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Chapter 2: Problem 97

The single cable supporting an unocceupied construction evator breaks when the elevator is at rest at the top of a \(120-\mathrm{m}\) -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

Understanding the problem

The elevator falls from rest from a height of 120 meters. We need to find the speed at impact, the time of the fall, the speed halfway down, and the time it takes to reach halfway. We'll use basic kinematic equations, assuming no air resistance. The acceleration due to gravity is taken as \( g = 9.8 \ \text{m/s}^2 \).
02

Determine the speed on impact

Using the kinematic equation \( v^2 = u^2 + 2as \), where \( u = 0 \ \text{m/s} \) (initial speed), \( a = 9.8 \ \text{m/s}^2 \) (acceleration), and \( s = 120 \ \text{m} \) (distance). We aim to find \( v \):\[ v^2 = 0 + 2 \times 9.8 \times 120 \]\[ v^2 = 2352 \]\[ v = \sqrt{2352} \approx 48.5 \ \text{m/s} \]
03

Calculate the time of fall

Using the equation \( s = ut + \frac{1}{2}at^2 \), solve for \( t \) with \( u = 0 \), \( a = 9.8 \ \text{m/s}^2 \), and \( s = 120 \ \text{m} \):\[ 120 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 120 = 4.9t^2 \]\[ t^2 = \frac{120}{4.9} \approx 24.49 \]\[ t = \sqrt{24.49} \approx 4.95 \ \text{s} \]
04

Find the speed halfway down

At halfway, the elevator has fallen 60 meters. Using \( v^2 = u^2 + 2as \) with \( s = 60 \ \text{m} \):\[ v^2 = 0 + 2 \times 9.8 \times 60 \]\[ v^2 = 1176 \]\[ v = \sqrt{1176} \approx 34.3 \ \text{m/s} \]
05

Determine time to halfway point

Using \( s = ut + \frac{1}{2}at^2 \) for \( s = 60 \ \text{m} \), solve for \( t \):\[ 60 = \frac{1}{2} \times 9.8 \times t^2 \]\[ 60 = 4.9t^2 \]\[ t^2 = \frac{60}{4.9} \approx 12.24 \]\[ t = \sqrt{12.24} \approx 3.5 \ \text{s} \]

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Key Concepts

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

Free Fall
Free fall describes the motion of an object solely under the influence of gravity. This means there are no other forces, like air resistance, acting on the object. When a construction elevator's cable breaks, the elevator is in free fall. It falls from the top of a building, accelerating under gravity alone.
This scenario helps us study how objects behave when only gravity affects them. During free fall, objects accelerate downwards at a constant rate. This is because gravity pulls them down with consistent force. Importantly, this acceleration, known as the acceleration due to gravity, is generally taken as 9.8 m/s虏 on Earth.
In the exercise described, understanding free fall helps calculate how fast the elevator travels as it hits the ground and how long it takes to fall. Both calculations come from the principles of free fall.
Kinematic Equations
Kinematic equations allow us to predict the future position or velocity of an object based on its current state of motion. These are especially useful in problems involving constant acceleration, such as in our free-falling elevator scenario. The main kinematic equations relate to elements like time, velocity, distance, and acceleration.
Let's look at the common ones used in free fall problems:
  • Equation for velocity: \[ v = u + at \]where \( v \) is the final velocity, \( u \) is the initial velocity (0 in free fall from rest), \( a \) is acceleration (gravity), and \( t \) is time.
  • Equation for distance: \[ s = ut + \frac{1}{2}at^2 \]This calculates the distance an object travels under constant acceleration.
  • Equation for final velocity (using distance):\[ v^2 = u^2 + 2as \]Simplifies finding the speed at a given distance.
By applying these equations, the elevator's speed and fall time were calculated, as seen in the solution steps.
Acceleration Due to Gravity
Acceleration due to gravity is the rate at which speed increases per second as an object falls freely under gravity. On Earth, this is roughly 9.8 m/s虏, represented by the symbol \( g \). This value is crucial in all free fall calculations.
Gravity impacts all free-falling objects, regardless of their mass, equally in a vacuum.In the elevator example, \( g \) played a core role in determining both the speed of the elevator at different stages of its fall and the total time it spent falling. It dictated how quickly the speed increased every second.
By assuming no air resistance, these calculations remain straightforward, allowing for clear problem-solving using the kinematic equations. Knowing \( g \) helps ensure predictions and calculations closely match reality, which is fundamental for successfully applying physics in real-life situations.

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Most popular questions from this chapter

Two trains, each having a speed of \(30 \mathrm{~km} / \mathrm{h},\) are headed at each other on the same straight track. A bird that can fly \(60 \mathrm{~km} / \mathrm{h}\) flies off the front of one train when they are \(60 \mathrm{~km}\) apart and heads directly for the other train. On reaching the other train, the (crazy) bird flics directly hack to the first train, and so forth. What is the total distance the bird travcls before the trains collide?

At a certain time a particle had a speed of \(18 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction, and \(2.4 \mathrm{~s}\) later its speed was \(30 \mathrm{~m} / \mathrm{s}\) in the opposite direction. What is the average acceleration of the particle during this \(2.4 \mathrm{~s}\) interval?

In an arcade video game, a spot is programmed to move across the screen according to \(x=9.00 t-0.750 t^{3}\), where \(x\) is distance in centimeters measured from the left edge of the screen and \(t\) is time in seconds. When the spot reaches a screen edge, at either \(x=0\) or \(x=15.0 \mathrm{~cm}, t\) is reset to 0 and the spot starts moving again according to \(x(t)\). (a) At what time after starting is the spot instantaneously at rest? (b) \(A t\) what value of \(x\) does this occur? (c) What is the spot's acceleration (including sign) when this occurs? (d) 1st it moving right or left just prior to coming to rest? (e) Just after? (f) At what time \(t>0\) docs it first reach an edge of the screen?

(a) If a particlc's position is given by \(x=4-12 t+3 t^{2}\) (where \(t\) is in seconds and \(x\) is in meters), what is its velocity at \(t=1 \mathrm{~s} ?(\mathrm{~b})\) Is it moving in the positive or negative direction of \(x\) just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (c) Is there cver an instant when the velocity is zero? If so. give the time \(r\), if not, answer no. (f) Is there a time after \(t=3 \mathrm{~s}\) when the particle is moving in the ncgative dircction of \(x ?\) If so, give the time \(t\), if not, answer no.

The Zero Gravity Research Facility at the NASA Glenn Research Center includes a 145 m drop tower. This is an evacuated vertical tower through which, among other possibilities, a 1 -m-diameter sphere containing an experimental package can he dropped. (a) How long is the sphere in free fall? (b) What is its speed just as it reaches a catching device at the bottom of the tower? (c) When caught, the sphere experiences an average deceleration of \(25 \mathrm{~g}\) as its speed is reduced to zero. Through what distance does it travel during the deceleration?
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