/*! 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 59 The cheetah is one of the fastes... [FREE SOLUTION] | 91Ó°ÊÓ

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The cheetah is one of the fastest-accelerating animals, because it can go from rest to \(27 \mathrm{~m} / \mathrm{s}\) (about 60 \(\mathrm{mi} / \mathrm{h}\) ) in \(4.0 \mathrm{~s}\). If its mass is \(110 \mathrm{~kg}\), determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower.

Short Answer

Expert verified
10023.75 Watts, 13.44 Horsepower

Step by step solution

01

Understand the Problem Statement

The problem involves calculating the average power developed by a cheetah as it accelerates from rest to a certain speed over a given time. We know its mass, final velocity, and time taken.
02

Calculate Acceleration

Use the formula for acceleration: \[ a = \frac{v_f - v_0}{t} \]where \(v_f = 27\,\text{m/s} \) is the final velocity, \( v_0 = 0\,\text{m/s} \) is the initial velocity, and \( t = 4.0\,\text{s} \) is the time period.Substituting the values:\[ a = \frac{27 - 0}{4.0} = 6.75 \text{ m/s}^2 \]
03

Calculate the Work Done

Use the work-energy principle which states that the work done is equal to the change in kinetic energy:\[ W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_0^2 \]where \( m = 110\,\text{kg} \), \( v_f = 27\,\text{m/s} \), and \( v_0 = 0\,\text{m/s} \).Substituting the values:\[ W = \frac{1}{2} \times 110 \times 27^2 - \frac{1}{2} \times 110 \times 0^2 \]\[ W = \frac{1}{2} \times 110 \times 729 \]\[ W = 40095 \text{ J} \]
04

Calculate Average Power in Watts

Average power \( P \) is given by the work done over time:\[ P = \frac{W}{t} = \frac{40095}{4} \]\[ P = 10023.75 \text{ W} \]
05

Convert Power from Watts to Horsepower

To convert power from watts to horsepower, use the conversion factor: \( 1 \text{ horsepower} = 746 \text{ watts} \).\[ P = \frac{10023.75}{746} \approx 13.44 \text{ hp} \]

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Key Concepts

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

Acceleration
Acceleration is a measure of how quickly an object's velocity changes. When we talk about acceleration, it's important to remember that it's a vector quantity. This means it has both magnitude and direction. For the cheetah problem, we're interested in how the cheetah speeds up from rest to its final velocity.To find acceleration, we use the formula:\[ a = \frac{v_f - v_0}{t} \]where:- \( v_f \) is the final velocity, \( 27 \, \text{m/s} \) for the cheetah.- \( v_0 \) is the initial velocity, \( 0 \, \text{m/s} \) because the cheetah starts from rest.- \( t \) is the time over which this change occurs, \( 4.0 \, \text{s} \).Plugging in these values, we find that the cheetah's acceleration is \( 6.75 \, \text{m/s}^2 \). This tells us how fast the cheetah increases its speed every second.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It's given by the formula:\[ KE = \frac{1}{2} m v^2 \]where:- \( m \) is the mass of the object.- \( v \) is the velocity of the object.For our cheetah, with a mass of \( 110 \, \text{kg} \) and a final velocity of \( 27 \, \text{m/s} \), we calculated:\[ KE = \frac{1}{2} \times 110 \times 27^2 = 40095 \, \text{J} \]This represents the kinetic energy when the cheetah reaches its top speed. Energy is a key factor when looking at the cheetah's performance, as it reflects how much work has been done to accelerate it to this speed.
Conversion between Units
In physics, being able to convert units correctly is crucial. In this problem, we needed to convert the cheetah's power from watts to horsepower.Remember that power, which is the rate of doing work, is initially given in watts \((W)\), the standard unit in the International System of Units. However, horsepower, another common unit, is often used in contexts involving engines or powerful animals like our cheetah.To convert between these units, use the conversion factor:- \( 1 \, \text{horsepower} = 746 \, \text{watts} \)Thus, if the cheetah generates \( 10023.75 \, \text{W} \), we follow:\[ \text{Horsepower} = \frac{10023.75}{746} \approx 13.44 \, \text{hp} \]This shows how much power the cheetah would need in terms of engines and other power sources, providing a tangible comparison.
Work-Energy Principle
The work-energy principle is pivotal in understanding how energy changes in a system. It states that the work done on an object is equal to the change in its kinetic energy.For the cheetah, this means as it accelerates, the work done by its muscles converts into kinetic energy. This can be calculated by:\[ W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_0^2 \]Given the initial and final velocity are \( 0 \, \text{m/s} \) and \( 27 \, \text{m/s} \) respectively, and the mass is \( 110 \, \text{kg} \), the work done becomes:\[ W = 40095 \, \text{J} \]This direct link between work and energy helps explain how the cheetah can achieve such rapid acceleration, transforming stored energy into the energy of motion effortlessly.

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Most popular questions from this chapter

A \(2.40 \times 10^{2}-\mathrm{N}\) force is pulling an 85.0 -kg refrigerator across a horizontal surface. The force acts at an angle of \(20.0^{\circ}\) above the surface. The coefficient of kinetic friction is 0.200 , and the refrigerator moves a distance of \(8.00 \mathrm{~m}\). Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

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Multiple-Concept Example 4 and Interactive LearningWare 6.2 at review the concepts that are important in this problem. \(\mathrm{A} 5.0 \times 10^{4}-\mathrm{kg}\) space probe is traveling at a speed of \(11000 \mathrm{~m} / \mathrm{s}\) through deep space. Retrorockets are fired along the line of motion to reduce the probe's speed. The retrorockets generate a force of \(4.0 \times 10^{5} \mathrm{~N}\) over a distance of \(2500 \mathrm{~km}\). What is the final speed of the probe?

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