An elevator cab of mass `m=500` kg is descending with speed `v_(i)=4.0m//s` when its supporting cable begins to slip, allowing it to fall with constant acceleration `vec(a)=vec(g)//5` (Fig. 8-11a).
(b) During the 12 m fall, what is the work `W_(T)` done on the cab by the upward pull `vec(T)` of hte elevator cable?

Explanation

KEY IDEA
We can calculate work `W_(T)` with Eq. 8-7 `(W=Fd cos phi)` by first writing `F_("net". y)=ma_(y)` for the components in Fig. 8-11b.
Calculations: We get
`T-F_(g)=ma`. (8-18)
Solving for .I., substituting mg for `F_(g)`, and then substituting the result in Eq. 8-7, we obtain
`W_(T)=Td cos phi = m (a+g)d cos phi`. (8-19)

Figure 8-11 An elevator cab, descending with speed `v_(i)`, suddenly begins to accelerate downward. (a) It moves through a displacement `vec(d)` with constant acceleration `vec(a)=g//5`. (b) A freebody diagram for the cab, displacement included.
Next, substituting `-g//5` for the (downward) acceleration a and then `180^(@)` for the angle `phi` between the direction of forces `vec(T)` and `m vec(g)`, we find
`W_(T)=m(- g/5 + g) d cos phi = 4/5 mgd cos phi`
`=4/5 (500 kg) (9.8 m//s^(2))(12m)cos 180^(@)`
`= -4.70xx10^(4) j~~-47 kJ`.
Caution: Note that `W_(T)` is not simply the negative of `W_(g)` because the cab accelerates during the fall. Thus, Eq. 8-16 (which assumes that the initial and final kinetic energies are equal) does not apply here.