Chapter 5.2, Problem 18PP

To determine

The coefficient of kinetic friction when a force is applied on an object with a given weight is moved on a surface with a constant speed.

Answer to Problem 18PP

  ${\mu }_{\text{K}}=0.692$

Explanation of Solution

Given data:

A horizontal force of 36 N is applied on a sled of weight 52-N to pull it across a cement sidewalk at a constant speed.

Formula Used:

Here, a force is applied horizontally on the sled to move it along the cement sidewalk with a constant speed. There will be a force opposing the applied force which is due to kinetic friction in this case since the body is in motion. According to Newton’s law, the frictional force $F_{f,kinetic}$ and the applied force $F_A$ are equal and they oppose each other.

This can be expressed as

  $F_A=-F_{f,kinetic}$   ...... (1)

Now, the kinetic friction force will be proportional to the normal force acting in the body. In this case, the normal force is equivalent 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

  $F_{f,kinetic}={\mu }_{\text{k}}{F}_{\text{N}}$   ...... (3)

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

Calculation:

The coefficient of kinetic friction can be deduced from (3) as

  ${\mu }_{\text{k}}=\frac{F_{f,kinetic}}{F_{\text{N}}}$

Given the weight of the sled is 52 N. Thus based on (2), the normal force is

  $F_{\text{N}}=52\text{N}$

Now, the applied force is $F_A=36\text{N}$ . Thus, based on (1), the magnitude of the kinetic friction force is

  $F_{f,kinetic}=36\text{N}$ .

Substituting for the kinetic force of friction and the normal force, the coefficient of kinetic friction is obtained as

  $\begin{array}{c}{\mu }_{\text{k}}=\frac{36}{52}\\ =0.692\end{array}$

Conclusion:

The coefficient of kinetic friction between the given surfaces is $0.692$ .