/*! 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 67 Multiple-Concept Example 13 pres... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 67

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest-accelerating animals, because it can go from rest to 27 m/s (about 60 mi/h) in 4.0 s. If its mass is 110 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
The average power is 10023.75 watts or approximately 13.44 horsepower.

Step by step solution

01

Understand the Known Values

Given: The initial velocity of the cheetah, \( v_i = 0 \) m/s (as it starts from rest), final velocity \( v_f = 27 \) m/s, time \( t = 4.0 \) s, and mass \( m = 110 \) kg. We need to determine the average power in watts and horsepower.
02

Calculate the Acceleration

Use the formula for acceleration \( a = \frac{v_f - v_i}{t} \). Substituting the known values, we have \( a = \frac{27 \, \text{m/s} - 0 \, \text{m/s}}{4.0 \, \text{s}} = 6.75 \, \text{m/s}^2 \).
03

Calculate the Force

Force can be calculated using Newton's second law \( F = ma \). Thus, \( F = 110 \, \text{kg} \times 6.75 \, \text{m/s}^2 = 742.5 \, \text{N} \).
04

Calculate the Work Done

Work done is given by the formula \( W = Fd \), where \( d \) is the distance covered. Since \( d \) can be calculated from the kinematic equation \( d = v_it + \frac{1}{2}at^2 \), we have \( d = 0 + \frac{1}{2} \times 6.75 \, \text{m/s}^2 \times (4.0 \, \text{s})^2 = 54 \, \text{m} \). Finally, \( W = 742.5 \, \text{N} \times 54 \, \text{m} = 40095 \, \text{J} \).
05

Calculate the Average Power in Watts

Average power \( P \) is given by \( P = \frac{W}{t} \). So, \( P = \frac{40095 \, \text{J}}{4.0 \, \text{s}} = 10023.75 \, \text{W} \).
06

Convert Power to Horsepower

To convert watts to horsepower, use the conversion factor: 1 horsepower = 746 watts. Thus, \( \text{Power in HP} = \frac{10023.75 \, \text{W}}{746 \, \text{W/HP}} \approx 13.44 \, \text{HP} \).

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.

Kinematics
Kinematics is the branch of physics that deals with the analysis of motion without considering the forces that cause it. In this problem, we focus on how the cheetah moves from a stationary position to a speed of 27 m/s in a set duration of 4 seconds. To analyze this motion, one crucial equation from kinematics connects initial velocity, final velocity, time, and acceleration:
  • Acceleration ( \( a \)) can be found using the formula \( a = \frac{v_f - v_i}{t} \) where \( v_f \) is the final velocity, \( v_i \) is the initial velocity, and \( t \) is the time.
    This allows us to calculate how quickly the velocity of the cheetah changes over time.
  • Another kinematic equation, \( d = v_i t + \frac{1}{2} a t^2 \), helps us find the distance covered.
    Here, it calculates the distance the cheetah travels while accelerating.
These equations are vital for understanding motion dynamics and allow us to determine variables like distance and acceleration during uniform motion phases.
Newton's Laws of Motion
Newton's laws of motion are fundamental principles that explain how objects interact and move. Newton's second law is particularly pivotal in this exercise. It defines the relationship between force and motion, stating that:
  • The acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass: \( F = ma \) where \( F \) is force, \( m \) is mass, and \( a \) is acceleration.
Applying this law helps us determine the force the cheetah generates to achieve its acceleration:
  • The calculated acceleration from kinematics helps us multiply it by the mass of the cheetah to find the force.
  • This force contributes significantly to understanding how the cheetah converts energy into motion.
Thus, Newton's second law becomes essential for solving complex problems involving force and motion.
Energy Conversion
Energy conversion plays a vital role in understanding how the cheetah maintains its fast acceleration. As the cheetah sprints, it converts potential energy stored within its muscles into kinetic energy. This is an example of mechanical energy transformation.
  • Calculating Work Done: Work is the energy transferred when a force is applied over a distance. For the cheetah, it can be calculated using \( W = Fd \).
  • Here, the distance it covers while accelerating provides the platform over which the force acts.
  • Consequently, calculating the work done gives insight into how much energy has been converted to bring the cheetah to its final velocity.
This transformation is crucial to compute average power and highlights the importance of energy dynamics in motion scenarios.
Acceleration
Acceleration refers to the rate at which velocity changes over time. In this problem, it is crucial to determining how quickly the cheetah can increase its speed. A higher acceleration value signifies faster changes in speed, which is what allows the cheetah to reach 27 m/s so rapidly.
  • Determining Acceleration: Using the kinematic equation: \( a = \frac{v_f - v_i}{t} \).
  • This indicates how the cheetah's velocity evolves from 0 to 27 m/s over 4 seconds, resulting in an acceleration of 6.75 m/s虏.
  • The calculated acceleration further aids in determining the force using Newton's second law, \( F = ma \).
Acceleration is not just about speed increments, but understanding it helps assess the cheetah's capabilities in energy conversion and force generation dynamics.

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

A truck is traveling at 11.1 m/s down a hill when the brakes on all four wheels lock. The hill makes an angle of \(15.0^{\circ}\) with respect to the horizontal. The coefficient of kinetic friction between the tires and the road is 0.750. How far does the truck skid before coming to a stop?

A basketball of mass 0.60 kg is dropped from rest from a height of 1.05 m. It rebounds to a height of 0.57 m. (a) How much mechanical energy was lost during the collision with the floor? (b) A basketball player dribbles the ball from a height of 1.05 m by exerting a constant downward force on it for a distance of 0.080 m. In dribbling, the player compensates for the mechanical energy lost during each bounce. If the ball now returns to a height of 1.05 m, what is the magnitude of the force?

The (non conservative) force propelling a \(1.50 \times 10^{3}-kg\) car up a mountain road does \(4.70 \times 10^{6} J\) of work on the car. The car starts from rest at sea level and has a speed of 27.0 \(m /s\) sat an altitude of \(2.00 \times 10^{2} m\) above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces.

Starting from rest, a \(1.9 \times 10^{-4}-kg\) flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.38 \(N\) on it. This force does \(+2.4 \times 10^{-4} J\) of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.

During a tug-of-war, team A pulls on team B by applying a force of 1100 N to the rope between them. The rope remains parallel to the ground. How much work does team A do if they pull team B toward them a distance of 2.0 m?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Electromagnetism

Read Explanation

Rotational Dynamics

Read Explanation

Magnetism

Read Explanation

Particle Model of Matter

Read Explanation

Scientific Method Physics

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.