/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Jumping up before the elevator h... [FREE SOLUTION] | 91Ó°ÊÓ

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Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height of \(36 \mathrm{~m}\). During the collision at the bottom of the elevator shaft, a \(90 \mathrm{~kg}\) passenger is stopped in \(5.0 \mathrm{~ms}\). (Assume that neither the passenger nor the cab rebounds.) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

Short Answer

Expert verified
(a) Impulse without jumping: 2391.3 kg·m/s (b) Average force without jumping: 478260 N (c) Impulse with jumping: 1761.3 kg·m/s (d) Average force with jumping: 352260 N.

Step by step solution

01

Understand the Problem

The problem involves calculating the impulse and average force on a passenger inside an elevator cab that is free-falling. First, we'll consider the passenger's situation without jumping, and then calculate if they jump upward with some speed relative to the cab before impact.
02

Determine Final Velocity of the Cab

The cab is in free-fall from a height of 36 m. We use the formula \( v = \sqrt{2gh} \) to find the velocity just before impact, where \( g = 9.8 \ m/s^2 \) is the acceleration due to gravity. So, \( v = \sqrt{2 \times 9.8 \times 36} \approx 26.57 \ m/s \).
03

Calculate Initial Scenario Impulse (Impulse without Jumping)

Impulse is given by the change in momentum, \( J = \Delta p = m(v_f - v_i) \). The final velocity \( v_f = 0 \) after the stopping, and the initial velocity \( v_i = 26.57 \ m/s \). So, \( J = 90 \times (0 - 26.57) = -2391.3 \ kg\cdot m/s \). Magnitude is \( 2391.3 \ kg\cdot m/s \).
04

Calculate Initial Scenario Average Force (Force without Jumping)

Average force can be calculated from the impulse using the formula \( F_{avg} = \frac{J}{\Delta t} \). Given \( \Delta t = 5 \ ms = 0.005 \ s \), \( F_{avg} = \frac{2391.3}{0.005} = 478260 \ N \).
05

Adjust for Jumping (Velocity Adjustment)

If the passenger jumps with a velocity of \( 7.0 \ m/s \) upwards just before the impact, we adjust the initial velocity relative to ground: \( v_i = 26.57 \ - 7.0 = 19.57 \ m/s \).
06

Calculate Jumping Scenario Impulse

Using the new initial velocity \( v_i = 19.57 \ m/s \), \( J = m(v_f - v_i) = 90 \times (0 - 19.57) = -1761.3 \ kg\cdot m/s \). Magnitude is \( 1761.3 \ kg\cdot m/s \).
07

Calculate Jumping Scenario Average Force

For the new impulse, \( F_{avg} = \frac{1761.3}{0.005} = 352260 \ N \).

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Key Concepts

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

Impulse and Momentum
In physics, impulse is an important concept that describes how much force is applied over a period of time to change an object's momentum. When solving problems about impulse, it helps to understand the relationship between impulse and momentum.
Impulse is calculated using the formula:
  • \[ J = \Delta p \]
  • Where \( J \) is the impulse, and \( \Delta p \) is the change in momentum
Momentum is a measure of an object's motion, and it depends on the object's mass and velocity. The change in momentum can be expressed using:
  • \[ \Delta p = m(v_f - v_i) \]
  • Where \( m \) is the mass, \( v_f \) is the final velocity, and \( v_i \) is the initial velocity.
For the elevator scenario, when the passenger is stopped, the final velocity is zero. This abrupt stop changes the passenger's momentum to zero as well, thus helping us calculate the total impulse experienced.
Free Fall
Free fall refers to the motion of an object under the sole influence of gravity. In this scenario, the elevator cab is in free fall after the safety cable snaps. Understanding free fall involves recognizing that objects accelerate towards the earth with a constant acceleration due to gravity, known as \( g \), which equals \( 9.8 \, m/s^2 \).
A key calculation in free fall problems is determining the velocity just before impact if the object starts from rest. The formula used is:
  • \[ v = \sqrt{2gh} \]
  • Where \( h \) is the height from which the object falls, and \( v \) is the final velocity just before impact.
For the elevator problem, it falls from a height of 36 meters. Plugging the values into the formula gives the cab a velocity of approximately 26.57 m/s at the moment before it hits the shaft, which we use to find the momentum of the passenger.
Average Force Calculation
Average force is another crucial concept linked to impulse in physics. It is the total impulse applied to an object divided by the time duration over which the force is applied. The formula for average force is:
  • \[ F_{avg} = \frac{J}{\Delta t} \]
  • Where \( F_{avg} \) is the average force, \( J \) is the impulse, and \( \Delta t \) is the time duration.
In the problem about the elevator, it states the stopping time is 5 milliseconds (converted to seconds as 0.005 s). Using this time and the calculated impulse, you can compute the average force exerted on the passenger. In cases where understanding this concept is key, it puts into perspective how sudden changes—like a short collision time—can result in large forces.
Physics Formulas
Physics problems often rely on applying basic formulas to find solutions. It's essential to be comfortable with using and interpreting these formulas correctly. In exercises involving impulse and free fall, several critical formulas often appear:
  • The free fall velocity formula: \( v = \sqrt{2gh} \)
  • The impulse-momentum theorem: \( J = m(v_f - v_i) \)
  • The formula for average force: \( F_{avg} = \frac{J}{\Delta t} \)
Each formula serves a specific role:
The free fall velocity helps determine how fast the object is moving before hitting the ground.
The impulse-momentum theorem links the effect of forces over time to resultant changes in motion.
Finally, the average force calculation conveys how significant force is applied during impacts or stops. Understanding each part of these formulas and how they interconnect is paramount to solving physics problems efficiently.

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Most popular questions from this chapter

Shows a \(0.300 \mathrm{~kg}\) baseball just before and just after it collides with a bat. Just before, the ball has velocity \(\vec{v}_{1}\) of magnitude \(12.0 \mathrm{~m} / \mathrm{s}\) and angle \(\theta_{1}=35.0^{\circ} .\) Just after, it is traveling directly upward with velocity \(\vec{v}_{2}\) of magnitude \(10.0 \mathrm{~m} / \mathrm{s}\). The duration of the collision is \(2.00 \mathrm{~ms}\). What are the (a) magnitude and (b) direction (relative to the positive direction of the \(x\) axis) of the impulse on the ball from the bat? What are the (c) magnitude and (d) direction of the average force on the ball from the bat?

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