## Fundamentals of Physics Extended (10th Edition)

$F_{max}=mg+\dfrac {mv^{2}_{1}}{L}=mg+\dfrac {2mgh}{L}=mg\left( 1+\dfrac {2h}{L}\right) \approx 932.62N$
Lets calculate the speed at the end of descend ( initial speed is zero so kinetic energy will be zero too) The total energy of tarzan initially is $E_{tot}=E_{k0}+E_{p0}=0+mgh(1)$ $E_{k0}$ and $E_{p0}$ are initial kinetic and potential energies of tarzan initially So after descend potential energy will be zero and all potential energy will be converted to kinetic energy so $E_{tot}=E_{k_{1}}+E_{p1}=\dfrac {mv^{2}_{1}}{2}+0\left( 2\right)$ So from (1) and (2) we get $E_{tot}=E_{k_{1}}+E_{po}=\dfrac {mv^{2}_{1}}{2}+0=mgh+0\Rightarrow v^{2}_{1}=2gh(3)$ lets assume vine doesnt break until tarzan reaches the lowest point (The greatest force tarzan create on vine will be when tarzan is at the lowest point. Becouse speed of tarzan will be greatest here so $\left( \dfrac {mv^{2}}{L}\right)$ will be greatest see equation (4)) The force applied to vine by tarzan at the lowest point (trajectory of tarzan will be circular) will be: $F_{max}=mg+\dfrac {mv^{2}_{1}}{L}\left( 4\right)$ where $L$ is the length of the vine lets calculate this using (3) we get: $F_{max}=mg+\dfrac {mv^{2}_{1}}{L}=mg+\dfrac {2mgh}{L}=mg\left( 1+\dfrac {2h}{L}\right) \approx 932.62N$