Answer
(a) The net work during the dive is greater.
(b) $\Delta W=6.83\times10^7J$
Work Step by Step
(a) $\vec{L}$ is always perpendicular with $\vec{s}$, so it contributes nothing to the net work. We can forget about it.
- During the dive, $T$ and $mg\cos75$ both support the plane in the direction of its movement. Therefore, $$\sum F=T+mg\cos75$$
The net work done by the net force $\sum F$ is $$W_d=(T+mg\cos75)s$$
- During the climb, $T$ supports the plane in the direction of its movement, while $mg\cos(180-115)=mg\cos65$ opposes it. Therefore, $$\sum F=T-mg\cos65$$
The net work done by the net force $\sum F$ is $$W_c=(T-mg\cos65)s$$
As we can easily see, the net work done during the dive is greater than the net work during the climb.
(b) $$\Delta W=W_d-W_c=(T+mg\cos75)s-(T-mg\cos65)s$$ $$\Delta W=mg(\cos75+\cos65)s=0.681mgs$$
We have $mg=5.9\times10^4N$ and $s=1.7\times10^3m$. Therefore, $$\Delta W=6.83\times10^7J$$
