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The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a 120 -m-high building. (a) With what speed does the elevator strike the ground? (b) How long is it falling? (c) What is its speed when it passes the halfway point on the way down? (d) How long has it been falling when it passes the halfway point?

Short Answer

Expert verified
(a) 48.5 m/s, (b) 4.95 s, (c) 34.3 m/s, (d) 3.5 s.

Step by step solution

01

Understand the Problem

The exercise involves a falling elevator. We need to calculate its speed when it hits the ground, the time it takes to fall, its speed halfway down, and the time it takes to reach halfway. Key information includes the height (120 m) and the elevator starting from rest. Acceleration due to gravity is assumed to be \( g = 9.8 \text{ m/s}^2 \).
02

Calculate Final Speed (Part a)

For part (a), we use the equation \( v^2 = u^2 + 2gh \) to find the final speed \( v \) of the elevator. The initial speed \( u = 0 \), and \( h = 120 \text{ m} \). Substitute these into the equation: \( v^2 = 0 + 2 \times 9.8 \times 120 \). Solve for \( v \).
03

Solve for Final Speed

Substitute the given values: \( v^2 = 2 \times 9.8 \times 120 = 2352 \). Taking the square root of both sides, \( v = \sqrt{2352} \approx 48.5 \text{ m/s} \). Thus, the speed of the elevator when it hits the ground is approximately 48.5 m/s.
04

Calculate Time of Fall (Part b)

For part (b), we use the equation \( v = u + gt \) to find the time \( t \). With \( u = 0 \) and \( v = 48.5 \text{ m/s} \), solve the equation: \( 48.5 = 0 + 9.8t \).
05

Solve for Time of Fall

Rearrange the equation to solve for \( t \): \( t = \frac{48.5}{9.8} \approx 4.95 \text{ s} \). So, the elevator falls for approximately 4.95 seconds.
06

Calculate Speed at Halfway (Part c)

For part (c), we use the same equation \( v^2 = u^2 + 2gh \) to find the speed at halfway, where \( h = 60 \text{ m} \). Substitute: \( v^2 = 0 + 2 \times 9.8 \times 60 \).
07

Solve for Speed at Halfway

Calculate: \( v^2 = 1176 \). Solving for \( v \), we have \( v = \sqrt{1176} \approx 34.3 \text{ m/s} \). Therefore, the speed at halfway is approximately 34.3 m/s.
08

Calculate Time to Halfway (Part d)

For part (d), we use the equation \( s = ut + \frac{1}{2}gt^2 \) to find time \( t \) when \( s = 60 \text{ m} \). Simplify: \( 60 = \frac{1}{2} \times 9.8 \times t^2 \). Solve for \( t \).
09

Solve for Time to Halfway

Rearrange the equation: \( 60 = 4.9t^2 \) which gives \( t^2 = \frac{60}{4.9} \approx 12.24 \). Taking the square root, \( t \approx 3.5 \text{ s} \). Therefore, it takes approximately 3.5 seconds to reach halfway.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration Due to Gravity
In the realm of physics, acceleration due to gravity is a fundamental concept that plays a crucial role in kinematics. Gravity induces a constant acceleration on objects falling freely under its influence, such as the elevator in our exercise.
This acceleration is denoted by the symbol "g" and has a standard value of approximately \(9.8\, \text{m/s}^2\) on Earth. This means that the velocity of any freely falling object increases by about 9.8 meters per second every second, regardless of its mass, assuming air resistance is negligible.
Understanding this concept is vital because it allows us to calculate how long an object will fall, how fast it will move during the fall, and other motion characteristics. It is essential when solving problems involving free-fall and projectile motion.
Free Fall Motion
Free fall is a specific type of motion where the only force acting on the falling object is gravity. In the context of the exercise provided, when the elevator cable breaks, the elevator undergoes free fall motion.
Key characteristics of free fall include:
  • Uniform acceleration due to gravity, \(g = 9.8\, \text{m/s}^2\).
  • Initial velocity \(u = 0\, \text{m/s}\) if the object starts from rest.
  • No air resistance or other forces involved aside from gravity.
In free fall, the equations of motion become particularly important as they simplify calculations involving time, velocity, and displacement. Additionally, the notable aspect of free fall is that all objects, irrespective of mass, fall at the same rate when only gravity acts upon them.
Projectile Motion
Although the exercise does not directly address horizontal motion, understanding projectile motion helps in grasping the broader concepts at play. Projectile motion refers to the motion of an object thrown into the air, subject only to the acceleration of gravity.
This type of motion has two components:
  • Horizontal motion, which is uniform and does not experience acceleration.
  • Vertical motion, which is subject to the acceleration due to gravity.
While the exercise deals primarily with vertical motion, comprehending projectile motion can greatly enhance one's ability to analyze motion paths more holistically. In projectile motion, the independence of vertical and horizontal components allows for a comprehensive application of kinematic equations.
Equations of Motion
Equations of motion are essential tools in kinematics, particularly when analyzing the motion of objects under the influence of constant acceleration, such as gravity. These equations allow us to compute various aspects of motion, like velocity, displacement, and time.
There are three primary kinematic equations used in problems involving constant acceleration, like our elevator scenario:
  • \(v = u + gt\): This equation calculates the final velocity \(v\) after time \(t\) with initial velocity \(u\).
  • \(s = ut + \frac{1}{2}gt^2\): This finds the displacement \(s\) after time \(t\).
  • \(v^2 = u^2 + 2gs\): This calculates the final velocity \(v\) based on initial velocity \(u\) and displacement \(s\).
Applying these equations helps tackle various segments of motion problems, from determining how fast an object is moving to how long it will take to reach a certain point.

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Most popular questions from this chapter

A steel ball is dropped from a building's roof and passes a window, taking \(0.125 \mathrm{~s}\) to fall from the top to the bottom of the window, a distance of \(1.20 \mathrm{~m} .\) It then falls to a sidewalk and bounces back past the window, moving from bottom to top in \(0.125\) s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is \(2.00 \mathrm{~s}\). How tall is the building?

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