Answer:
600 kJ
Solution:
As per the question:
Kinetic energy of the car, [tex]E'_{k}[/tex] = 200 kJ
[tex]\eta = 0.43[/tex]
[tex]E_{k} = \frac{1}{2}mv^{2}[/tex] (1)
Now,
The speed of the car reduces to half of its initial speed
If the initial speed is 'v' m/s, then the [tex]v' = \frac{v}{2}[/tex]
Also,
KE varies in proportion to the square of the velocity of the car.
Thus
[tex]E_{k} = \frac{1}{2}mv'^{2}[/tex]
[tex]E_{k} = \frac{1}{2}m(\frac{v}{2})^{2} = \frac{1}{2}m(\frac{v^{2}}{4}[/tex] (2)
Thus from eqn (1) and (2):
[tex]E_{k} = \frac{1}{4}E'_{k}[/tex]
[tex]E_{k} = 4E'_{k} = 4\times 200 = 800 kJ[/tex]
Therefore,
[tex]\Delta E = E'_{k} - E_{k} = 800 - 200 = 600 kJ[/tex]
