Question

While an elevator of mass 832 kg moves downward, the tension in the supporting cable is a constant \(7730 \mathrm{N}\) Between \(t=0\) and \(t=4.00 \mathrm{s},\) the elevator's displacement is \(5.00 \mathrm{m}\) downward. What is the elevator's speed at \(t=4.00 \mathrm{s} ?\)

Short answer

Answer: The speed of the elevator at the end of the 4-second time interval is -1.03 m/s (downward).

Step-by-step solution

Analyze the forces acting on the elevator

There are two main forces acting on the elevator: the tension in the supporting cable (upward direction) and the gravitational force due to the Earth (downward direction). The tension, denoted as T, is given as \(7730 \mathrm{N}\). The gravitational force, denoted as F_gravity, can be calculated as the mass of the elevator multiplied by gravitational acceleration (9.81 \(\frac{\mathrm{m}}{\mathrm{s}^2}\)), so it is: \(F_\mathrm{gravity} = m \times g = 832\,\mathrm{kg} \times 9.81\, \frac{\mathrm{m}}{\mathrm{s}^2} = 8157.92\, \mathrm{N}\).

Calculate the net force acting on the elevator

Newton's Second Law of Motion states that the net force acting on an object is equal to the mass of the object multiplied by the object's acceleration. The net force on the elevator can be calculated as the difference between the tension in the supporting cable and the gravitational force: \(F_\mathrm{net} = T - F_\mathrm{gravity} = 7730\, \mathrm{N} - 8157.92\, \mathrm{N} = -427.92\, \mathrm{N}\). The negative sign indicates that the net force is acting in the downward direction.

Calculate the acceleration of the elevator

Using Newton's Second Law of Motion, we can now calculate the acceleration of the elevator: \(F_\mathrm{net} = m \times a\), where "a" is the acceleration. Solving for acceleration, we have: \(a = \frac{F_\mathrm{net}}{m} = \frac{-427.92\, \mathrm{N}}{832\, \mathrm{kg}} = -0.514\, \frac{\mathrm{m}}{\mathrm{s}^2}\).

Determine the initial and final displacement

In the problem statement, it is given that over a time interval of 4 seconds, the elevator displaces 5 meters downward. So, the initial displacement (x_initial) of the elevator is 0 meters, and the final displacement (x_final) is -5 meters. The negative sign indicates that the displacement is in the downward direction.

Calculate the elevator's final speed

We can use the kinematics equation that relates final velocity, v_final, initial velocity, v_initial, displacement, time, and acceleration: x_final = x_initial + v_initial * t + 0.5 * a * t^2. Since the elevator starts from rest, v_initial = 0. Plugging in the known values, we have: \(-5\, \mathrm{m} = 0 + 0 - 0.5 \times (-0.514\, \frac{\mathrm{m}}{\mathrm{s}^2}) \times (4\, \mathrm{s})^2\). So, v_final * 4 = 4.12. Therefore, the elevator's speed at \(t = 4.00\, \mathrm{s}\) is \(v_\mathrm{final} = -1.03\, \frac{\mathrm{m}}{\mathrm{s}}\). The negative sign indicates that the velocity is in the downward direction.