Inertia of an elevator

Question

Imagine a situation where we have a relatively light empty elevator moving up at a pretty high constant velocity say something above $10\ m/s$, just so that it has a considerable amount of momentum upwards. It is being pulled by a string upwards by a motor with a gearbox that wont allow the elevetor to fall downwards even if power cuts. Now What happens if we cut power that pulls the elevator upwards? There is some inertia that the elevetor has that should make it want to go upwards. Also consider there is some friction involved that further wants to slow down its motion going upwards.

Can we calculate that inertia and predict at what distance (from the moment we cut power) the elevator would stop its motion upwards? If so, how would you calculate the inertia of an elevator going upwards? how will you account for the friction?

Answer 1

When the elevator of mass $m$ is pulled by the 'string' then after some time it will acquire some velocity, say $v$. At that point it will have a kinetic energy $K$:

$$K=\frac12 mv^2$$

Now you cut the power. Inertia demands that the elevator will continue to travel upwards for 'a bit'.

During this period, kinetic energy $K$ is converted to potential energy $U$, in such a way that:

$$\frac12 mv^2=mg\Delta h\tag{1}$$

where $\Delta h$ is the height it will travel before it comes to a standstill ($v=0$). After that, the elevator begins to free fall.

We can also find the deceleration $a$ during that first phase by taking the time derivative of $(1)$:

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac12 mv^2\right)=\frac{\mathrm{d}}{\mathrm{d}t}\left(mg\Delta h\right)$$

which yields:

$$a=-g$$

(the sign is minus because $a$ is a deceleration)