The velocity in meter per second is \begin{align*} v & =228\left ( \frac{1000}{km}\right ) \left ( \frac{h}{3600}\right ) \\ & =\left ( 228\right ) \frac{1000}{3600}\\ & =63.333\text{ m/s} \end{align*}
In normal direction, the acceleration is \(a_{n}=\frac{v^{2}}{\rho }=\frac{\left ( 63.333\right ) ^{2}}{115}=\allowbreak 34.879\) m/s\(^{2}\) or in terms of \(g\), it becomes \(\frac{34.879}{9.81}=3.555\) g.
Now, force balance in vertical direction gives\begin{align*} L-mg & =ma_{n}\\ L & =m\left ( g+a_{n}\right ) \\ & =970\left ( 9.81+34.879\right ) \\ & =43348.33\text{ N} \end{align*}
For maximum speed, friction acts downwards as car assumed to be just about to slide upwards. Resolving forces in normal and tangential gives \begin{align} \mu _{s}N\cos \theta +N\sin \theta & =m\frac{v_{\max }^{2}}{\rho }\tag{1}\\ -mg+N\cos \theta -\mu _{s}N\sin \theta & =0\tag{2} \end{align}
From (2)\[ N=\frac{mg}{\cos \theta -\mu _{s}\sin \theta }\] From (1)\begin{align*} v_{\max }^{2} & =\frac{\rho }{m}\left ( \mu _{s}N\cos \theta +N\sin \theta \right ) \\ & =\frac{\rho }{m}\left ( \mu _{s}\left ( \frac{mg}{\cos \theta -\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{mg}{\cos \theta -\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho \left ( \mu _{s}\left ( \frac{g}{\cos \theta -\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{g}{\cos \theta -\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho g\left ( \frac{\mu _{s}}{1-\mu _{s}\tan \theta }+\frac{\tan \theta }{1-\mu _{s}\tan \theta }\right ) \\ & =\rho g\left ( \frac{\mu _{s}+\tan \theta }{1-\mu _{s}\tan \theta }\right ) \\ & =324\left ( 9.81\right ) \left ( \frac{0.2+\tan \left ( 31\frac{\pi }{180}\right ) }{1-0.2\tan \left ( 31\frac{\pi }{180}\right ) }\right ) \\ & =2893.165\text{ m/s} \end{align*}
Hence\begin{align*} v_{\max } & =53.788\text{ m/s}\\ & =53.788\left ( 0.001\right ) \left ( 3600\right ) \\ & =193.637\text{ km/hr} \end{align*}
For minimum speed, friction acts upwards as car assumed to be just about to slide downwards. Resolving forces in normal and tangential gives \begin{align} -\mu _{s}N\cos \theta +N\sin \theta & =m\frac{v_{\min }^{2}}{\rho }\tag{3}\\ -mg+N\cos \theta +\mu _{s}N\sin \theta & =0\tag{4} \end{align}
From (4)\[ N=\frac{mg}{\cos \theta +\mu _{s}\sin \theta }\] From (3)\begin{align*} v_{\min }^{2} & =\frac{\rho }{m}\left ( -\mu _{s}N\cos \theta +N\sin \theta \right ) \\ & =\frac{\rho }{m}\left ( -\mu _{s}\left ( \frac{mg}{\cos \theta +\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{mg}{\cos \theta +\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho g\left ( -\mu _{s}\left ( \frac{1}{\cos \theta +\mu _{s}\sin \theta }\right ) \cos \theta +\left ( \frac{1}{\cos \theta +\mu _{s}\sin \theta }\right ) \sin \theta \right ) \\ & =\rho g\left ( \frac{-\mu _{s}}{1+\mu _{s}\tan \theta }+\frac{\tan \theta }{1+\mu _{s}\tan \theta }\right ) \\ & =\rho g\left ( \frac{\tan \theta -\mu _{s}}{1+\mu _{s}\tan \theta }\right ) \\ & =324\left ( 9.81\right ) \left ( \frac{\tan \left ( 31\frac{\pi }{180}\right ) -0.2}{1+0.2\tan \left ( 31\frac{\pi }{180}\right ) }\right ) \\ & =1137.425\text{ m/s} \end{align*}
Hence\begin{align*} v_{\min } & =33.726\text{ m/s}\\ & =33.726\left ( 0.001\right ) \left ( 3600\right ) \\ & =121.414\text{ km/hr} \end{align*}
Hence\[ 121.414\leq v\leq 193.637 \]
