Question

A traveler at an airport takes an escalator up one floor. The moving staircase would itself carry him upward with vertical velocity component v, between entry and exit points separated by height h. However, while the escalator is moving, the hurried traveler climbs the steps of the escalator at a rate of n steps/sec. Assume that the height of each step is $h_s$.

• A. Amount of work done by the traveler during his escalator ride, given that his mass is m, is $\frac{mghnh_s}{(v+nh_s)}$
• B. Amount of work done by the traveler during his escalator ride, given that his mass is m, is $\frac{mghnh_s}{(v-nh_s)}$
• C. The work the escalator motor does on this person is $\frac{mgvh}{(v+nh_s)}$
• D. The work the escalator motor does on this person is $\frac{mgvh}{(v-nh_s)}$

Answer 1

Answer: (A), (C)

A, C

Answer 2

To solve this problem, we need to understand the concepts of work done by a force and relative motion.

Let h be the total vertical height of the escalator.
Let v be the vertical velocity component of the escalator.
Let n be the rate at which the traveler climbs steps (steps/sec).
Let h_s be the height of each step.
Let m be the mass of the traveler.

1. Calculate the total vertical speed of the traveler relative to the ground:

The escalator provides a vertical speed v.
The traveler climbing the steps provides an additional vertical speed $v_{traveler\_relative\_escalator} = n \times h_s$.
So, the total vertical speed of the traveler relative to the ground is:
$V_{total} = v + nh_s$

2. Calculate the time taken to reach the top:

The total vertical distance to be covered is h.
Time taken $T = \frac{h}{V_{total}} = \frac{h}{v + nh_s}$

3. Calculate the work done by the traveler:

Work done by the traveler is the force exerted by the traveler multiplied by the vertical distance the traveler moves relative to the escalator.
The force exerted by the traveler to lift himself against gravity is mg.
The vertical distance the traveler moves relative to the escalator is $d_{traveler\_relative} = (nh_s) \times T$.
Substitute the value of T:
$d_{traveler\_relative} = nh_s \times \frac{h}{v + nh_s} = \frac{nh_s h}{v + nh_s}$
Work done by the traveler $W_{traveler} = mg \times d_{traveler\_relative}$
$W_{traveler} = mg \times \frac{nh_s h}{v + nh_s} = \frac{mghnh_s}{v + nh_s}$
This matches option (A).

4. Calculate the work done by the escalator motor on the person:

Work done by the escalator motor on the person is the force exerted by the escalator on the person multiplied by the vertical distance the escalator moves relative to the ground.
The force exerted by the escalator on the person to lift them against gravity is mg.
The vertical distance the escalator moves relative to the ground is $d_{escalator} = v \times T$.
Substitute the value of T:
$d_{escalator} = v \times \frac{h}{v + nh_s} = \frac{vh}{v + nh_s}$
Work done by the escalator motor $W_{escalator} = mg \times d_{escalator}$
$W_{escalator} = mg \times \frac{vh}{v + nh_s} = \frac{mgvh}{v + nh_s}$
This matches option (C).

Consistency Check (Work-Energy Theorem):

The total work done on the person by all non-conservative forces (traveler's effort, escalator motor) must equal the change in potential energy, assuming no change in kinetic energy ($W_{net} = \Delta PE$).
$\Delta PE = mgh$
$W_{traveler} + W_{escalator} = \frac{mghnh_s}{v + nh_s} + \frac{mgvh}{v + nh_s}$
$= \frac{mgh(nh_s + v)}{v + nh_s} = mgh$
This confirms that our calculations for $W_{traveler}$ and $W_{escalator}$ are consistent.

Therefore, options (A) and (C) are correct.