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6.5 HW 5

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6.5.1 Problem 1

Landing speed is $v_{2}=4$ m/sec.

$$\begin{align*} U_{12} & =T_{2}-T_{1}\\ & =\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}\\ & =\frac{1}{2}m\left ( 4^{2}-\left ( 241\frac{1000}{km}\frac{hr}{3600}\right ) ^{2}\right ) \\ & =\frac{1}{2}70\left ( \left ( 4\right ) ^{2}-\left ( 241\left ( \frac{1000}{3600}\right ) \right ) ^{2}\right ) \\ & =-156294.6 \end{align*}$$

Hence work on person is $-156.295$ kJ

6.5.2 Problem 2

Force in the $x$ direction is

$$\begin{align*} F & =F_{p}-F_{friction}\\ & =\left ( 20+12x\right ) -\mu _{k}N\\ & =\left ( 20+12x\right ) -\left ( 0.4\right ) \left ( 32\right ) \end{align*}$$

Hence

$$\begin{align*} U_{12} & =T_{2}-T_{1}\\ \int _{0}^{15}\bar{F}\cdot d\bar{r} & =\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}\\ \int _{0}^{15}\left ( \left ( 20+12x^{\frac{1}{2}}\right ) -\left ( 0.4\right ) \left ( 32\right ) \right ) dx & =\frac{1}{2}\frac{32}{32.2}v_{2}^{2} \end{align*}$$

Since $v_{1}=0$ then above becomes

$$\begin{align*} \int _{0}^{15}\left ( \left ( 20+12x^{\frac{1}{2}}\right ) -\left ( 0.4\right ) \left ( 32\right ) \right ) dx & =\frac{1}{2}\left ( \frac{32}{32.2}\right ) v_{2}^{2}\\ \int _{0}^{15}12x^{\frac{1}{2}}+7.2dx & =0.49689v_{2}^{2}\\ \left ( \frac{\left ( 12\right ) \left ( 2\right ) }{3}x^{\frac{3}{2}}+7.2x\right ) _{0}^{15} & =0.49689v_{2}^{2}\\ \frac{\left ( 12\right ) \left ( 2\right ) }{3}\left ( 15\right ) ^{\frac{3}{2}}+7.2\left ( 15\right ) & =0.49689v_{2}^{2}\\ 572.758 & =0.49689v_{2}^{2}\\ v_{2}^{2} & =\frac{572.758}{0.49689}\\ & =1152.686 \end{align*}$$

Hence

$$\begin{align*} v_{2} & =\sqrt{1152.686}\\ & =33.951\text{ ft/sec} \end{align*}$$

6.5.3 Problem 3

Since all forces we can use conservation of energy $T_{1}+V_{1}=T_{2}+V_{2}$ Where $V_{1}=0$ since spring is not compressed yet and $T_{2}=0$ since the car would be stopped by then. Hence

$$\begin{align*} \frac{1}{2}mv_{2}^{2} & =\frac{1}{2}k\Delta ^{2}\\ \Delta ^{2} & =\frac{mv_{2}^{2}}{k}\\ & =\frac{\left ( 1324\right ) \left ( 5.5\left ( \frac{1000}{km}\right ) \left ( \frac{hr}{3600}\right ) \right ) ^{2}}{9.4\times 10^{4}}\\ & =\frac{\left ( 1324\right ) \left ( 5.5\left ( \frac{1000}{3600}\right ) \right ) ^{2}}{9.4\times 10^{4}}\\ \Delta ^{2} & =0.033\text{ } \end{align*}$$

Hence

$$\Delta =0.182\,\text{\ meter}$$ We could also have solved this using work-energy. Force on car is $-kx$, hence $U_{12}=\int _{0}^{x}\bar{F}\cdot d\bar{r}$ and therefore $$\begin{align*} \int _{0}^{x}\bar{F}\cdot d\bar{r} & =\frac{1}{2}mv_{2}^{2}-\frac{1}{2}mv_{1}^{2}\\ \int _{0}^{x}-kxdx & =-\frac{1}{2}mv_{1}^{2}\\ \frac{1}{2}k\left ( x^{2}\right ) _{0}^{x} & =\frac{1}{2}1324\left ( 5.5\left ( \frac{1000}{3600}\right ) \right ) ^{2}\\ 9.4\times 10^{4}x^{2} & =1324\left ( 5.5\left ( \frac{1000}{3600}\right ) \right ) ^{2}\\ x^{2} & =\frac{1324\left ( 5.5\left ( \frac{1000}{3600}\right ) \right ) ^{2}}{9.4\times 10^{4}}\\ x & =0.182\text{ meter} \end{align*}$$

6.5.4 Problem 4

Distance is $L=51$ meter (not clear in problem image).

