tex]² )
we substitute
Δk = [tex]\frac{1}{2}[/tex]×1814.37( (26.8224)² - (13.4112)² )
Δk = [tex]\frac{1}{2}[/tex] × 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
