Question

4. . When you push a 1.80-kg book resting on a tabletop, it takes
2.25 N to start the book sliding. Once it is sliding, however, it
takes only 1.50 N to keep the book moving with constant speed.
What are the coefficients of static and kinetic friction between
the book and the tabletop?

Answer 1

Answer:

The coefficient of static friction is $\mu_s =0.1276$

The coefficient of kinetic friction is $\mu_k = 0.085$

Explanation:

From the question we are told that

   The mass of the book is  m =  1.80 kg

    The force required by the book to start sliding is $F_s = 2.25 \ N$

    The force required to keep the booking moving at constant speed is $F_c = 1.50 \ N$

Generally the static  frictional force is mathematically represented as  

           $F_f = \mu_s * m * g$

This static frictional  force is equivalent to the force required for the book to start sliding

So

          $\mu_s * m * g = 2.25$

=>       $\mu_s * 1.80 * 9.8 = 2.25$

=>      $\mu_s =0.1276$

Generally the kinetic  frictional force is mathematically represented as  

           $F_F = \mu_k * m * g$

This static frictional  force is equivalent to the force required for the book to  slide at constant velocity

So

         $F_c = F_F$

=>       $1.50 = \mu_k * m * g$

=>       $1.50 = \mu_k *1.80 * 9.8$

=>       $\mu_k = 0.085$

Answer 2

The coefficients of static and kinetic friction between  the book and the tabletop are 0.127 and 0.085.

Given data:

The mass of book is, m = 1.80 kg.

The magnitude of force required to start sliding is, f = 2.25 N.

The magnitude of force during the motion is, f' = 1.50 N.

The force of friction comes to play when the object is on the verge of motion is known as static force of friction, then the expression is,

$f = \mu_{s} \times m\times g$

here, $\mu_{s}$  is the coefficient of static friction.

Solving as,

$2.25 = \mu_{s} \times 1.80\times 9.8\\\\\mu_{s} = \dfrac{2.25}{1.80 \times 9.8}\\\\\mu_{s} = 0.127$

And, the force of friction when the object is already in the motion is known as kinetic force of friction. And its expression is,

$f' = \mu_{k} \times m\times g\\\\\mu_{k}=\dfrac{f'}{m\times g}\\\\\mu_{k}=\dfrac{1.50}{1.80 \times 9.8}\\\\\mu_{k}=0.085$

Thus, we can conclude that the coefficients of static and kinetic friction between  the book and the tabletop are 0.127 and 0.085.

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