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A person is standing at rest on level ground. How fast would she have to run to (a) double her total energy and (b) increase her total energy by a factor of 10?

Short Answer

Expert verified
Running at \( \sqrt{10} \times v_i \) increases energy by a factor of ten.

Step by step solution

01

Understanding Total Energy

Total energy in this context refers to the sum of potential and kinetic energy. Since the person is initially at rest and on level ground, we primarily consider kinetic energy, which is given by \( KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity. Initially, the kinetic energy is zero because the velocity is zero.
02

Define Initial Kinetic Energy

Let's assume the person starts with a velocity \( v_i = 0 \). For a simple situation, we assume the person's mass is \( m \). As they are at rest, their initial kinetic energy \( KE_i = \frac{1}{2} m (0)^2 = 0 \).
03

Express Energy Doubling

Doubling the total energy from rest on level ground means the kinetic energy should become \( KE_f = 2 \times KE_i = 2 \times 0 = 0 \). Therefore, theoretically, this scenario does not require running since a value from zero has no double.
04

Factor of 10 Increase in Total Energy

For the total energy to increase by a factor of 10, we use \( KE_f = 10 \times KE_i \). Initially with \( KE_i = 0 \), this equation implies \( KE_f = 10 \times 0 = 0 \). Therefore, any non-zero initial energy corresponds to a scenario that requires further clarification due to zero initial energies causing undefined behavior.
05

Clarifying Intent of Kinetic Change (Factor Increase)

Given the theoretical context applies poorly to zero energy levels, reinterpret assumptions by assuming initial kinetic energy represents a minimum threshold where \( v_i eq 0 \). Consider non-zero but miniscule initial velocity \( v_i eq 0 \). So \( KE_i = \frac{1}{2} mv_i^2 \). Doubling this requires solving new \( v_f \), ensuring calculations yield new velocity accurately within practical non-zero estimates.
06

Calculate New Velocity for Energy Increase

Set \( KE_f = 10\times KE_i = \frac{1}{2} m v_f^2\). Solve for \( v_f \) using non-zero assumptions: \[ v_f^2 = v_i^2 \times (\text{factoritude}) = v_i^2 \times 10 \]. Thus, as kinetic energy scales with \( v^2 \), the factor means \( v_f = v_i \times \sqrt{10} \).

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Key Concepts

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

Potential Energy
Potential energy is the energy stored in an object due to its position or state. In our exercise, since the person is standing on level ground, we focus on potential energy related to height and gravity. However, because they are not elevated, the potential energy from gravitational height is zero.

Potential energy can come from various sources, such as:
  • Gravitational potential energy – determined by height and weight.
  • Elastic potential energy – found in stretched or compressed springs.
  • Chemical potential energy – stored within chemical bonds.
Since the exercise focuses on someone standing on level ground, there is no height to contribute to gravitational potential energy. Thus, any increase in energy due to running must originate from kinetic energy derived from velocity.
Velocity
Velocity describes how fast an object is moving in a particular direction. In physics, it is a vector quantity, which means it has both magnitude and direction. For our exercise, the person is initially at rest with a velocity of zero.

As the person begins to run, their velocity changes, which directly affects their kinetic energy. The relationship between velocity and kinetic energy is given by:
  • Formula: \( KE = \frac{1}{2} mv^2 \)
  • \( m \) represents mass, while \( v \) is velocity.
To double or increase kinetic energy, the velocity must change. Doubling the energy means, theoretically, involving modifications to velocity. Yet, if initial kinetic energy is zero, the notion of doubling initiates from non-zero minimal starting points to assure practical calculations. Thus, velocity changes directly influence total energy calculations.
Total Energy
Total energy is the sum of potential energy and kinetic energy. For this setup, as the person is on level ground and at rest initially, total energy consists predominantly of kinetic energy once motion begins.

In this exercise, we explore changes to total energy through adjustments in kinetic energy, because potential energy remains constant at zero. The goal is to either:
  • Double the total energy, which theoretically remains unaltered from zero.
  • Increase the total energy by a factor of 10, achieved by raising kinetic energy through increased velocity.
When examining the factor increase, velocity plays a vital role due to its squared relation in kinetic energy. Thus, solving for new velocities involves calculations based on assumed non-zero beginnings, navigating from negligible starting points to accurately quantify energy increments. The exercise calculations involve aligning with real-world scenarios to ensure solutions reflect meaningful increases in total energy.

