Question

An elevator cabin is moving up, being pulled by an elevator cable under a tension of 7,250 Newtons. The combined mass of four people in an elevator is 165 kg, while the mass of an elevator cabin itself is 190 kg. Find the acceleration ( in m/s2 ) of the elevator cabin in the system of coordinate that has a positive vertical direction upward.

Answer 1

The acceleration of the elevator cabin, with the combined mass of the passengers and cabin being 355 kg and the tension in the cable being 7250 N, is 10.62 m/s² upwards.

Step-by-step explanation:

The question involves calculating the acceleration of an elevator cabin using Newton's second law of motion. Given the tension in the elevator cable and the masses of the passengers and the cabin, we need to determine the upward acceleration of the entire system.

Let's define the combined mass of the system (M) as the sum of the mass of the passengers (165 kg) and the mass of the cabin (190 kg), which gives us M = 165 kg + 190 kg = 355 kg. The net force exerted on the system in the upward direction equals the tension in the cable minus the weight of the system (Fnet = T - Mg), where g is the acceleration due to gravity (9.81 m/s2).

Using Newton's second law Fnet = Ma, where a is the acceleration we want to find, we can substitute Fnet and rearrange to solve for a:

• Fnet = T - Mg
• Ma = T - Mg
• a = (T - Mg) / M

Plugging in the values:

a = (7250 N - (355 kg × 9.81 m/s2)) / 355 kg

a = (7250 N - 3483.55 N) / 355 kg

a = 3766.45 N / 355 kg

a = 10.62 m/s2

So, the acceleration of the elevator cabin is 10.62 m/s2 in the upward direction.