/*! 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 72 The Genesis Crash. When the 210 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 72

The Genesis Crash. When the 210 -kg Genesis Mission capsule crashed (see Exercise 5.17 in Chapter 5 ) with a speed of 311 \(\mathrm{km} / \mathrm{h}\) , it buried itself 81.0 \(\mathrm{cm}\) deep in the desert floor. Assuming constant acceleration during the crash, at what average rate did the capsule do work on the desert?

Short Answer

Expert verified
The average rate at which the capsule did work on the desert was 41.764 MW.

Step by step solution

01

Convert units

First, we need to convert the speed from kilometers per hour to meters per second. The initial speed is given as 311 km/h. We can use the conversion factor 1 km/h = 0.27778 m/s to solve this: \[ 311 \text{ km/h} \times 0.27778 \text{ m/s per km/h} = 86.389 \text{ m/s} \] Also, convert the distance from centimeters to meters: \[ 81 \text{ cm} = 0.81 \text{ m} \]
02

Use the work-energy principle

According to the work-energy principle, the work done on an object is equal to the change in its kinetic energy. Since the capsule comes to a stop, its final kinetic energy is zero. The initial kinetic energy \( KE_{initial} \) is given by: \[ KE_{initial} = \frac{1}{2} mv^2 \] where \( m = 210 \text{ kg} \) and \( v = 86.389 \text{ m/s} \). So, \[ KE_{initial} = \frac{1}{2} \times 210 \text{ kg} \times (86.389 \text{ m/s})^2 = 783,820.87785 \text{ J} \]
03

Calculate work done

The work done on the capsule by the desert equals the change in kinetic energy (since the final kinetic energy is zero, this is just the initial kinetic energy lost). Therefore, the work done is \[ W = 783,820.87785 \text{ J} \]
04

Determine the average force

To find the average force \( F \), we use the work formula through distance: \[ W = F \times d \] where \( d = 0.81 \text{ m} \). Rearranging gives: \[ F = \frac{W}{d} = \frac{783,820.87785 \text{ J}}{0.81 \text{ m}} = 967,674.967 \text{ N} \]
05

Calculate the average rate of work done

When computing the average rate \( P \) at which work was done, we need to use the formula: \[ P = F \times v_{average} \] The average speed in this scenario can be considered as roughly half of the initial speed because we are dealing with constant acceleration: \[ v_{average} = \frac{86.389 \text{ m/s}}{2} = 43.1945 \text{ m/s} \] Now calculate the power: \[ P = 967,674.967 \text{ N} \times 43.1945 \text{ m/s} = 41,764,495.2 \text{ W} = 41.764 \text{ MW} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. It depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy (\( KE \)) is\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass in kilograms (kg) and \( v \) is the velocity in meters per second (m/s). For example, if a capsule has a mass of 210 kg and a velocity of 86.389 m/s, you would plug these values into the formula to get the kinetic energy:
  • Mass (\( m \)): 210 kg
  • Velocity (\( v \)): 86.389 m/s
By calculating, the kinetic energy for the formula above, the capsule has approximately 783,820.88 Joules (J) of energy before impact. This energy plays a crucial role in understanding how much work was done when the capsule came to rest.
Average Rate of Work
The average rate of work refers to the amount of energy transferred or converted per unit of time during a process. In physics, work is often associated with the application of force directly through a distance. However, when looking at average rate, it's about how quickly that work is completed鈥攚hich is effectively the power. The formula to calculate work (\( W \)) is given as:\[ W = F \times d \]where \( F \) is the force in newtons and \( d \) is the distance in meters. When working with constant forces, the work done can be reflected as energy change.Average rate of work (or power, \( P \)) can be calculated using:\[ P = F \times v_{average} \]where \( v_{average} \) is the average speed over the distance. In this context, the capsule's average speed being about half of its initial velocity shows how effectively energy dissipates into the desert dry during its brief but drastic deceleration.
Constant Acceleration
Constant acceleration implies a steady change in velocity over time. This is a key concept when analyzing motion, especially during events like the Genesis crash, where the capsule buries itself into the ground. In contexts of motion such as the capsule crash, using constant acceleration allows the simplification of calculations concerning velocity and displacement because it assumes the speed decreases uniformly. When applying the work-energy principle, the constant acceleration simplifies to calculate average speed, which in this case is taken as half the initial speed, due to the linear deceleration during impact. This results in a straightforward computation of the work done over the distance until it halted completely.
Unit Conversion
Unit conversion is the process of converting one unit of measurement into another. It is essential in ensuring all measurements are expressed in the same units for accurate calculations. For the average rate of work calculation:
  • Speed Conversion: Transforming speed from kilometers per hour (km/h) to meters per second (m/s) is necessary as many physical equations, like kinetic energy and power, use standard SI units.
  • Distance Conversion: Converting centimeters (cm) to meters (m) enables coherent calculation with standard SI units.
  • Conversion factors ensure precision: 1 km/h = 0.27778 m/s and 1 cm = 0.01 m.
Such conversions are crucial not just for accuracy, but to maintain consistency across measurements, thereby avoiding any inaccuracies in the results derived.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Chin-Ups. While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m}\) . (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-k g\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.

Proton Bombardment. A proton with mass \(1.67 \times10^{-27} \mathrm{kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) directiy toward a uranium nucleus 5.00 \(\mathrm{m}\) away. The proton is repelled by the uranium nucleus with a force of magnitude \(F=\alpha / x^{2}\) , where \(x\) is the separation between the two objects and \(\alpha=2.12 \times\) \(10^{-26} \mathrm{N} \cdot \mathrm{m}^{2} .\) Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10} \mathrm{m}\) from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the pro- ton get? (c) What is the speed of the proton when it is again 5.00 \(\mathrm{m}\) away from the uranium nucleus?

Meteor Crater. About \(50,000\) years ago, a meteor crashed into the earth near present-day Flagstaff, Arizona. Recent \((2005)\) measurements estimate that this meteor had a mass of about \(1.4 \times 10^{3} \mathrm{kg}\) (around \(150,000\) tons) and hit the ground at 12 \(\mathrm{km} / \mathrm{s}\). (a) How much kinetic energy did this meteor deliver to the ground? (b) How does this energy compare to the energy released by a 1.0 -megaton nuclear bomb? (A megaton bomb releases the same energy as a million tons of TNT, and 1.0 ton of TNT releases \(4.184 \times 10^{9} \mathrm{J}\) of energy.)

A physics professor is pushed up a ramp inclined upward at \(30.0^{\circ}\) above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 85.0 \(\mathrm{kg}\) . He is pushed 2.50 \(\mathrm{m}\) along the incline by a group of students who together exert a constant horizontal force of 600 \(\mathrm{N}\) . The professor's speed at the bottom of the ramp is 2.00 \(\mathrm{m} / \mathrm{s}\) . Use the work-energy theorem to find his speed at the top of the ramp.

A force in the \(+x\) -direction has magnitude \(F=b / x^{n}\) , where \(b\) and \(n\) are constants. (a) For \(n>1\) , calculate the work done on a particle by this force when the particle moves along the \(x\) -axis from \(x=x_{0}\) to infinity. (b) Show that for \(0< n< 1,\) even though \(F\) becomes zero as \(x\) becomes very large, an infinite amount of work is done by \(F\) when the particle moves from \(x=x_{0}\) to infinity.
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation

Kinematics Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Rotational Dynamics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.