/*! 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 10 A 60 kg runner in a sprint moves... [FREE SOLUTION] | 91Ó°ÊÓ

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A 60 kg runner in a sprint moves at 11 m/s. A 60 kg cheetah in a sprint moves at \(33 \mathrm{m} / \mathrm{s} .\) By what factor does the kinetic energy of the cheetah exceed that of the human runner?

Short Answer

Expert verified
The kinetic energy of the cheetah is approximately 9 times larger than the kinetic energy of the human runner.

Step by step solution

01

Calculate the kinetic energy of the human runner

The kinetic energy formula is \(0.5 \times \text{mass} \times (\text{speed})^2\). Substituting values: E_human = \(0.5 \times 60 \mathrm{kg} \times (11 \mathrm{m/s})^2 =\) 3630 J
02

Calculate the kinetic energy of the cheetah

Similarly, substituting values for the cheetah in the kinetic energy formula, E_cheetah = \(0.5 \times 60 \mathrm{kg} \times (33 \mathrm{m/s})^2 =\) 32670 J
03

Calculate the ratio of cheetah's kinetic energy to the human's kinetic energy

The ratio which shows by how much factor the cheetah's energy exceeds the human's energy can be calculated as: \( \frac{E_{\text{cheetah}}}{E_{\text{human}}} = \frac{32670 \mathrm{J}}{3630\mathrm{J}} \approx 9\)

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Key Concepts

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

Physics Education
Understanding kinetic energy is a fundamental concept in physics education. Kinetic energy is the energy that an object possesses due to its motion, often represented by the symbol KE. It's important for students to comprehend how different variables such as mass and velocity impact this energy. According to the kinetic energy formula:
\[ KE = 0.5 \times \text{mass} \times (\text{velocity})^2 \]The mass of an object is directly proportional to its kinetic energy, meaning more mass results in more energy if the speed remains constant. Velocity plays an even more significant role as it is squared in the formula, making the object’s speed a crucial factor in energy calculations.
These insights are essential for students to visualize real-world scenarios, like comparing the kinetic energy of different sprints involving a runner and a cheetah. Applying these concepts helps in analyzing how variations in speed lead to significant changes in energy.
Energy Calculation
Calculating kinetic energy involves substituting the known values into the kinetic energy formula. For energy calculations in real-life scenarios, such as the sprint of a runner and a cheetah, the accurate application of the formula is key.
Following the exercise, the formula for the human runner is calculated as:
  • Mass = 60 kg
  • Velocity = 11 m/s
  • KE = 0.5 x 60 x (11)^2 = 3630 J
Similarly, for the cheetah:
  • Mass = 60 kg
  • Velocity = 33 m/s
  • KE = 0.5 x 60 x (33)^2 = 32670 J
These calculations illustrate how much greater kinetic energy the faster-moving cheetah possesses compared to the human. Such problem-solving exercises enhance students' practical understanding of energy calculations, boosting their analytical skills in evaluating motion dynamics.
Motion Analysis
Motion analysis refers to the study and evaluation of the movement dynamics of objects. When analyzing motion, students can assess how velocity and mass contribute to the overall energy. In this scenario, the focus is on determining the factor by which the cheetah's kinetic energy exceeds the runner's.
By computing the ratio of their kinetic energies:\[\text{Factor} = \frac{KE_{\text{cheetah}}}{KE_{\text{human}}} = \frac{32670}{3630} \approx 9\]This comparison reveals that the cheetah’s kinetic energy is approximately nine times that of the runner. Such exercises are fundamental in motion analysis as they illuminate the pronounced effect velocity has on kinetic energy, especially when the same mass is involved.
By understanding these principles, students better appreciate how speed can drastically affect the energy profile of moving entities, whether they are athletes or animals.

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Most popular questions from this chapter

A 50 g ball of clay traveling at \(6.5 \mathrm{m} / \mathrm{s}\) hits and sticks to a \(1.0 \mathrm{kg}\) block sitting at rest on a frictionless surface. a. What is the speed of the block after the collision? b. Show that the mechanical energy is not conserved in this collision. What percentage of the ball's initial kinetic energy is "lost"? Where did this kinetic energy go?

When you ride a bicycle at constant speed, almost all of the energy you expend goes into the work you do against the drag force of the air. In this problem, assume that all of the energy expended goes into working against drag. As we saw in Section \(5.6,\) the drag force on an object is approximately proportional to the square of its speed with respect to the air. For this problem, assume that \(F \propto v^{2}\) exactly and that the air is motionless with respect to the ground unless noted Suppose a cyclist and her bicycle have a combined mass of \(60 \mathrm{kg}\) and she is cycling along at a speed of \(5 \mathrm{m} / \mathrm{s}\). If the drag force on the cyclist is \(10 \mathrm{N},\) how much energy does she use in cycling \(1 \mathrm{km} ?\) A. \(6 \mathrm{kJ}\) B. \(10 \mathrm{kJ}\) C. \(50 \mathrm{kJ}\) D. \(100 \mathrm{kJ}\)

Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-section area of \(0.92 \mathrm{m}^{2}\) and a drag coefficient of \(0.0091 \longrightarrow\) xceptionally low due to a number of adaptations. Such a fish can sustain a speed of \(30 \mathrm{m} / \mathrm{s}\) for a few seconds. Assume seawater has a density of \(1026 \mathrm{kg} / \mathrm{m}^{3}\). What is the specific power for motion at this high speed?

In an amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is \(3.0 \mathrm{m}\) above the bottom where they exit the slide, moving horizontally, \(1.2 \mathrm{m}\) above a swimming pool. What horizontal distance do they travel from the exit point before hitting the water? Does the mass of the person make any difference?

The famous cliff divers of Acapulco leap from a perch \(35 \mathrm{m}\) above the ocean. How fast are they moving when they reach the water surface? What happens to their kinetic energy as they slow to a stop in the water?

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