Question

A block of mass $m$is at rest at the origin at $t=0$. It is pushed with constant force $F_0$from $x=0$to $x=L$across a horizontal surface whose coefficient of kinetic friction is ${\mu }_{\mathrm{k}}={\mu }_0(1-x/L)$. That is, the coefficient of friction decreases from ${\mu }_0$at $x=0$to zero at $x=L$.

a. Use what you've learned in calculus to prove that

$a_{\mathrm{r}}=v_{\mathrm{r}}\frac{d{v}_x}{dx}$

b. Find an expression for the block's speed as it reaches position $L$.

Short answer

Part (a) It is proved that $a_x=v_x\frac{d{v}_x}{dx}.$

Part (b) The speed at $x=L$is $v_L={\left\{2[\frac{FL}{m}-\frac{gl{\mu }_0}{2}]\right\}}^{1/2}.$.

Step-by-step solution

Part (a) Step 1: Given.

$\begin{array}{rl}& {\mu }_k={\mu }_0(1-\frac{x}{L})\\ & \text{At}x=0,v=0.\end{array}$

Part (a): Step 2: Explanation of solution.

The acceleration $a_x$is given by,

$\begin{array}{c}{a}_x=\frac{d^{2}x}{d{t}^2}\\ =\frac{d}{dt}\left(\frac{dx}{dt}\right)\\ =\frac{d}{dx}\left(\frac{dx}{dt}\right)\frac{dx}{dt}\end{array}$

Now, substitute $\frac{dx}{dt}=v_x$in the above expression.

$\begin{array}{c}{a}_x=\frac{d}{dx}\left(v_x\right)v_x\\ =v_x\frac{d{v}_x}{dx}\end{array}$

It is proved that $a_x=v_x\frac{d{v}_x}{dx}$.

Part (b) Step 3: Given.

The frictional co-efficient of friction between the surface is${\mu }_k={\mu }_0(1-\frac{x}{L})$.

Part (b) Step 4: Calculation.

Weight of the block is$w=mg,g$is the acceleration due to gravity. Hence, the normal force on the block is$n=w=mg$. So, the frictional force is $f_k={\mu }_{k}n={\mu }_0(1-\frac{x}{L})mg$. The applied force on the block is $F$.

Hence, the total force on the block is$f=F-{\mu }_0(1-\frac{x}{L})mg$.

So, acceleration is given by$a_x=\frac{f}{m}=\frac{F}{m}-{\mu }_0(1-\frac{x}{L})g$.

Use equation (1) to write

$v_{x}d{v}_x=a_{x}dx$

Now substitute the value of $a_x$

$v_{x}d{v}_x=\frac{F}{m}-{\mu }_0(1-\frac{x}{L})gdx$

Integrate the above equation.

$\begin{array}{rl}& {\int }_{v_0}^{v_L}{v}_{x}d{v}_x={\int }_0^L\frac{F}{m}-{\mu }_0(1-\frac{x}{L})gdx\\ & {\left[\frac{v_x^2}{2}\right]}_{v_0}^{v_L}={[\frac{Fx}{m}-{\mu }_0(x-\frac{x^2}{2L})g]}_0^L\end{array}$

Now, put the value $v_0=0$, in the above equation to get:$\frac{v_L^2}{2}=[\frac{FL}{m}-\frac{g{L}_{{\mu }_0}}{2}]$.

Hence, the speed is $v_L={\left\{2[\frac{FL}{m}-\frac{gL{\mu }_0}{2}]\right\}}^{1/2}$.

Conclusion.

Part (a): It is proved that $a_x=v_x\frac{d{v}_x}{dx}.$

Part (b): The speed at$X=L$is$v_L={\left\{2\left[\frac{FL}{m}-\frac{gl{\mu }_0}{2}\right]\right\}}^{1/2}$.