Q.54 A large box of mass M聽is moving... [FREE SOLUTION]

Chapter 6: Q.54 (page 157)

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 $渭_{s}$ and $渭_{1}$ , respectively. Find an expression for the shortest distance $d_{min}$ in which the large box can stop without the small box slipping.

Short Answer

Expert verified

The expression for the minimum distance $d_{min}$ for which the large box can stop without the slipping of a small box $d_{min} = \frac{{v_{0}}^{2}}{2 (9.8) 渭_{渭}}$ .

Step by step solution

Given.

Speed of the large box: $v_{0}$

Coefficient of static friction: $渭_{s}$

Coefficient of static friction: $渭_{k}$

Formula used.

$\begin{aligned} F = 渭 n \\ v^{2} = u^{2} + 2 a s \end{aligned}$

Calculation.

There are 3 forces acting on the small box

Normal force, $n = m g$
Force due to gravity.

Friction force $F_{f}$

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

$\begin{aligned} F_{f} = m a \\ = m g 渭_{s 鈰�} . \end{aligned}$ ( $1$ )

According to the equation of kinematics, $v^{2} = v_{0}^{2} + 2 a d$

The final velocity of the large box is zero because the box stops. So the equation become ${v_{0}}^{2} = 2 a d$ .

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

We get

$\begin{aligned} v_{0}^{2} = 2 a d \\ d = \frac{v_{0}^{2}}{2 (9.8) 渭_{x}} \end{aligned}$

The minimum distance is $d_{m i n} = \frac{v_{0}^{2}}{2 (9.8) 渭_{x}}$ .

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Step by step solution

Given.

Formula used.

Calculation.

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