Chapter 2, Problem 2.14P

(a)

To determine

The velocity of the mass with respect to time.

Answer to Problem 2.14P

The velocity of the mass with respect to time is $v\left(t\right)=-v\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)$.

Explanation of Solution

Given, the force on the mass $m$ is $F=-F_0{e}^{v\text{'}/v}$

Write the expression for $t$ from problem 2.7

    $t=m{\displaystyle \underset{v_0}{\overset{v}{\int }}\frac{dv\text{'}}{F\left(v\text{'}\right)}}$        (I)

Substituting $F=-F_0{e}^{v\text{'}/v}$ in equation (I) and solving

$\begin{array}{c}t=m{\displaystyle \underset{v_0}{\overset{v}{\int }}\frac{dv\text{'}}{-F_0{e}^{v\text{'}/v}}}\\ =\frac{m}{-F_0}{\displaystyle \underset{v_0}{\overset{v}{\int }}{e}^{-v\text{'}/v}dv\text{'}}\\ =\frac{m}{-F_0}{\left[\frac{e^{-v\text{'}/v}}{\frac{-1}{v}}\right]}_{v_0}^v\\ =\frac{m}{-F_0}{\left[\frac{e^{-v\text{'}/v}}{\frac{-1}{v}}\right]}_{v_0}^v\\ =\frac{mv}{F_0}{\left[e^{-v\text{'}/v}\right]}_{v_0}^v\end{array}$

$\begin{array}{c}t=\frac{mv}{F_0}\left[e^{-v/v}-e^{-v_0/v}\right]\\ \frac{F_{0}t}{mv}=e^{-v/v}-e^{-v_0/v}\\ e^{-v/v}=\frac{F_{0}t}{mv}+e^{-v_0/v}\end{array}$

Taking log on both sides

 $\begin{array}{c}\frac{-v}{v}=\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\\ v\left(t\right)=-v\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\end{array}$

Conclusion:

Thus, the velocity of the mass with respect to time is $v\left(t\right)=-v\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)$.

(b)

To determine

The time that takes to for the mass to come instantaneously to rest.

Answer to Problem 2.14P

The time that takes to for the mass to come instantaneously to rest is $t^{*}=\frac{mv}{F_0}\left(1-e^{-v_0/v}\right)$.

Explanation of Solution

From part (a), the velocity of the mass with respect to time is $v\left(t\right)=-v\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)$. When the mass comes to rest the velocity is zero, $v\left(t\right)=0$.

Therefore,

$\begin{array}{c}-v\ln \left(\frac{F_0{t}^{*}}{mv}+e^{-v_0/v}\right)=0\\ \ln \left(\frac{F_0{t}^{*}}{mv}+e^{-v_0/v}\right)=0\\ \frac{F_0{t}^{*}}{mv}+e^{-v_0/v}=e^0\\ \frac{F_0{t}^{*}}{mv}+e^{-v_0/v}=1\end{array}$

$\begin{array}{c}\frac{F_0{t}^{*}}{mv}=1-e^{-v_0/v}\\ t^{*}=\frac{mv}{F_0}\left(1-e^{-v_0/v}\right)\end{array}$

Conclusion:

Thus, the time that takes to for the mass to come instantaneously to rest is $t^{*}=\frac{mv}{F_0}\left(1-e^{-v_0/v}\right)$.

(c)

To determine

The distance the mass travels before coming instantaneously to rest.

Answer to Problem 2.14P

The distance the mass travels before coming instantaneously to rest is $x\left(t^{*}\right)=\frac{m{v}^2}{F_0}\left[1-e^{\frac{-v_0}{v}}\left(\frac{-v_0}{v}+1\right)\right]$.

Explanation of Solution

From part (a), the velocity of the mass with respect to time is $v\left(t\right)=-v\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)$. Since velocity is the rate of change of displacement by time, $v\left(t\right)=\frac{d}{dt}x\left(t\right)$.

Therefore, integrating the velocity will gives displacement of the mass.

$\begin{array}{c}x\left(t\right)={\displaystyle {\int }_0^{t}v\left(t\text{'}\right)dt\text{'}}\\ ={\displaystyle {\int }_0^t-v\ln \left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)}\end{array}$        (I)

For integration, consider $z=\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}$. Thus, $dz=\frac{F_0}{mv}dt\text{'}$  and $dt\text{'}=\frac{mv}{F_0}dz$.

