Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 4, Problem 43P
To determine

The coefficient of kinetic friction between the drag-race tires in contact with an asphalt surface.

Expert Solution & Answer
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Answer to Problem 43P

Solution:

The coefficient of kinetic friction between the drag-race tires and asphalt was found to be 1.4.

Explanation of Solution

During a drag-race, the vehicle exerts a very strong force on the ground. This in turn exerts a very large acceleration on the vehicle. The force of kinetic friction between the tires and the surface provides the acceleration for the vehicle.

Given:

The distance covered by the drag-racer s=1 km=1000 m

The time taken to cover the distance t=12 s

The initial speed of the drag-racer is v0=0

Formula used:

Frictional force can be obtained by:

  Ffr=μkFN

According to Newton’s second law of motion:

  Ffr=ma

Second equation of motion is:

  s=v0t+12at2

Here, the force of kinetic friction is Ffr , acceleration due to gravity is g , the mass of the car is m , its acceleration is a , initial speed is v0 , its final speed is v and the time taken to travel this distance is t .

The weight W of the car acts vertically downwards and due to Newton’s third law, the car experiences an upward force called normal force FN .Since, there is no motion either in the upward and the downward directions,

  FN+W=0

Therefore,

  FN=W

The normal force has the same magnitude as that of the weight and it acts opposite to the direction of the weight.

As the magnitude of the normal force is FN=mg .

The force of kinetic friction is related to the normal force as,

  Ffr=μkFN

Therefore Ffr=μkmg   ......(1)

This force provides accelerationa on the vehicle.

  Ffr=ma   .......(2)

From equations (1) and (2),

  μkmg=ma

The acceleration experienced by the car is given by,

  a=μkg   ......(3)

Calculation:

The acceleration of the vehicle can be calculated using the equation of motion,

  s=v0t+12at2

Since, the vehicle starts from rest, v0=0 .

Therefore,

  s=12at2a=2st2   ......(4)

From equations (3) and (4),

  a=2st2=μkgμk=2sgt2

Using the given values of s , g and t in the equation

  μk=2sgt2=2( 1000 m)( 9.8  m/s 2 ) ( 12 s )2=1.417

Conclusion:

The drag-racer would cover a distance of 1 km in 12 s starting from rest if the coefficient of kinetic friction between the tires and the asphalt surface is 1.4.

Chapter 4 Solutions

Physics: Principles with Applications

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