Taking zero PE at horizontal datum when car comes to a stop at the bottom of hill, then using

$$T_{1}+V_{1}+U_{12}=T_{2}+V_{2}$$ Where $T_{1}$ is KE in state 1, when the car just hit the brakes, and $V_{1}$ is its gravitational PE and $U_{12}$ is work by non-conservative forces, which is friction here. $T_{2}=0$ since car stops in state 2 and $V_{2}$ is PE in state 2 which is zero. Hence we have $$\begin{align*} \frac{1}{2}mv_{1}^{2}+mgL\sin \theta +\int _{0}^{L}-\mu Ndx & =0\\ \frac{1}{2}mv_{1}^{2}+mgL\sin \theta -\int _{0}^{L}\mu \left ( mg\cos \theta \right ) dx & =0\\ \frac{1}{2}mv_{1}^{2}+mgL\sin \theta -L\mu mg\cos \theta & =0\\ \mu & =\frac{\frac{1}{2}v_{1}^{2}+gL\sin \theta }{Lg\cos \theta } \end{align*}$$

Hence

$$\begin{align*} \mu & =\frac{\frac{1}{2}\left ( 58\left ( \frac{1000}{3600}\right ) \right ) ^{2}+9.81\left ( 51\right ) \sin \left ( 21\left ( \frac{\pi }{180}\right ) \right ) }{\left ( 51\right ) \left ( 9.81\right ) \cos \left ( 21\left ( \frac{\pi }{180}\right ) \right ) }\\ & =\frac{129.784\,0+179.295\,1}{467.\,0796}\\ & =0.662 \end{align*}$$

6.5.5 Problem 5

$$T_{1}+V_{1}=T_{2}+V_{2}$$ Where state $1$ is initial state, and state $2$ is when bob at bottom. Datum is taken when bob at bottom. Hence $$\begin{align*} 0+mg\left ( L-L\cos \theta \right ) & =\frac{1}{2}mv_{2}^{2}+0\\ Lmg\left ( 1-\cos \theta \right ) & =\frac{1}{2}mv_{2}^{2} \end{align*}$$

Now let state $3$ be when bob is up again on the other side. Hence we have

$$T_{2}+V_{2}=T_{3}+V_{3}$$ But $T_{3}=0$ and $V_{3}=mgh_{\max }$, therefore $$\frac{1}{2}mv_{2}^{2}=mgh_{\max }$$ Or, since $\frac{1}{2}mv_{2}^{2}=Lmg\left ( 1-\cos \theta \right )$, then the above becomes $$\begin{align*} h_{\max } & =L\left ( 1-\cos \theta _{i}\right ) \\ & =1.86\left ( 1-\cos \left ( 16\left ( \frac{\pi }{180}\right ) \right ) \right ) \\ & =0.0721\text{ m} \end{align*}$$

6.5.6 Problem 6

6.5.6.1 Part (a)

$$\begin{align*} V & =\int ^{x}k\delta -\beta \delta ^{3}d\delta \\ & =\frac{kx^{2}}{2}-\frac{\beta x^{4}}{4} \end{align*}$$

6.5.6.2 Part (b)

Let datum be at top. Hence

$$\begin{align*} T_{1}+V_{1,gravity}+V_{1,rope} & =T_{2}+V_{2,gravity}+V_{2,rope}\\ 0+0+0 & =\frac{1}{2}mv_{2}^{2}-mgh+\left ( \frac{k\delta ^{2}}{2}-\frac{\beta \delta ^{4}}{4}\right ) \\ v & =\sqrt{2gh-\frac{2}{m}\left ( \frac{k\delta ^{2}}{2}-\frac{\beta \delta ^{4}}{4}\right ) }\\ & =\sqrt{\frac{\beta \delta ^{4}-2kv^{2}+4mgh}{2m}}\\ & =\sqrt{\frac{\left ( 0.000014\right ) \left ( 150\right ) ^{4}-2\left ( 2.6\right ) \left ( 150\right ) ^{2}+4\left ( 170\right ) \left ( 150\right ) }{2\left ( \frac{170}{32.2}\right ) }}\\ & =\sqrt{-749.360\,3}\\ & =\sqrt{\frac{\left ( 0.000013\right ) \left ( 250\right ) ^{4}-2\left ( 2.58\right ) \left ( 250\right ) ^{2}+4\left ( 170\right ) \left ( 250\right ) }{2\left ( \frac{170}{32.2}\right ) }} \end{align*}$$

But $\delta =h-150=400-150=250$, hence

$$\begin{align*} v & =\sqrt{\frac{\left ( 0.000014\right ) \left ( 250\right ) ^{4}-2\left ( 2.6\right ) \left ( 250\right ) ^{2}+4\left ( 170\right ) \left ( 400\right ) }{2\left ( \frac{170}{32.2}\right ) }}\\ & =12.64184\text{ ft/sec} \end{align*}$$

6.5.6.3 Part (c)

$$\delta =\sqrt{\frac{k}{3\beta }}=\sqrt{\frac{2.6}{3\left ( 0.000014\right ) }}=248.806\,7$$ Hence $$\begin{align*} a & =\left \vert g\left ( 1-\frac{k\delta -\beta \delta ^{3}}{W}\right ) \right \vert \\ & =\left \vert 32.2\left ( 1-\frac{\left ( 2.6\right ) \left ( 250\right ) -\left ( 0.000014\right ) \left ( 250\right ) ^{3}}{170}\right ) \right \vert \\ & =49.48382\text{ ft/s}^{2}\\ & =\frac{49.48382}{32.2}\\ & =1.537\text{ g} \end{align*}$$