1) The braking force is provided by the frictional force, which is given by:
[tex] F_f=\mu m g [/tex]
where
[tex] \mu=0.7 [/tex] is the coefficient of friction
m=1500 kg is the mass of the car
[tex] g=9.81 m/s^2 [/tex] is the gravitational acceleration
Substituting numbers into the equation, we find
[tex] F_f= (0.7)(1500 kg)(9.81 m/s^2)=10301 N [/tex]
2) The work done by the frictional force to stop the car is equal to the product between the force and the distance d:
[tex] W=-F_fd [/tex] (1)
where we put a negative sign because the force is in the opposite direction of the motion of the car.
3) For the work-energy theorem, the work done by the frictional force is equal to the variation of kinetic energy of the car:
[tex] \Delta K=K_f -K_i =W [/tex] (2)
The final kinetic energy is zero, so the variation of kinetic energy is just equal to the initial kinetic energy of the car:
[tex] \Delta K=-K_i=-\frac{1}{2}mv^2=-\frac{1}{2}(1500 kg)(20 m/s)^2=-300000 J [/tex]
4) By equalizing eq. (1) and (2), we find the distance, d:
[tex] -K_i = -Fd [/tex]
[tex] d=\frac{K_i}{F}=\frac{300000 J}{10301 N}=29.1 m [/tex]
