Solution

\section{Understanding the motion}
$$\textbf{Solution}$$

The elevator descends a distance $S=400\,\mathrm{m}$ in $t=40\,\mathrm{s}$.
During the first half of the descent it accelerates with acceleration $a$,
and during the second half it decelerates with the same magnitude.

The total distance traveled is

$$S= \frac{1}{2}a\left(\frac{t}{2}\right)^2+ \frac{1}{2}a\left(\frac{t}{2}\right)^2 =\frac{at^2}{4}.$$

Therefore,

$$a=\frac{4S}{t^2} =\frac{4\cdot400}{40^2} =1\,\mathrm{m/s^2}.$$

During the first half of the descent, the effective gravitational acceleration
for the pendulum is

$$g_{\mathrm{eff},1}=g-a,$$

so its period is

$$T_1=2\pi\sqrt{\frac{l}{g-a}}.$$

During the second half,

$$g_{\mathrm{eff},2}=g+a,$$

and therefore

$$T_2=2\pi\sqrt{\frac{l}{g+a}}.$$

Hence, during the first and second halves of the descent, the numbers of
oscillations are

$$n_1=\frac{t/2}{T_1} =\frac{t}{4\pi}\sqrt{\frac{g-a}{l}},$$

and

$$n_2=\frac{t/2}{T_2} =\frac{t}{4\pi}\sqrt{\frac{g+a}{l}}.$$

Thus the total number of oscillations during one descent is

\[
n_{\mathrm{d}}

\frac{t}{4\pi\sqrt{l}}
\left(\sqrt{g-a}+\sqrt{g+a}\right).
\]

If the elevator were not accelerating, the pendulum would make

$$n_0=\frac{t}{2\pi}\sqrt{\frac{g}{l}}$$

oscillations in the same time.

The number of oscillations lost is therefore

\[
n_0-n_{\mathrm{d}}

\frac{t\sqrt{g}}{4\pi\sqrt{l}}
\left[
2-\sqrt{1-\frac{a}{g}}
-\sqrt{1+\frac{a}{g}}
\right].
\]

The normal period of the pendulum is

$$T_0=2\pi\sqrt{\frac{l}{g}}.$$

Therefore, the time lost by the clock during one descent is

$$\Delta t=(n_0-n_{\mathrm{d}})T_0.$$

After cancellation,

$$\boxed{ \Delta t= \frac{t}{2} \left[ 2-\sqrt{1-\frac{a}{g}} -\sqrt{1+\frac{a}{g}} \right] }.$$

For $t=40\,\mathrm{s}$, $a=1\,\mathrm{m/s^2}$, and
$g=9.8\,\mathrm{m/s^2}$,

\[
\Delta t

20
\left[
2-\sqrt{1-\frac{1}{9.8}}
-\sqrt{1+\frac{1}{9.8}}
\right]
\approx 0.054\,\mathrm{s}.
\]

The same time loss occurs during an ascent, since the two effective
accelerations $g-a$ and $g+a$ simply occur in the opposite order.

In $5$ hours,

$$5\,\mathrm{h}=18000\,\mathrm{s},$$

so the number of ascents or descents is

$$N=\frac{18000}{40}=450.$$

Consequently, the total time lost is

$$\Delta T=N\Delta t =450(0.054) \approx24.3\,\mathrm{s}.$$

Therefore,

$$\boxed{\Delta T\approx24\,\mathrm{s}}.$$