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Chapter 7: Problem 16

\(\cdot\) \(\cdot\) If a running house cat has 10.0 \(\mathrm{J}\) of kinetic energy at speed \(v,\) (a) At what speed (in terms of \(v )\) will she have 20.0 \(\mathrm{J}\) of kinetic energy? (b) What would her kinetic energy be if she ran half as fast as the speed in part (a)?

Short Answer

Expert verified
Speed is \( \sqrt{2}v \) for 20 J; kinetic energy is 5.0 J at half this speed.

Step by step solution

01

Understanding Kinetic Energy

The kinetic energy (KE) of an object is given by the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity or speed. In this problem, we know the initial kinetic energy of the cat is 10.0 J.
02

Setting Up the Speed Ratio

Since the cat's kinetic energy is directly proportional to the square of its speed, if we want to double the kinetic energy from 10.0 J to 20.0 J, the equation \( \frac{1}{2}m(2v')^2 = 20 \) must hold. We need to solve for the new speed \( v' \) in terms of \( v \).
03

Solving for New Speed \(v'\)

Doubling the kinetic energy, we set \( 20 = \frac{1}{2}m(v')^2 \). Since we know that initially \( 10 = \frac{1}{2}mv^2 \), we equate \( v' = \sqrt{2}v \). The new speed \( v' \) must rise by a factor of \( \sqrt{2} \).
04

Calculating Half-Speed Kinetic Energy

In part (b), running at half the speed found in part (a), the new speed is \( v' = \frac{1}{2}\sqrt{2}v = \frac{\sqrt{2}}{2}v = \frac{v}{\sqrt{2}} \). Substitute \( \frac{v}{\sqrt{2}} \) back into the kinetic energy formula: \( KE = \frac{1}{2}m\left(\frac{v}{\sqrt{2}}\right)^2 \).
05

Kinetic Energy Calculation for Half-Speed

Now calculate: \( KE = \frac{1}{2}m\left(\frac{v^2}{2}\right) = \frac{1}{4}mv^2 \). Given that \( \frac{1}{2}mv^2 = 10.0 \) J, \( \frac{1}{4}mv^2 \) is 5.0 J. Therefore, the kinetic energy when running at this speed is 5.0 Joules.

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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 motion of objects without considering the forces that cause the motion. It involves the description of an object's movement through basic terms like position, velocity, and acceleration.
Kinematics helps us understand how an object moves, how fast it goes, and how its speed changes over time. By learning about kinematics, we can predict how an object will behave in different physical scenarios.
For instance, if we want to know how a cat's speed affects its kinetic energy, understanding kinematic principles can let us calculate changes in velocity and how that translates to energy, as seen in the exercise about the running cat.
Motion Equations
Equations of motion are mathematical formulas that describe the relationship between an object's velocity, acceleration, time, and displacement. These equations can be used to predict future motion based on current movement.
In our exercise, the motion equation for kinetic energy is used: - \( KE = \frac{1}{2}mv^2 \)In this equation,:- \( KE \) represents kinetic energy- \( m \) is the mass of the object- \( v \) is the velocity or speed.This equation shows how an object's kinetic energy is proportional to its velocity squared. If we double the speed of the cat, its kinetic energy would quadruple because energy depends on \( v^2 \). This fundamental relationship helps us understand how speed changes affect energy levels.
Velocity Calculation
Velocity refers to the speed of an object in a specific direction. Calculating velocity involves determining both its magnitude (how fast the object is moving) and its direction.
In the original exercise, velocity calculation plays a crucial role in determining the cat's kinetic energy at different speeds. When the speed is required to double the kinetic energy from 10 J to 20 J, we solve for the velocity factor. This process involves:- Recognizing kinetic energy's proportionality to \( v^2 \)- Using the motion equation to find the new speed required to achieve doubled energy.As solved in the exercise, by doubling kinetic energy, new speed \( v' \) is calculated as \( v' = \sqrt{2}v \). This demonstrates how velocity calculations are key to understanding variations in kinetic energy.

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Most popular questions from this chapter

\(\bullet\) A force of magnitude 800.0 \(\mathrm{N}\) stretches a certain spring by 0.200 \(\mathrm{m}\) from its equilibrium position. (a) What is the force constant of this spring? (b) How much elastic potential energy is stored in the spring when it is: (i) stretched 0.300 \(\mathrm{m}\) from its equilibrium position and (ii) compressed by 0.300 \(\mathrm{m}\) from its equilibrium position? (c) How much work was done in stretch- ing the spring by the original 0.200 \(\mathrm{m} ?\)

\(\bullet\) \(\bullet\) You and three friends stand at the corners of a square whose sides are 8.0 \(\mathrm{m}\) long in the mid- dle of the gym floor, as shown in the accompanying figure. You take your physics book and You take your physics book and push it from one person to the other. The book has a mass of \(1.5 \mathrm{kg},\) and the coefficient of kinetic friction between the book and the floor is \(\mu_{k}=0.25\) . (a) The book slides from you to Beth and then from Beth to Carlos, along the lines connect- ing these people. What is the work done by friction during this displacement? (b) You slide the book from you to Carlos along the diagonal of the square. What is the work done by friction during this displacement? (c) You slide the book to Kim who then slides it back to you. What is the total work done by fric- tion during this motion of the book? (d) Is the friction force on the book conservative or nonconservative? Explain.

\(\cdot\) \(\cdot\) An 8.00 kg package in a mail-sorting room slides 2.00 \(\mathrm{m}\) down a chute that is inclined at \(53.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40 . Calculate the work done on the package by (a) friction (b) gravity, and (c) the normal force. (d) What is the net work done on the package?

\(\cdot\) \(\cdot\) Animal energy. Adult cheetahs, the fastest of the great cats, have a mass of about 70 \(\mathrm{kg}\) and have been clocked at up to 72 mph \((32 \mathrm{m} / \mathrm{s})\) . (a) How many joules of kinetic energy does such a swift cheetah have? (b) By what factor would its kinetic energy change if its speed were doubled?

\(\bullet\) \(\bullet\) Energy requirements of the body. A 70 \(\mathrm{kg}\) human uses energy at the rate of \(80 \mathrm{J} / \mathrm{s},\) on average, for just resting and sleeping. When the person is engaged in more strenuous activities, the rate can be much higher. (a) If the individual did nothing but rest, how many food calories per day would she or he have to eat to make up for those used up? (b) In what forms is energy used when a person is resting or sleep- ing? In other words, what happens to those 80 \(\mathrm{J} / \mathrm{s} ?\) Hint: What kinds of energy, mechanical and otherwise, do our body components have?) (c) If an average person rested and did other low-level activity for 16 hours (which consumes 80 \(\mathrm{J} / \mathrm{s} )\) and did light activity on the job for 8 hours (which consumes \(200 \mathrm{J} / \mathrm{s} ),\) how many calories would she or he have to con- sume per day to make up for the energy used up?
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