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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? |
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Answer» Solution :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. |
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