Since acceleration is constant along barrel, then using\[ v_{f}^{2}=v_{0}^{2}+2\ddot{r}L \] We will solve for \(\ddot{r}\). Assuming \(v_{0}=0\) then\begin{align*} 1681^{2} & =2\ddot{r}\left ( 4.2\right ) \\ \ddot{r} & =\frac{1681^{2}}{2\left ( 4.2\right ) }\\ & =336400.1\text{ m/s}^{2} \end{align*}
But\[ \bar{a}=\left ( \ddot{r}-L\dot{\theta }^{2}\right ) \hat{u}_{r}+\left ( L\ddot{\theta }+2\dot{r}\dot{\theta }\right ) \hat{u}_{\theta }\] But \(\ddot{\theta }=0\), hence the above becomes\begin{align*} \bar{a} & =\left ( 336400.1-\left ( 4.2\right ) \left ( 0.18\right ) ^{2}\right ) \hat{u}_{r}+\left ( 2\left ( 1681\right ) \left ( 0.18\right ) \right ) \hat{u}_{\theta }\\ & =336400\ \hat{u}_{r}+605.16\ \hat{u}_{\theta } \end{align*}
Free body diagram for bullet gives\[ P=ma_{r}+mg\sin \theta \] but \(a_{r}=336400\) m/s\(^{2}\) and \(m=16.9\ \)kg and \(\theta =23^{0}\), hence\begin{align*} P & =\left ( 16.9\right ) \left ( 336400\right ) +\left ( 16.9\right ) \left ( 9.81\right ) \sin \left ( \left ( 23\right ) \frac{\pi }{180}\right ) \\ & =5685225\text{ N}\\ & =5.685\times 10^{6}\text{ N} \end{align*}
Free body diagram for bullet gives\begin{align*} N & =ma_{\theta }+mg\cos \theta \\ & =\left ( 16.9\right ) \left ( 605.16\right ) +\left ( 16.9\right ) \left ( 9.81\right ) \cos \left ( \left ( 23\right ) \frac{\pi }{180}\right ) \\ & =10379.81\text{ N}\\ & =10.379\times 10^{3}\text{ N} \end{align*}
The free body diagram is
For \(A\), \begin{align} \sum F_{x} & =m_{A}a_{A_{x}}\nonumber \\ F_{A} & =m_{A}a_{A_{x}}\tag{2} \end{align}
\begin{align} \sum F_{y} & =m_{A}a_{A_{y}}\nonumber \\ N_{A}-W_{A} & =0\tag{3} \end{align}
For \(B\)\begin{align} \sum F_{x} & =m_{B}a_{B_{x}}\nonumber \\ F_{o}-F_{A}-F_{B} & =m_{B}a_{B_{x}}\tag{4} \end{align}
\begin{align} \sum F_{y} & =m_{B}a_{B_{y}}\nonumber \\ N_{B}-W_{B}-N_{A} & =0\tag{5} \end{align}
Hence (3) becomes \begin{align} N_{A} & =W_{A}\tag{6A}\\ & =m_{A}g\nonumber \end{align}
And (5) becomes\begin{align} N_{B} & =W_{B}+N_{A}\nonumber \\ & =m_{B}g+m_{A}g\nonumber \\ & =\left ( m_{B}+m_{A}\right ) g\tag{6B} \end{align}
But \(F_{A}=N_{A}\mu _{1}\) then (2) becomes\[ N_{A}\mu _{1}=m_{A}a_{A_{x}}\] But from (6) the above reduces to (since \(N_{A}=m_{A}g\))\[ m_{A}g\mu _{1}=m_{A}a_{A_{x}}\] Hence\begin{align*} a_{A_{x}} & =g\mu _{1}\\ & =\left ( 32.2\right ) \left ( 0.26\right ) \\ & =8.372\text{ ft/s}^{2} \end{align*}
Similarly \(F_{B}=N_{B}\mu _{2}\) hence (4) becomes\[ F_{o}-N_{A}\mu _{1}-N_{B}\mu _{2}=m_{B}a_{B_{x}}\] But \(N_{B}=\left ( m_{A}+m_{B}\right ) g\), then above becomes\begin{align*} F_{o}-m_{A}g\mu _{1}-\left ( m_{A}+m_{B}\right ) g\mu _{2} & =m_{B}a_{B_{x}}\\ a_{B_{x}} & =\frac{F_{o}-m_{A}g\mu _{1}-\left ( m_{A}+m_{B}\right ) g\mu _{2}}{m_{B}}\\ a_{B_{x}} & =\frac{438-\left ( 50\right ) \left ( 0.26\right ) -\left ( 50+71\right ) \left ( 0.46\right ) }{\frac{71}{32.2}}\\ & =167.504\text{ ft/sec}^{2} \end{align*}
Free body diagram for block \(B\) results in\[ 2g-\frac{8T}{m_{B}}=a_{A}\] Free body diagram for block \(A\) resuts in\[ -m_{A}g\sin \theta -\mu m_{A}g\cos \theta +2T=m_{A}a_{A}\] In addision, since rope length is fixed, we find that \(a_{A}=2a_{B}\). The above are 3 equations in 3 unknowns \(T,a_{A},a_{B}\). Solving gives\begin{align*} a_{A} & =4.439\text{ m/s}^{2}\\ a_{B} & =2.2197\text{ m/s}^{2}\\ T & =37.95\text{ N} \end{align*}
maximum displacement is \(2\) ft. Maximum speed is \(9.404\) ft/sec.