Answer
a) $v_f=49.7m/s$
b) $W_{nc}=-4.89\times10^4J$
Work Step by Step
a) In the absence of nonconservative forces, the principle of energy conservation applies: $$E_f-E_0=0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v_f^2-v_0^2)+g(h_f-h_0)=0$$
We have $v_0=0$, $h_0=0$, and $h_f=-126m$, so $$\frac{1}{2}v_f^2+9.8(-126)=0$$ $$v_f=49.7m/s$$
b) According to the work-energy theorem, $$\sum W_{nc}=E_f-E_0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=\sum W_{nc}$$
We have $v_0=0$ and $h_0=0$. From these, we can simplify the equation $$\frac{1}{2}mv_f^2+mgh_f=\sum W_{nc}$$
We also have $m=118kg$, $h_f=-126m$, $v_f=40.5m/s$. Therefore, $$W_{nc}=-4.89\times10^4J$$
