Answer
a) $T=829.5N$
b) $W=9124.5J$
c) $W=-8516.2J$
d) $v_f=3.92m/s$
Work Step by Step
a) There are 2 forces influencing the person's upward motion: tension $T$ in the cable and the person's weight $mg$. From Newton's 2nd law, $$T-mg=ma$$ $$T=m(g+a)$$
We know $m_{person}=79kg$, $g=9.8m/s^2$ and $a=0.7m/s^2$. Therefore, $$T=829.5N$$
b) We know the movement distance $s=11m$. Since $T$ pulls the person upward along the same direction as the displacement, $\theta=0^o$
$$W=(T\cos\theta)s=9124.5J$$
c) Distance $s=11m$, the person's weight $mg=79\times9.8=774.2N$. The person's weight points downward while her motion points upward, so $\theta=180^o$
$$W=(mg\cos\theta)s=-8516.2J$$
d) The net work done by all forces is $\sum W=W_T+W_{mg}=608.3J$
According to the work-energy theorem, $$\sum W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2$$
The person is pulled from rest, so $v_0=0$. Therefore, $$\sum W=\frac{1}{2}mv_f^2$$ $$v_f=\sqrt{\frac{2\sum W}{m}}$$
We also know $m=79kg$ and $W=9124.5J$, so $$v_f=3.92m/s$$
