Question

in what distance can a 1500kg automobile be stopped if the brake is applied when the speed is 20m/s and the coefficient of sliding friction is 0.7 between the tyres and the ground?please I need a full step by step explanation

Answer 1

1) The braking force is provided by the frictional force, which is given by:

$F_f=\mu m g$

where

$\mu=0.7$ is the coefficient of friction

m=1500 kg is the mass of the car

$g=9.81 m/s^2$ is the gravitational acceleration

Substituting numbers into the equation, we find

$F_f= (0.7)(1500 kg)(9.81 m/s^2)=10301 N$

2) The work done by the frictional force to stop the car is equal to the product between the force and the distance d:

$W=-F_fd$ (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:

$\Delta K=K_f -K_i =W$ (2)

The final kinetic energy is zero, so the variation of kinetic energy is just equal to the initial kinetic energy of the car:

$\Delta K=-K_i=-\frac{1}{2}mv^2=-\frac{1}{2}(1500 kg)(20 m/s)^2=-300000 J$

4) By equalizing eq. (1) and (2), we find the distance, d:

$-K_i = -Fd$

$d=\frac{K_i}{F}=\frac{300000 J}{10301 N}=29.1 m$