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An elevator is moving upward at a constant speed of2.50 \(\mathrm{m} / \mathrm{s} .\) A bolt in the elevator ceiling 3.00 \(\mathrm{m}\) above the elevator floor works loose and falls. (a) How long does it take for the bolt to fall to the elevator floor? What is the speed of the bolt just as it hits the elevator floor (b) according to an observer in the elevator? (c) According to an observer standing on one of the floor landings of the building? (d) According to the observer in part (c), what distance did the bolt travel between the ceiling and the floor of the elevator?

Short Answer

Expert verified
(a) 0.782 s, (b) 7.67 m/s, (c) 10.17 m/s, (d) 4.955 m.

Step by step solution

01

Understanding the Problem

We are given that the elevator is moving upward at a constant speed of \(2.50\, \text{m/s}\). A bolt falls from a ceiling inside the elevator 3.00 meters above the floor. We need to calculate the time it takes for the bolt to fall down and hit the floor, the speed of the bolt just before impact according to different observers, and the distance traveled by the bolt as seen by an observer outside the elevator.
02

Time of Fall for the Bolt (Observer in Elevator)

Inside the elevator, the bolt is in freefall with no initial vertical velocity relative to the elevator itself. The distance \( s \) it falls is 3.00 m, and it accelerates at \( g = 9.81\, \text{m/s}^2 \) under gravity. We use the equation \( s = \frac{1}{2} g t^2 \) to find the time \( t \):\[3.00 = \frac{1}{2} \times 9.81 \times t^2\]Solving for \( t \), we get:\[t = \sqrt{\frac{2 \times 3.00}{9.81}} \approx 0.782 \text{ s}\]
03

Speed of the Bolt Before Impact (Observer in Elevator)

For the observer inside the elevator, the bolt starts from rest and accelerates due to gravity. We use the formula \( v = gt \) to find the speed \( v \) of the bolt just before it impacts:\[v = 9.81 \times 0.782 \approx 7.67\, \text{m/s}\]
04

Speed of the Bolt Before Impact (Observer on Floor Landing)

For this observer, the bolt has an initial upward velocity of \(2.50\, \text{m/s}\) (the same as the elevator) and the gravitational acceleration affects it as well. The final speed \( v \) is given by:\[v = 2.50 + 9.81 \times 0.782 \approx 10.17\, \text{m/s}\]
05

Distance Traveled by the Bolt (Observer on Floor Landing)

The distance the bolt travels relative to the observer outside is more complex because it includes the initial velocity of the elevator. We use the equation \( s = v_0 t + \frac{1}{2} a t^2 \):\[s = 2.50 \times 0.782 + \frac{1}{2} \times 9.81 \times (0.782)^2\]Calculating each term:- \(2.50 \times 0.782 = 1.955\, \text{m}\)- \(\frac{1}{2} \times 9.81 \times 0.782^2 = 3.00\, \text{m}\)Adding these gives \(4.955\, \text{m}\).

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Key Concepts

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

Kinematics
Kinematics is the branch of classical mechanics that studies the motion of objects without considering the causes of this motion, such as forces. It focuses on the description of motion, expressed mathematically through parameters like displacement, velocity, and acceleration.
Kinematics equations allow us to calculate various motion characteristics when certain variables are known. These equations can evaluate how objects move along straight paths, such as the path of a free-falling object from rest. In such problems, we often consider factors like initial velocity, often designated as \( v_0 \), final velocity \( v \), time \( t \), displacement \( s \), and acceleration \( a \), typically due to gravity in free-fall scenarios.
When you know three of these factors, you can find the fourth. For instance, if you know the height from which an object falls and the gravitational acceleration, you can determine how long it takes for that object to hit the ground.
Free Fall
Free fall refers to the motion of an object where gravity is the only influence acting upon it. This type of motion is a special case of kinematics, where the object experiences constant acceleration. Earth's gravitational acceleration \( g \) is approximately \( 9.81 \, \text{m/s}^2 \).
In our exercise, the bolt begins from rest relative to the elevator, so its initial velocity \( v_0 \) in the frame of the elevator is zero. The only acceleration affecting it is gravitational, causing the bolt to speed up as it falls to the elevator floor. By using the kinematic equation \( s = \frac{1}{2} g t^2 \), we can determine the time \( t \) it takes to fall a distance \( s \).
This equation illustrates how free-falling objects undergo uniform acceleration, simplifying the prediction of their motion when air resistance is negligible.
Relative Motion
Relative motion examines how the movement of one object seems different when viewed from various frames of reference. This principle allows us to relate motions observed in different coordinate systems relative to each other.
In the given problem, the bolt's motion appears different to observers moving with the elevator and those standing still on a floor landing. For an observer inside the elevator, the bolt seems to start from rest and accelerate straight downward, whereas an outside observer sees it initially moving upward with the elevator's speed and then influenced by gravity.
This difference arises because the relative motion includes the velocity of the reference frame itself; the external observer must consider the elevator's motion in addition to the bolt's subsequent downward acceleration.
Observation Frames
Observation frames refer to specific views or locations from which motion is observed. Identifying the frame is crucial to understand and calculate the motion characteristics accurately.
In classical mechanics, the choice of observation frames affects how we perceive the kinematics of an object. In the exercise, different frames lead to distinct calculations. For the observer in the elevator, the falling bolt's motion and speed are measured without considering the elevator's constant upward speed. Meanwhile, someone on a fixed floor experiences a frame that does take this motion into account, versus only the acceleration due to gravity.
The use of observation frames thus highlights the importance of perspective in analyzing physical phenomena like motion, helping us understand the varied experiences of observers in different contexts.

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Most popular questions from this chapter

The coordinates of a bird flying in the \(x y\) -plane are given by \(x(t)=\alpha t\) and \(y(t)=3.0 \mathrm{m}-\beta t^{2},\) where \(\alpha=2.4 \mathrm{m} / \mathrm{s}\) and \(\beta=1.2 \mathrm{m} / \mathrm{s}^{2}\) (a) Sketch the path of the bird between \(t=0\) and \(t=2.0 \mathrm{s} .\) (b) Calculate the velocity and acceleration vectors of the bird as functions of time. (c) Calculate the magnitude and direction of the bird's velocity and acceleration at \(t=2.0 \mathrm{s}\) . (d) Sketch the velocity and acceleration vectors at \(t=2.0 \mathrm{s}\) . At this instant, is the bird speeding up, is it slowing down, or is its speed instantaneously not changing? Is the bird turning? If so, in what direction?

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