Q69P - University Physics with Modern Physics

Question

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficient of friction there is not constant; it starts at 0.100 at P and increases linearly with distance past P, reaching a value of 0.600 at 12.5 m past point  . (a) Use the work-energy theorem to find how far this box slides before stopping. (b) What is the coefficient of friction at the stopping point? (c) How far would the box have slid if the friction coefficient didn’t increase but instead had the constant value of 0.100?

Step-by-Step Solution

Answer

(a) The box will slide for 5.11 m before stopping.

(b) The coefficient of friction at the stopping point is 0.304. 

(c) The box will slide a distance of 10.3 m before stopping.

Step 1: Definition of friction and coefficient of friction

The resistance or opposition an object experiences due to the motion on a surface is called friction.

 

The coefficient of friction is equal to the ratio of the normal force and friction force.

Step 2: Given data

The initial speed of the box is, ${\mathrm{v}}_1=4.50\hspace{0.33em}\mathrm{m}/\mathrm{s}$

The final speed of the box is, ${\mathrm{v}}_2=0\hspace{0.33em}\mathrm{m}/\mathrm{s}$ 

Let  be the distance past P increase the coefficient of kinetic friction in a linear

fashion is,

${\mathrm{\mu }}_{\mathrm{k}}=0.100+\mathrm{kx}$ …………..(1)

Here, k is a positive constant.

Step 3: (a) Find how far this box slides before stopping

when, x=12.5 m and ${\mathrm{u}}_{\mathrm{k}}=0.600$ 

Substituting the given data in equation (1), we get,

$$\begin{gathered} 0.600=0.100+k\times (12.5\hspace{0.33em}m) \\ k=\frac{0.500}{12.5 m} \\ k=\frac{0.040}{m} \end{gathered}$$ 

 

Now, using the work-energy theorem, the work done by the frictional force is,

 

$$\begin{gathered} {\mathrm{w}}_{\mathrm{f}}={\mathrm{k}}_2-{\mathrm{k}}_1 \\ -\int {\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgdx}=0-\frac{1}{2}{\mathrm{mv}}_1^2 \end{gathered}$$  

 

The distance, of the box slide, is determined by integrating the above equation from 0 to $x_2$, then we get 

$$\begin{gathered} \mathrm{g}{\int }_0^{{\mathrm{x}}_2}\left(0.100+\mathrm{kx}\right)\mathrm{dx}=\frac{1}{2}{\mathrm{v}}_1^2 \\ \mathrm{g}\left[\left(0.100\right){\mathrm{x}}_2+\mathrm{k}\frac{{\mathrm{x}}_2^2}{2}\right]=\frac{1}{2}{\mathrm{v}}_1^2 \\ \left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\left[\left(0.100\right){\mathrm{x}}_2+\left(0.040/\mathrm{m}\right)\frac{{\mathrm{x}}_2^2}{2}\right]=\frac{1}{2}{\left(4.50\mathrm{m}/\mathrm{s}\right)}^2 \\ 0.196{{\mathrm{x}}_2}^2+0.98{\mathrm{x}}_2-10.125=0 \end{gathered}$$ 

 

 This is the form of a quadratic equation, on solving it, we get ${\mathrm{x}}_2=5.11\hspace{0.33em}\mathrm{m}$ 

 

Hence, the box is a slide 5.11 m before stopping. 

Step 4: (b) Find the coefficient of friction at the stopping point

The coefficient of friction at the stopping point is,

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{k}}=0.100+\mathrm{Ax} \\ =0.100+\left(0.040/\mathrm{m}\right)\left(5.11\mathrm{m}\right) \\ =0.304 \end{gathered}$$  

Hence, the coefficient of friction at the stopping point is 0.304 . 

Step 5: (c) Find how far would the box have slid

If the coefficient of friction is, ${\mu }_k=0.100$ , then from the work-energy theorem, we get the distance the box slide with a constant coefficient of friction

$$\begin{gathered} {\mathrm{w}}_{\mathrm{f}}={\mathrm{k}}_2-{\mathrm{k}}_1 \\ -{\mathrm{\mu }}_{\mathrm{k}}\mathrm{mgx}\text{'}=0-\frac{1}{2}{\mathrm{mv}}_1^2 \\ \mathrm{x}\text{'}=\frac{{\left(4.50\mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.100\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)} \\ \mathrm{x}\text{'}=10.3 \mathrm{m} \end{gathered}$$ 

 

 

Hence, the box will slide a distance of 10.3 m before stopping.