Question

Review: A uniform board of length L is sliding along a smooth, frictionless, horizontal plane as shown in Figure P8.79a. The board then slides across the boundary with a rough horizontal surface. The coefficient of kinetic friction between the board and the second surface is. ${\mathit{\mu }}_{\mathbf{k}}$ (a) Find the acceleration of the board at the moment its front end has traveled a distance x beyond the boundary. (b) The board stops at the moment its back end reaches the boundary as shown in Figure P8.79b. Find the initial speed v of the board

Short answer

(a) The acceleration of the board at the moment its front end has traveled a distance x beyond the boundary is $a=\frac{{\mu }_{k}xg}{L}$.

(b) The initial speed v of the board just before crossing the boundary is $v=\sqrt{{\mu }_{k}gL}$.

Step-by-step solution

Introduction

The dissipative force acting between two bodies in contact, when they are in relative motion, is called friction. It is given as-

$f_k={\mu }_{k}N$

Where

$f_k$is the frictional force

${\mu }_k$is the coefficient of friction

$N$is the normal reaction force

Explanation for part (a)

If m is the mass of the whole board. Then, mass of the portion of board on the rough surface is- $\frac{mx}{L}$. So, weight of the portion of board on the rough surface is$\frac{mxg}{L}$ and as weight and normal reaction are equal in magnitude, so, the friction force is given as-

.$\begin{array}{c}{f}_k=\frac{{\mu }_{k}mxg}{L}\\ =ma\end{array}$

Then,

$a=\frac{{\mu }_{k}xg}{L}$

The acceleration is acting in the direction opposite to motion.

Explanation for part (b)

let the board covers a distance dx in forward direction. As the board moves forward, its kinetic energy gets converted into internal energy due to friction. The whole kinetic energy of the board is converted to internal energy by the time its back reaches the boundary. So the total work done during this time is

$\begin{array}{c}W=\underset{0}{\overset{L}{\int }}{f}_{x}dx\\ =\underset{0}{\overset{L}{\int }}\frac{{\mu }_{k}mxg}{L}dx\\ ={\left.\frac{{\mu }_{k}mg}{L}\frac{x^2}{2}\right|}_o^L\\ =\frac{{\mu }_{k}mLg}{2}\end{array}$

As total work done is equal to change in kinetic energy of the body.

$\begin{array}{c}W=\frac{1}{2}m{v}^2\\ \frac{{\mu }_{k}mLg}{2}=\frac{1}{2}m{v}^2\\ v=\sqrt{{\mu }_{k}gL}\end{array}$

This is the velocity that the board has just before crossing the boundary.