91影视

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

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 $a_x=v_x\frac{d{v}_x}{dx}$.

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

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 $f_k={渭}_{k}n={渭}_0(1鈭�\frac{x}{L})mg$. The applied force on the block is $F$.

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

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

Use equation (1) to write

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

Now substitute the value of $a_x$

$v_{x}d{v}_x=\frac{F}{m}鈭�{渭}_0(1鈭�\frac{x}{L})gdx$

Integrate the above equation.

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

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

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

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

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