Let state 1 be when the crate is thrown at the platform. Let the crate by body \(A\) and the platform be body \(B\). We will use work-energy to solve this. \[ T_{1}+V_{1}+U_{12}^{internal}+U_{12}^{external}=T_{2}+V_{2}\] Where \(U_{12}^{internal}\) is work done due to internal forces between the two bodies, which is the friction. We will use notation \(v_{A_{1}}\) to mean velocity of \(A\) in state \(1\) and \(v_{A_{2}}\) to mean velocity of \(A\) in state \(2\) and the same for body \(B\). Therefore the above equation becomes\[ \frac{1}{2}m_{A}v_{A_{1}}^{2}+\frac{1}{2}m_{A}v_{B_{1}}^{2}-\int _{0}^{d}\mu _{k}m_{A}gdx=\frac{1}{2}m_{A}v_{A_{2}}^{2}+\frac{1}{2}m_{A}v_{B_{2}}^{2}\] Notice that \(V_{1}=V_{2}\) and hence they cancel. Also since in state 2 both body \(A\) and \(B\) move with same speed \(v\), and also \(v_{B_{1}}=0\), then the above simplifies to\[ \frac{1}{2}m_{A}v_{A_{1}}^{2}-\mu _{k}m_{A}gd=\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}\] We now solve for \(d\), the distance that body \(A\) (the crate) slides. The above is one equation with one unknown.\begin{align*} d & =\frac{\frac{1}{2}m_{A}v_{A_{1}}^{2}-\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}}{\mu _{k}m_{A}g}\\ & =\frac{1}{2}\frac{\left ( 210\right ) \left ( 15\right ) ^{2}-\left ( 210+741\right ) \left ( 2.652\right ) ^{2}}{\left ( 0.28\right ) \left ( 210\right ) \left ( 32.2\right ) }\\ & =10.712\text{ ft} \end{align*}
Let state 1 be just before release and state 2 after it moves by \(1.5\) meter
Therefore\[ T_{1}+V_{1}+U_{12}^{internal}+U_{12}^{external}=T_{2}+V_{2}\] \(T_{1}=0,V_{1}=0\) and \(U_{12}=0\) since there is no friction and no external force. \(T_{2}=\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}\) since both bodies will have same speed. \(V_{2}\) comes from spring and gravity. The distance \(h=1.5\) meter. Therefore \(V_{2}=\frac{1}{2}kh^{2}-m_{B}gh\) since spring extend by same amount \(m_{B}\) and \(m_{A}\) moves. The above becomes\[ 0=\frac{1}{2}\left ( m_{A}+m_{B}\right ) v^{2}+\frac{1}{2}kh^{2}-m_{B}gh \] Where \(v\) above is \(1.3\) since both bodies move with same speed. We want to solve for \(m_{B}\) the only unknown in this equation\begin{align*} 0 & =\frac{1}{2}m_{A}v^{2}+\frac{1}{2}m_{B}v^{2}+\frac{1}{2}kh^{2}-m_{B}gh\\ & =m_{B}\left ( \frac{1}{2}v^{2}-gh\right ) +\frac{1}{2}m_{A}v^{2}+\frac{1}{2}kh^{2}\\ m_{B} & =\frac{m_{A}v^{2}+kh^{2}}{2gh-v^{2}} \end{align*}
Plug-in numerical values gives\begin{align*} m_{B} & =\frac{\left ( 4\right ) \left ( 1.3\right ) ^{2}+\left ( 9\right ) \left ( 1.5\right ) ^{2}}{2\left ( 9.81\right ) \left ( 1.5\right ) -\left ( 1.3\right ) ^{2}}\\ & =0.974\text{ kg} \end{align*}
Let state 1 be just before release and state 2 after it rotation.
