91影视

Force on the particle, $F_x=F_o{e}^{鈭�t/T}$.

Velocity at time $t=0$.

Position at time$v=0$.

Force on a object is given by the equation:

$F=ma$

Here,

$m$is the mass.

$a$is the acceleration.

Force on the particle is given as:

$F_x=F_o{e}^{鈭�t/T}鈥�鈥�\text{(1)}$

Force on an object is given by the equation

$F=ma鈥�..\left(2\right)$

On comparing Eq. $\left\{1\right\}$and Eq.{$2$}

role="math" localid="1647718262854" $\begin{array}{r}ma=F_o{e}^{鈭�t/T}\\ a=\frac{F_s{e}^{鈭�t/T}}{m}\end{array}$

The velocity of a particle at a time $t$is obtain by integrating the above equation,

role="math" localid="1647718224750" $\begin{array}{r}鈭�adt=鈭�\frac{F_o{e}^{鈭�t/T}}{m}dt\\ 鈭�\frac{dv}{d}dt=\frac{F_0}{w_0}\left[鈭�T{e}^{鈭�t/T}\right]鈥�鈥�\text{(3)}\end{array}$

$v=-\frac{F_{0}T}{m}\left[e^{-t/T}\right]+C$

Here,

$C$is the arbitrary constant of integration

Apply the boundary condition that at time $t=0$the velocity of the particle is $0$.

Plugging the values in the above equation

$\begin{array}{rl}& v=鈭�\frac{F_{0}T}{m}\left[e^{鈭�t/T}\right]+C\\ & 0=鈭�\frac{F_{0}T}{m}\left(e^0\right)+C\\ & C=\frac{F_{0}T}{m}\end{array}$

Substitute the value of $C$in Eq. ($3$

$\begin{array}{rl}v=& 鈭�\frac{F_{0}T}{m}\left[e^{鈭�t/T}\right]+\frac{F_{0}T}{m}\\ & =\frac{F_{0}T}{m}[1鈭�e^{鈭�t/T}]\end{array}$

The expression of particle velocity at any time $t$is$v\left(t\right)=\frac{F_{0}T}{m}[1鈭�e^{鈭�t/T}]$.

Expression of particle velocity at any time$t$ is:

$v\left(t\right)=\frac{F_{0}T}{m}[1鈭�e^{鈭�t/T}]$

The velocity of a particle is an exponent function of time, therefore as the time elapsed the magnitude of exponent term will be $0$.

And the velocity of the particle is given as:

$\begin{array}{c}v\left(t\right)=\frac{F_{0}T}{m}[1鈭�e^{鈭�鈭�/T}]\\ =\frac{F_{0}T}{m}[1鈭�0]\\ =\frac{F_{0}T}{m}\end{array}$

Part (a): Expression of particle velocity at any time$t$is $v\left(t\right)=\frac{F_{0}T}{m}[1鈭�e^{鈭�t/T}]$.

Part (b): Objects velocity after a very long time is$\frac{F_{0}T}{m}$.