Acar of mass 1,200 kg is travelling along a straight horizontal road at a speed of 20 m/s when it brakes sharply then skids. friction brings the car to rest. if the coefficient of friction between the tires and road is 0.8, calculate: a. the deceleration ( -7.84 m/s2)

The deceleration of the car is

Explanation:

It is given that,

Mass of the car, m = 1200 kg

Initial speed of the car, u = 20 m/s

Friction brings the car to rest, final speed, v = 0

The coefficient of friction between the tires and road is 0.8

Let a is the deceleration of car due to force of friction. The frictional force is given by :

So, the deceleration of the car is . Hence, this is the required solution.

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.

the closer the planet is to the sun, the larger the gravity