Question

Traveling at a speed of 58 km/h, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficent of kinetic friction between the tires and the road is 0.720. How far does the car skid before coming to a halt? Ignore the effects of air resistance.

Answer 1

Friction force is
F = μ m g
and the braking work is therefore
W = - F s = - μ m g s
with the minus sign because force and displacement go opposite.

But W = - 1/2 m v², so
μ g s = 1/2 v²
s = v² / 2μg = (58.0/3.6)² / (2 * 0.720 * 9.81) = 18.4 m

Answer 2

is that right or nah

Answer 3

its right. i cant understand the question mark

Answer 4

yes

Answer 5

Weight will be the same as Normal Contact force (N). So N=W=mg= 10m Newtons.
Frictional force (R)= $N, where $ is the mu or coefficient of friction, having R=0.72* 10m = 7.2m Newtons.
Then we recoil the 2nd Law of Newton's formula, F-R=ma, but since there is no driving force, then your formula simplifies to -R=ma, or -7.2m=ma, so a is -7.2m/s^2. Convert toKm/h^2
Now you use the formula v^2=u^2-2as, where v is final speed (0m/s), u is initial speed and s is distance, making s subject of the formula s= u^2/2a