Chapter 5.2, Problem 23PP

To determine

The coefficient of kinetic friction between the two given surfaces.

Answer to Problem 23PP

The coefficient of kinetic friction between the given surfaces is ${\mu }_{\text{k}}=0.1495$.

Explanation of Solution

Given:

A bookcase of mass 41-kg is to be moved from one point to another and a force of 65 N is applied in the horizontal direction causing an acceleration 0.12 m/s² .

Formula Used:

Here, a force is applied horizontally on the bookcase to move it along the carpet . The entire distribution of forces on the bookcase in shown in Figure 1.

  

Figure 1

There will be a force opposing the applied force which is due to kinetic friction in this case since the body is in motion. Here, the body is in acceleration and hence there will be a net force in horizontal direction given by,

  $F_{net}=F_A-F_f$ …… (1)

The net force is given by

  $F_{net}=ma$

Now, the kinetic friction force will be proportional to the normal force acting in the body. In this case, the normal force is equal to the weight of the body as there are no other forces along the normal.

  $F_{\text{N}}=mg$ …… (2)

Thus, the kinetic friction force can be expressed as,

  $F_f={\mu }_{\text{k}}{F}_{\text{N}}$ …… (3)

Here ${\mu }_{\text{k}}$ represents the coefficient of kinetic friction.

Based on this, the magnitude of horizontal force to be applied can be deduced as,

  $F_A=ma+{\mu }_{\text{k}}\left(mg\right)$ …… (4)

And, the coefficient of kinetic friction can be calculated as,

  ${\mu }_{\text{k}}=\frac{F_A-ma}{mg}$ …… (5)

Calculation:

Given the mass of bookcase as 41-kg, and horizontal force as 65 N.

Now, by substituting the acceleration due to gravity and given parameters, the coefficient of kinetic friction can be calculated based on (5) as

  $\begin{array}{c}{\mu }_{\text{k}}=\frac{65-41\times 0.12}{41\times 9.8}\\ =0.1495\end{array}$

Conclusion:

Thus, the coefficient of kinetic friction between the given surfaces is 0.1495.