Question

Workers are loading equipment into a freight elevator at the top floor of a building. However, they overload the elevator and the worn cable snaps. The mass of the loaded elevator at the time of the accident is \(1600 \mathrm{~kg}\). As the elevator falls, the guide rails exert a constant retarding force of \(3700 \mathrm{~N}\) on the elevator. At what speed does the elevator hit the bottom of the shaft \(72 \mathrm{~m}\) below?

Short answer

The speed at which the elevator hits the bottom of the shaft is approximately \(32.86 \mathrm{m/s}\).

Step-by-step solution

Calculate the gravitational force on the elevator

The force due to gravity can be calculated with the equation: \(F_{g} = m \cdot g\), where \(m\) is the mass of the elevator, \(1600 \mathrm{~kg}\), and \(g\) is the acceleration due to gravity \(9.81 \mathrm{~m/s^{2}}\). Therefore, \(F_{g} = 1600 \mathrm{~kg} \cdot 9.81 \mathrm{~m/s^{2}} = 15700 \mathrm{~N}\).\n

Calculate the net force on the elevator

The net force on the elevator will be the difference between the gravitational force and the retarding force. This can be written as: \(F_{n} = F_{g} - F_{r}\), where \(F_{g} = 15700 \mathrm{~N}\) and \(F_{r} = 3700 \mathrm{~N}\). Substituting these values, we get \(F_{n} = 15700 \mathrm{~N} - 3700 \mathrm{~N} = 12000 \mathrm{~N}\).

Calculate the work done by the net force

The work done by the net force can be calculated using the equation: \(W_{n} = F_{n} \cdot d\), where \(F_{n}\) is the net force on the elevator and \(d\) is the distance the elevator falls. Substituting the values \(F_{n} = 12000 \mathrm{~N}\) and \(d = 72 \mathrm{~m}\), we get \(W_{n} = 12000 \mathrm{~N} \cdot 72 \mathrm{~m} = 864000 \mathrm{J}\).

Calculate the speed of the elevator using the work-energy theorem

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. From this, we can solve for the final speed of the elevator: \(W = \Delta KE = \frac{1}{2} m (v_{f}^{2} - v_{i}^{2})\). Here, as the elevator started from rest, the initial velocity \(v_{i}\) is 0. Therefore, \(v_{f}^{2} = \frac{2W}{m}\). Substituting \(W = 864000 \mathrm{~J}\) and \(m = 1600 \mathrm{~kg}\), we get \(v_{f}^{2} = \frac{2 \cdot 864000 \mathrm{~J}}{1600 \mathrm{~kg}} = 1080 \mathrm{m^{2}/s^{2}}\). The final velocity \(v_{f}\) will be the square root of this, giving us \(v_{f} = \sqrt{1080} = 32.86 \mathrm{m/s}\).\n

Key concepts

Gravitational Force

Imagine an object, like an elevator, free-falling due to gravity. An unseen force pulls it straight towards the Earth's center - this is what we call gravitational force. It's an essential concept in physics, particularly when studying motion and mechanisms like elevators.
This force is calculated with a very straightforward formula: \( F_g = m \times g \), where \( m \) represents the mass of the object, and \( g \) represents the acceleration due to gravity, which on the surface of the Earth is roughly \( 9.81 \text{m/s}^2 \). The result is a force measured in newtons (N). For our elevator with a mass of 1600 kg, the gravitational force comes out to 15700 N. This force is what initially sets the elevator in motion once the cable snaps. Such knowledge is not only fundamental in physics but also critical for engineering safe systems - like elevators in buildings!

Net Force Calculation

When multiple forces act on an object, as in our scenario, the resultant effect is due to the net force. It's quite like summing up the influence of pushing a box from opposite sides. The direction and magnitude of each force determine where and how fast the box moves.
In the elevator example, the net force calculation takes into account both the gravitational pull downwards and the retarding force exerted by the guide rails. The formula used is: \( F_n = F_g - F_r \). Here, \( F_g \) is the gravitational force and \( F_r \) is the retarding or opposing force. The net force determines the acceleration of the elevator as it descends, and in our case, it's 12000 N. Calculating net force correctly is crucial, as it relates directly to the motion of the elevator - and ensuring our understanding aligns with the physical behavior of objects in real-world applications.

Work-Energy Theorem

In physics, we often link the concept of work to mechanical energy changes, and this linkage forms the basis of the work-energy theorem. The theorem tells us that the work done by the net force on an object results in a change in the object's kinetic energy. The formula \( W = \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2) \) quantifies this concept, stating that the work \( W \) equals the change in kinetic energy \( \Delta KE \).
The elevator's falling scenario provides a clear example. The work done by the net force is the force times the distance the elevator falls. This work leads to a change in kinetic energy, which equals the energy the elevator has due to its motion right before hitting the bottom. For the elevator, this work totaled 864000 J, all of which converted into kinetic energy. Applying the work-energy theorem lets us calculate the final speed of the elevator as it crashes. Understanding this theorem paints a full picture of how forces influence motion, which has practical implications in designing anything from elevators to roller coasters.