\[ T_{1}+V_{1}+U_{12}^{internal}+U_{12}^{external}=T_{2}+V_{2}\] But \(U_{12}=0\) since there is no friction and no external forces. Now \(T_{1}=0\) since at rest. And \[ V_{1}=-m_{B}gL_{1}\sin \alpha +m_{A}gL_{2}\sin \alpha \] Where \(\alpha =36^{o}\). In state 2\[ T_{2}=\frac{1}{2}m_{B}\left ( L_{1}\omega \right ) ^{2}+\frac{1}{2}m_{A}\left ( L_{2}\omega \right ) ^{2}\] Where \(\omega \) is the angular velocity, which we do not know, but will solve for. And \[ V_{2}=m_{B}gL_{1}\sin \beta -m_{A}gL_{2}\sin \beta \] Therefore (1) becomes\begin{align*} -m_{B}gL_{1}\sin \alpha +m_{A}gL_{2}\sin \alpha & =\frac{1}{2}m_{B}\left ( L_{1}\omega \right ) ^{2}+\frac{1}{2}m_{A}\left ( L_{2}\omega \right ) ^{2}+m_{B}gL_{1}\sin \beta -m_{A}gL_{2}\sin \beta \\ -m_{B}gL_{1}\sin \alpha +m_{A}gL_{2}\sin \alpha -m_{B}gL_{1}\sin \beta +m_{A}gL_{2}\sin \beta & =\omega ^{2}\left ( \frac{1}{2}m_{B}L_{1}^{2}+\frac{1}{2}m_{A}L_{2}^{2}\right ) \\ \omega ^{2} & =\frac{m_{A}gL_{2}\left ( \sin \alpha +\sin \beta \right ) -m_{B}gL_{1}\left ( \sin \alpha +\sin \beta \right ) }{\frac{1}{2}m_{B}L_{1}^{2}+\frac{1}{2}m_{A}L_{2}^{2}}\\ \omega & =\sqrt{\frac{2\left ( m_{A}gL_{2}-m_{B}gL_{1}\right ) \left ( \sin \alpha +\sin \beta \right ) }{m_{B}L_{1}^{2}+m_{A}L_{2}^{2}}} \end{align*}
Now we solve for \(\omega \) and use it to find speed of \(B\) from \(v_{B}=L_{1}\omega \). Since \(m_{A}=814,m_{B}=129,L_{1}=10,L_{2}=5,\alpha =36,\beta =110-36=74^{o}\) then\begin{align*} \omega & =\sqrt{2\frac{\left ( \left ( 814\right ) \left ( 32.2\right ) \left ( 5\right ) -\left ( 129\right ) \left ( 32.2\right ) \left ( 10\right ) \right ) \left ( \sin \left ( 36\left ( \frac{\pi }{180}\right ) \right ) +\sin \left ( 74\left ( \frac{\pi }{180}\right ) \right ) \right ) }{\left ( 129\right ) 10^{2}+\left ( 814\right ) 5^{2}}}\\ & =2.888\,\text{\ rad/sec} \end{align*}
Hence \begin{align*} v_{B} & =L_{1}\omega \\ & =10\left ( 2.888\right ) \\ & =28.88\text{ ft/sec} \end{align*}
Power \(P\) is
\[ P=Fv \]
Where \(F\) is force generated by cyclist. From force balance we see that \(F=mg\sin \theta \). Hence for constant power, we want
\[ \left ( mg\sin 13^{0}\right ) v_{1}=\left ( mg\sin 20^{0}\right ) v_{2}\]
Solving for \(v_{2}\)
\begin{align*} v_{2} & =\frac{\left ( mg\sin 13^{0}\right ) v_{1}}{\left ( mg\sin 20^{0}\right ) }\\ & =\frac{\sin \left ( 13\left ( \frac{\pi }{180}\right ) \right ) }{\sin \left ( 20\left ( \frac{\pi }{180}\right ) \right ) }21\\ & =13.812\text{ mph} \end{align*}
\[ \varepsilon =\frac{P_{out}}{P_{in}}\]
Where \(\varepsilon \) is the efficiency and \(P_{out}\) is power out and \(P_{in}\) is power in. But \(P_{out}=Fv_{c}\). So we just need to find force in the cable that the motor is pulling with. This force is \(\frac{W}{4}\), since there are \(4\) cables and hence the weight is distributed over them, and then the tension in the one cable attached to the motor is \(\frac{W}{4}\). Now we have all the information to find \(\varepsilon \)
\begin{align*} P_{out} & =\left ( 4.5\right ) \left ( \frac{460}{4}\right ) \\ & =517.\,\allowbreak 5\text{ lb-ft/sec} \end{align*}
But \(hp=550\) lb-ft/sec, therefore in hp the above is \(\frac{517.\,\allowbreak 5}{550}=\allowbreak 0.941\), hence
\[ \varepsilon =\frac{\allowbreak 0.941}{1.37}=0.687 \]
\[ \varepsilon =\frac{P_{out}}{P_{in}}\]
Mass of crate is \(\frac{216}{32.2}\) slug. (note, number given \(216\) is weight) These problem should make it more clear if \(lb\) given is meant to be weight or mass.
We need to find \(P_{in}\). We are given \(\varepsilon \). We now calculate \(P_{out}\) and then will be able to find \(P_{in}\). But \(P_{out}=Fv\), where \(F\) is force given by motor to pull the crate. From free body diagram, we see that this force is \(F=\mu mg\cos \theta +mg\sin \theta \). Hence
\begin{align*} P_{out} & =\left ( \mu mg\cos \theta +mg\sin \theta \right ) v\\ & =mg\left ( \mu \cos \theta +\sin \theta \right ) v\\ & =\left ( \frac{216}{32.2}\right ) \left ( 32.2\right ) \left ( 0.24\cos \left ( 28\left ( \frac{\pi }{180}\right ) \right ) +\sin \left ( 28\left ( \frac{\pi }{180}\right ) \right ) \right ) \left ( 6.6\right ) \\ & =971.374\text{ lb-ft/sec}\\ & =\frac{971.374}{550}=1.\,766\text{ hp} \end{align*}
Hence
\begin{align*} P_{in} & =\frac{P_{out}}{\varepsilon }\\ & =\frac{1.\,766}{0.78}\\ & =2.264\text{ hp} \end{align*}