A1100 kg car moves along a horizontal road at speed v0 = 21.7 m/s. the road is wet, so the static friction coefficient between the tires and the road is only µs = 0.144 and the kinetic friction coefficient is even lower, µk = 0.1008. the acceleration of gravity is 9.8 m/s 2 . what is the shortest possible stopping distance for the car under such conditions? use g = 9.8 m/s 2 and neglect the reaction time of the driver. answer in units of m.

the shortest possible distance to stop is 238m

Explanation:

The car is moving at a constant velocity, so it's not accelerating, on the other side we have a friction force, this force is decelerating the car, so let's analyze it:

in this case, we need to use the kinetic friction coefficient because the car is moving:

The car is decelerating at

to obtain the distance we can use:

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.