Answer:
[tex]-3.8\cdot 10^5 J[/tex]
Explanation:
According to the work-energy theorem, the work done on the car is equal to its variation of kinetic energy:
[tex]W=\Delta K=K_f -K_i = \frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2[/tex]
where
m = 1900 kg is the mass of the car
vf = 15 m/s is the final speed
vi = 25 m/s is the initial speed
Substituting the data, we find
[tex]W=\frac{1}{2}(1900 kg)(15 m/s)^2-\frac{1}{2}(1900 kg)(25 m/s)^2 =-3.8\cdot 10^5 J[/tex]
and the work is negative because it is done against the motion of the car (in fact, the car slows down, so it loses kinetic energy).
