Question

A baseball player with mass$\mathit{m}\mathbf{}\mathbf{=}\mathbf{79}\mathbf{}\mathit{k}\mathit{g}$, sliding into second base, is retarded by a frictional force of magnitude$\mathbf{470}\mathbf{}\mathit{N}$.What is the coefficient of kinetic friction${\mu }_k$between the player and the ground?

Short answer

The coefficient of kinetic friction between the player and the ground is $0.61$.

Step-by-step solution

Write the given information:

Mass,$m=79\text{kg}$

Frictional force,$f_s=470\text{N}$

Determine the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.

Consider the formula for the second law of motion:

$F_{net}=\sum ma$

Here, F is the net force, mis mass and ais an acceleration

Determine thefree body diagram

Free body diagram of basketball player:

Determine the coefficient of kinetic friction

By using Newton’s 2nd law along y direction,

$\sum _{}^{}{F}_y=m{a}_y$

Since player is not moving along y.$a_y=0$

$$\begin{gathered} N-mg=0 \\ N=mg \end{gathered}$$

Relation between frictional force and Normal force is as follows:

$f_s={\mu }_{k}N$

Rewrite the formulas, substitute the values and solve as:

$\begin{array}{rcl}{\mu }_k& =& \frac{f_s}{N}\\ & =& \frac{f_s}{mg}\\ & =& \frac{470}{\left(79\right)\left(9.81\right)}\\ & =& 0.61\\ & & \end{array}$

Hence, the coefficient of kinetic friction between the player and the ground is 0.61.