Answer:
The final kinetic energy is four times of initial kinetic energy when speed of the car doubles.
[tex]K_{2}[/tex] = 4 [tex]K_{1}[/tex]
Explanation:
Mass of the car = 770 kg
initial speed [tex]V_{1}[/tex] = 23.1 [tex]\frac{m}{s}[/tex]
Final speed [tex]V_{2}[/tex] = 46.2 [tex]\frac{m}{s}[/tex]
Initial kinetic energy [tex]K_{1}[/tex] = [tex]\frac{1}{2} m V_{1} ^{2}[/tex]
⇒ [tex]K_{1}[/tex] = 0.5 × 770 × [tex]23.1^{2}[/tex]
⇒ [tex]K_{1}[/tex] = 205440 J
Final kinetic energy [tex]K_{2}[/tex] = [tex]\frac{1}{2} m V_{2} ^{2}[/tex]
⇒ [tex]K_{2}[/tex] = 0.5× 770 × [tex]46.2^{2}[/tex]
⇒ [tex]K_{2}[/tex] = 821760 J
Thus [tex]K_{2}[/tex] = 4 [tex]K_{1}[/tex]
Therefore the final kinetic energy is four times of initial kinetic energy when speed of the car doubles.