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Most popular questions from this chapter

A nuclear bomb containing 12.0 \(\mathrm{kg}\) of plutonium explodes. The sum of the rest masses of the products of the explosion is less than the original rest mass by one part in \(10^{4} .\) (a) How much energy is released in the explosion? (b) If the explosion takes place in 4.00\(\mu \mathrm{s}\) , what is the average power developed by the bomb? (c) What mass of water could the released energy lift to a height of 1.00 \(\mathrm{km} ?\)

An imperial spaceship, moving at high speed relative to the planet Arrakis, fires a rocket toward the planet with a speed of 0.920\(c\) relative to the spaceship. An observer on Arrakis measures that the rocket is approaching with a speed of 0.360\(c .\) What is the speed of the spaceship relative to Arrakis? Is the spaceship moving toward or away from Arrakis?

A spaceship moving at constant speed \(u\) relative to us broadcasts a radio signal at constant frequency \(f_{0 .}\) As the spaceship approaches us, we receive a higher frequency \(f ;\) after it has passed, we receive a lower frequency. (a) As the spaceship passes by, so it is instantaneously moving neither toward nor away from us, show that the frequency we receive is not \(f_{0}\) , and derive an expression for the frequency we do receive. Is the frequency we receive higher or lower than \(f_{0} ?\) (Hint: In this case, successive wave crests move the same distance to the observer and so they have the same transit time. Thus \(f\) equals 1\(/ T .\) Use the time dilation formula to relate the periods in the stationary and moving frames.) (b) A spaceship emits electromagnetic waves of frequency \(f_{0}=345 \mathrm{MHz}\) as measured in a frame moving with the ship. The spaceship is moving at a constant speed 0.758\(c\) relative to us. What frequency \(f\) do we receive when the spaceship is approaching us? When it is moving away? In each case what is the shift in frequency, \(f-f_{0} ?(\mathrm{c})\) Use the result of part (a) to calculate the frequency \(f\) and the frequency shift \(\left(f-f_{0}\right)\) we receive at the instant that the ship passes by us. How does the shift in frequency calculated here compare to the shifts calculated in part (b)?

Two events observed in a frame of reference \(S\) have positions and times given by \(\left(x_{1}, t_{1}\right)\) and \(\left(x_{2}, t_{2}\right),\) respectively. (a) Frame \(S^{\prime}\) moves along the \(x\) -axis just fast enough that the two events occur at the same position in \(S^{\prime} .\) Show that in \(S^{\prime},\) the time interval \(\Delta t^{\prime}\) between the two events is given by $$\Delta t^{\prime}=\sqrt{(\Delta t)^{2}-\left(\frac{\Delta x}{c}\right)^{2}}$$ where \(\Delta x=x_{2}-x_{1}\) and \(\Delta t=t_{2}-t_{1}\) . Hence show that if \(\Delta x>c \Delta t,\) there is \(n o\) frame \(S^{\prime}\) in which the two events occur at the same point. The interval \(\Delta t^{\prime}\) is sometimes called the proper time interval for the events. Is this term appropriate? (b) Show that if \(\Delta x>c \Delta t,\) there is a different frame of reference \(S^{\prime}\) in which the two events occur simultaneously. Find the distance between the two events in \(S^{\prime} ;\) express your answer in terms of \(\Delta x, \Delta t,\) and \(c\) This distance is sometimes called a proper length. Is this term appropriate? (c) Two events are observed in a frame of reference \(S^{\prime}\) to occur simultaneously at points separated by a distance of 2.50 \(\mathrm{m} .\) In a second frame \(S\) moving relative to \(S^{\prime}\) along the line joining the two points in \(S^{\prime},\) the two events appear to be separated by 5.00 \(\mathrm{m} .\) What is the time interval between the events as measured in \(S ?[\)Hint : Apply the result obtained in part (b).]

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