Solving the integration part in the above equation,

$\begin{array}{c}{\displaystyle \int -v\ln \left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)}={\displaystyle \int \ln z\frac{mv}{F_0}dz}\\ =\frac{mv}{F_0}{\displaystyle \int \ln zdz}\end{array}$        (II)

Let $u=\ln z$ and $dv=dz$. Using the formula, ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$. Here, $du=\frac{1}{z}dz$ and $v=z$.

The integration part becomes,

$\begin{array}{c}{\displaystyle \int \ln zdz}=\left(\ln z\right)z-{\displaystyle \int z\left(\frac{1}{z}\right)dz}\\ =z\ln z-z\end{array}$

Substitute the above equation in equation (II)

$\begin{array}{c}{\displaystyle \int -v\ln \left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)}=\frac{mv}{F_0}\left(z\ln z-z\right)\\ =\frac{mv}{F_0}z\left(\ln z-1\right)\end{array}$

Substitute the value of $z$ in the above equation

${\displaystyle \int -v\ln \left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)}=\frac{mv}{F_0}\left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)\left[\ln \left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)-1\right]$

Substitute the above integral in equation (I) and solving

$\begin{array}{c}x\left(t\right)={\displaystyle {\int }_0^{t}v\left(t\text{'}\right)dt\text{'}}\\ =-v\frac{mv}{F_0}{\left[\left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)\left[\ln \left(\frac{F_{0}t\text{'}}{mv}+e^{-v_0/v}\right)-1\right]\right]}_0^t\\ =-\frac{m{v}^2}{F_0}\left[\left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\left[\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)-1\right]-\left(e^{-v_0/v}\right)\left[\ln \left(e^{-v_0/v}\right)-1\right]\right]\\ =-\frac{m{v}^2}{F_0}\left[\left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)-\frac{F_{0}t}{mv}-e^{-v_0/v}-\left(e^{-v_0/v}\right)\ln \left(e^{-v_0/v}\right)+e^{-v_0/v}\right]\end{array}$

$\begin{array}{c}=-\frac{m{v}^2}{F_0}\left[\left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)-\frac{F_{0}t}{mv}-\left(\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}\right)\right]\\ =-\frac{m{v}^2}{F_0}\left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)+\left(\frac{m{v}^2}{F_0}\right)\frac{F_{0}t}{mv}+\left(\frac{m{v}^2}{F_0}\right)\left(\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}\right)\\ =-\frac{m{v}^2}{F_0}\left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)+vt+\left(\frac{m{v}^2}{F_0}\right)\left(\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}\right)\\ x\left(t\right)=-\frac{m{v}^2}{F_0}\left[\left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)\ln \left(\frac{F_{0}t}{mv}+e^{-v_0/v}\right)+\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}\right]+vt\end{array}$

From part (b), the time that takes to for the mass to come instantaneously to rest is $t^{*}=\frac{mv}{F_0}\left(1-e^{-v_0/v}\right)$.

Therefore,

$\begin{array}{c}x\left(t^{*}\right)=-\frac{m{v}^2}{F_0}\left[\left(\frac{F_0}{mv}\left(\frac{mv}{F_0}\left(1-e^{-v_0/v}\right)\right)+e^{-v_0/v}\right)\ln \left(\frac{F_0}{mv}\left(\frac{mv}{F_0}\left(1-e\frac{-v_0}{v}\right)\right)+e^{-v_0/v}\right)+\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}\right]+v\left(\frac{mv}{F_0}\left(1-e^{-v_0/v}\right)\right)\\ =-\frac{m{v}^2}{F_0}\left[\left(1-e^{-v_0/v}+e^{-v_0/v}\right)\ln \left(1-e^{-v_0/v}+e^{-v_0/v}\right)+\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}-1+e^{\frac{-v_0}{v}}\right]\\ =-\frac{m{v}^2}{F_0}\left[\left(1\right)\ln \left(1\right)+\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}-1+e^{\frac{-v_0}{v}}\right]\\ =\frac{m{v}^2}{F_0}\left[1-\frac{-v_0}{v}{e}^{\frac{-v_0}{v}}-e^{\frac{-v_0}{v}}\right]\end{array}$$x\left(t^{*}\right)=\frac{m{v}^2}{F_0}\left[1-e^{\frac{-v_0}{v}}\left(\frac{-v_0}{v}+1\right)\right]$

Conclusion:

Thus, the distance the mass travels before coming instantaneously to rest is $x\left(t^{*}\right)=\frac{m{v}^2}{F_0}\left[1-e^{\frac{-v_0}{v}}\left(\frac{-v_0}{v}+1\right)\right]$.