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*μk*= 0.54. How fast was the car traveling if it leaves 60-m-long skid marks?

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Friction force that is applied to the object slows down the application of force on the object consequently stops the motion in the direction of the object. The force of friction is determined by the friction coefficient between the object’s surface and the surface as well as how much reaction force is normal.

GivenMass of the car is m=1500 kg

Coefficient of kinetic friction is μk=0.54

Length of the skid marks is l=60 m

Now for the weight of the car :

ω=mg

ω= 1500 × 9.81

ω= 14715 N

Normal reaction force applied to the car is :

N=14715 N Now for the friction force;

f = μN

f = 0.54 × (14715)

f = -7946.1 N

Now for the acceleration of the car:

a=f/m

a=(−/7946).1500

a = -5.29 m/s

^{2}Now for the velocity:

v

^{2}= u^{2 }+ 2as0 = u

^{2}+ 2× -5.29× 600 = u

^{2}– 634.8u = 25.19 m/s

Thus, the velocity of the car will be equal to 25.19 m/sThe key concept to solving the problem is the work energy theorem.

Initially, the frictional force equals the net force that causes retardation. Acceleration can be obtained by comparing the two force equations. The third law of motion calculates the initial velocity using this value.

A coefficient of friction describes the relationship between friction force between two objects and normal reaction between them.

It can be defined as the ratio between friction and the normal force.

[katex]\mu = \frac{F}{N}[/katex]

Here [katex]\mu[/katex]

Newton’s second law is that the acceleration of an object caused by a net force directly proportional the magnitude of that force in the same direction. It is also inversely proportional the object’s mass.

[katex]F = ma[/katex]

Here [katex]F[/katex]

The equation for velocity distance is given below.

[katex]{v^2} – {u^2} = 2as[/katex]

Here [katex]v[/katex]

Normal reaction is equal to car weight.

[katex]N = mg[/katex]

mrepresents the mass,_gthe gravity, andNthe normal force.The friction force is created by

[katex]F = \mu N[/katex]

Here F is the force that causes friction. [katex]\mu [/katex]

Substitute [katex]mg[/katex]

[katex]\mu[/katex]0

It can be used to calculate Newton’s second law equation.

[katex]\mu[/katex]1

The opposite forces create a negative sign. If the vehicle is slowing down, then acceleration will be negative.

Substitute [katex]\mu[/katex]2

[katex]\mu[/katex]3

Substitute [katex]\mu[/katex]4

[katex]\mu[/katex]5Ans:

The vehicle was traveling at [katex]\mu[/katex]6 speed