Question

A large box of mass $M$is moving on a horizontal surface at speed $V$0 . A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are ${\mu }_s$and ${\mu }_1$, respectively. Find an expression for the shortest distance$d_{min}$ in which the large box can stop without the small box slipping.

Short answer

The expression for the minimum distance $d_{min}$for which the large box can stop without the slipping of a small box$d_{min}=\frac{{v_0}^2}{2\left(9.8\right){\mu }_{\mu }}$.

Step-by-step solution

Given.

Speed of the large box: $v_0$

Coefficient of static friction: ${\mu }_s$

Coefficient of static friction: ${\mu }_k$

Formula used.

$\begin{array}{r}F=\mu n\\ v^2=u^2+2as\end{array}$

Calculation.

There are 3 forces acting on the small box

1. Normal force, $n=mg$
2. Force due to gravity.

Friction force $F_f$

We have to find the forces in horizontal direction, so forces in $x$- direction would be

$\begin{array}{r}{F}_f=ma\\ =mg{\mu }_{s\cdots }.\end{array}$ ($1$)

According to the equation of kinematics, $v^2=v_0^2+2ad$

The final velocity of the large box is zero because the box stops. So the equation become ${v_0}^2=2ad$.

Now solving for $d$by substituting the value of $\text{'}a\text{'}$ from the equation ($1$).

We get

$\begin{array}{r}{v}_0^2=2ad\\ d=\frac{v_0^2}{2\left(9.8\right){\mu }_x}\end{array}$

The minimum distance is $d_{min}=\frac{v_0^2}{2\left(9.8\right){\mu }_x}$.