Chapter 5.2, Problem 22PP

To determine

The coefficient of kinetic friction between the two given surfaces.

Answer to Problem 22PP

The coefficient of kinetic friction between the given surfaces is ${\mu }_{\text{k}}=0.1275$.

Explanation of Solution

Given:

A block of mass 1.4-kg slides along a rough surface and it slows down with a acceleration $-1.25{\text{ m/s}}^{\text{2}}$ .

Formula Used:

The distribution of forces on the block in shown in Figure 1.

  

Figure 1

Here, the block is moving freely with a deceleration and hence the net force acting on it is the kinetic frictional force. This can be expressed as

  $F_{net}=-F_f$ …… (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, netforce can be expressed as

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

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

  $\begin{array}{c}{\mu }_{\text{k}}=\frac{-ma}{mg}\\ =-\frac{a}{g}\end{array}$ …… (5)

Calculation:

From the equation for coefficient of kinetic friction, it can be observed that under free sliding, the coefficient of friction is independent of mass of the object and it depends on the acceleration values.

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

  $\begin{array}{c}{\mu }_{\text{k}}=-\frac{-1.25}{9.8}\\ =0.1275\end{array}$

Conclusion:

Thus, the coefficient of kinetic friction between the given surfaces is 0.1275.