Question

24. Review. A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a. The student gives her backpack a quick kick at t = 0 , imparting to it speed v and causing it to slide across the elevator floor. At time t, the backpack hits the opposite wall a distance L away from the student. Find the coefficient of kinetic friction ${\mathit{\mu }}_{\mathbf{k}}$ between the backpack and the elevator floor.

Short answer

The coefficient of the kinetic friction is ${\mu }_k=\frac{2(vt-L)}{(a+g)t^2}$.

Step-by-step solution

Concept and explanation

The elevator's width is equal to = L .

If a student is standing in an elevator with an upward acceleration of a, the normal force exerted on the student is,

$\begin{array}{c}n=mg+ma\\ =m(a+g)\end{array}$

Let ${\mu }_k$the kinetic friction coefficient between the backpack and the elevator floor be at t = 0 , the student kicks her rucksack, giving it a speed of v and causing it to slide over the elevator floor.

The backpack collides with the opposing wall at time t. That is, the distance her backpack has travelled is L.

Determining the kinetic coefficient of friction

Consider the forces at work on the backpack as it moves through the Earth's frame of reference

$\begin{array}{c}\Sigma F_y=m{a}_y\\ n-mg=ma\\ n=ma+mg\\ =m(a+g)\end{array}$

Then, frictional force data-custom-editor="chemistry" $f_k={\mu }_{k}n$

$\begin{array}{c}\Sigma F_x={\mu }_{k}m(a+g)\\ =m{a}_x\\ -{\mu }_{k}m(a+g)=m{a}_x\\ a_x=-{\mu }_k(a+g)\end{array}$

As a result, the equations below describe the motion across the floor.

$\begin{array}{c}S=vt+\frac{1}{2}{a}_x{t}^2\\ L=vt-\frac{1}{2}{\mu }_k(a+g)t^2\\ \frac{1}{2}{\mu }_k(a+g)t^2=vt-L\\ {\mu }_k(a+g)t^2=2(vt-L)\\ {\mu }_k=\frac{2(vt-L)}{(a+g)t^2}\\ \end{array}$

Thus, the coefficient of the kinetic friction is ${\mu }_k=\frac{2(vt-L)}{(a+g)t^2}$.