Friction in an Elevator You are riding in  an elevator on the way to the 18th floor of your dormitory. The elevator is accelerating upward with $a=1 m/s^2$. Beside you is the box containing your new computer; the box and its contents have a total mass of $25 kg$. While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is $\mu =0.4$, What magnitude of force must you apply ?

Given in the equation,

$\to$Acceleration of the elevator $=a=1 m/s^2$ (upward)

$\to$Coefficient of kinetic friction between the box and the elevator floor$={\mu }_R=0.4$

$\to$Mass of box $= m =25kg$

$\to$Let the force applied to the box horizontally be F as the box slides with constant speed 

$\Rightarrow$The acceleration of box is zero.

$\to$F.B.D of box

$\to$By equation of equilibrium for box, we get:-

$\Rightarrow$In vertical direction, $N_1=W^1$

Where $N_1=$Normal force acting on box

 $W^1=$apparent weight of the box

$$\begin{gathered} \Rightarrow W^1=m(a+g) \\ \Rightarrow W^1=25(1+10) \\ \Rightarrow W^1=275 N \\ \Rightarrow N_1=275 N \end{gathered}$$

$\Rightarrow$In horizontal direction, $F=f$

Where $f =$Frictional force acting on box

$$\begin{gathered} \Rightarrow f={\mu }_k{N}_1=0.4\times 275=110 N \\ \Rightarrow F=110 N \end{gathered}$$