Question

A warehouse worker exerts a constant horizontal force of magnitude 85 Non a 40 kgbox that is initially at rest on the horizontal floor of the warehouse. When the box has moved a distance of 1.4 m, its speed is 1.0 m/s. What is the coefficient of kinetic friction between the box and the floor?

Short answer

$\mu =0.18$

Step-by-step solution

Given                      

$$\begin{gathered} force=85Nonmass=40kg \\ {v}_i=0m/sand{v}_f=1.0m/s \\ x=1.4m \end{gathered}$$

Understanding the concept

The problem deals with the frictional force. Frictional force is given by the product of coefficient of friction and the normal reaction. It also involves the kinematic equations of motion in which the motion of an object is described with constant acceleration.

Formula:

${\mathbf{v}}^{\mathbf{2}}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{a}}_{\mathbf{x}}\mathbf{∆}\mathbf{x}$

Calculate the coefficient of kinetic friction between the box and the floor

$F-f_k=m{a}_x,F_N-mg=0$

using $v^2=V_0^2+2a_x∆\mathrm{x}$we find the acceleration be

$$\begin{gathered} a_x=\frac{v^2-v_0^2}{2∆x} \\ =\frac{{\left(1m/s\right)}^2-0}{2\left(1.4m\right)} \\ =0.357m/s^2 \end{gathered}$$

The above equations can be combined to give

$$\begin{gathered} f_k={\mu }_k{F}_N \\ {\mu }_k=\frac{f_k}{F_N} \\ =\frac{F-m{a}_x}{mg} \\ =\frac{85N-\left(40kg\right)\left(0.357m/s^2\right)}{\left(40kg\right)\left(9.8m/s^2\right)} \\ =0.18 \end{gathered}$$

Thus,${\mu }_k=0.18$