Question

The height difference between the top and bottom of a downward-moving escalator is h = 20 m. A mischievous boy of mass m = 50 kg runs up from the bottom to the top at an (average) speed, relative to the steps, that is one-and-a-half times their translational speed. Find the work done by the boy ($g = 9.8 m/s^2$)

• A. 14.7 kJ
• B. 19.6 kJ
• C. 24.5 kJ
• D. 29.4 kJ

Answer 1

Answer: (D)

29.4 kJ

Answer 2

Let the length of the escalator be $L$ and its inclination be $\theta$ such that

$$\sin\theta = \frac{h}{L}.$$

The escalator moves downward with speed $u$ (along its length). The boy runs upward relative to the escalator with speed $1.5u$. Thus, his speed relative to the ground is

$$1.5u - u = 0.5u.$$

Since the vertical height the boy gains is $h$, the escalator (of length $L$) satisfies:

$$h = L\sin\theta.$$

Step 1. Find time to reach the top:

The boy covers the entire length $L$ (along the escalator) with his ground speed (along the escalator) $0.5u$:

$$T = \frac{L}{0.5u} = \frac{2L}{u}.$$

Step 2. Find the distance he runs relative to the escalator:

Relative to the moving steps, he runs:

$$\text{Distance } d = \text{(relative speed)} \times T = 1.5u \times \frac{2L}{u} = 3L.$$

Step 3. Determine the force along the escalator:

To overcome gravity, the component of weight along the escalator is:

$$F = mg\sin\theta = mg\left(\frac{h}{L}\right).$$

Step 4. Compute the work done:

The work done by the boy is the force times the distance traveled relative to the escalator:

$$W = F \times d = \left(mg\frac{h}{L}\right) \times (3L) = 3mg h.$$

Substitute $m = 50\,\text{kg}$, $g = 9.8\,\text{m/s}^2$, and $h = 20\,\text{m}$:

$$W = 3 \times 50 \times 9.8 \times 20 = 29400\,\text{J} = 29.4\,\text{kJ}.$$