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(I) An \(85-\mathrm{kg}\) football player traveling \(5.0 \mathrm{~m} / \mathrm{s}\) is stopped in \(1.0 \mathrm{~s}\) by a tackler. ( \(a\) ) What is the original kinetic energy of the player? \((b)\) What average power is required to stop him?

Short Answer

Expert verified
(a) 1062.5 J, (b) 1062.5 W

Step by step solution

01

Calculate the Original Kinetic Energy

The original kinetic energy (KE) of the player can be calculated using the formula for kinetic energy: \[ KE = \frac{1}{2}mv^2 \]where \( m = 85 \, \text{kg} \) is the mass of the player and \( v = 5.0 \, \text{m/s} \) is the velocity of the player.\[ KE = \frac{1}{2} \times 85 \, \text{kg} \times (5.0 \, \text{m/s})^2 \]\[ KE = \frac{1}{2} \times 85 \times 25 \]\[ KE = \frac{1}{2} \times 2125 \]\[ KE = 1062.5 \, \text{J} \]
02

Calculate the Average Power Required

Average power (P) required to stop the player is given by the formula:\[ P = \frac{\Delta E}{\Delta t} \]where \( \Delta E \) is the change in energy (which is equal to the original kinetic energy since the player is stopped) and \( \Delta t = 1.0 \, \text{s} \) is the time over which the energy change happens.\[ P = \frac{1062.5 \text{ J}}{1.0 \text{ s}} \]\[ P = 1062.5 \, \text{W} \]

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Key Concepts

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

Power
Power in physics is the rate at which work is done or energy is transferred over time. When it comes to stopping moving objects, like the football player in this exercise, power is crucial because it determines how quickly kinetic energy is removed from the system. In scenarios like this, power can be calculated using the formula:
  • \( P = \frac{\Delta E}{\Delta t} \), where \( \Delta E \) is the change in energy and \( \Delta t \) is the time interval.
Think of power as the intensity of energy use. The more power applied, the quicker the kinetic energy is used up, stopping the player faster. The time to stop, in this case, is 1 second, making the calculated power \( 1062.5 \, \text{W} \). Power is measured in watts, which equals one joule per second. It's essential to understand the context, as higher power implies a more intense or rapid process of coming to a stop.
Physics
Physics is a branch of science that involves the study of matter and its motion through space and time, along with related concepts like energy and force. In this exercise, physics is at play when considering how an 85-kg football player, moving with a velocity of 5 m/s, comes to a stop. This encompasses concepts like motion, force, work, and ultimately how these translate into formulating kinetic energy and power equations. Essentially, physics provides the tools to understand and predict these real-world interactions, allowing us to calculate terms like kinetic energy and power. Physics is grounded in precise mathematical formulations, providing reliable predictions that help us make sense of motion, from mundane tasks to complex systems.
Mechanics
Mechanics, a subfield of physics, deals specifically with the motion and behavior of physical bodies when exposed to forces or displacements. It's the essential backdrop for any problem involving kinetic energy and power. In our case, the mechanics of how a football player's kinetic energy is calculated hinges on his mass and velocity. Mechanics is often divided into statics (studying bodies at rest), dynamics (studying bodies in motion), and kinematics (the geometry of motion without considering forces), among others. For the football player:
  • Dynamics comes into play as we consider how he comes to a stop through external force.
  • Kinematics might help us describe the player's movement before being tackled.
Mechanics offers the rules and laws—including Newton's laws of motion—that govern how energy transformations and motions occur.
Energy Conservation
Energy conservation in physics is the principle stating that energy cannot be created or destroyed, only transformed from one form to another. For the football player, his kinetic energy—due to his motion—can be transformed into other energy forms until he comes to rest. This principle implies that the kinetic energy initially present in the moving player must be accounted for. Whether it translates into thermal energy due to friction, sound, or other forms, the total energy remains constant even as it changes form. Understanding energy conservation helps us grasp how the player's movement energy is expended. It highlights why calculating kinetic energy and power is crucial since they inform us about energy changes. This insight not only helps in solving such physics problems but also in optimizing real-life systems for energy efficiency.

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Most popular questions from this chapter

A particle moves where its potential energy is given by \(U(r)=U_{0}\left[\left(2 / r^{2}\right)-(1 / r)\right] .(a)\) Plot \(U(r)\) versus \(r .\) Where does the curve cross the \(U(r)=0\) axis? At what value of \(r\) does the minimum value of \(U(r)\) occur? (b) Suppose that the particle has an energy of \(E=-0.050 U_{0} .\) Sketch in the approximate turning points of the motion of the particle on your diagram. What is the maximum kinetic energy of the particle, and for what value of \(r\) does this occur?

(II) A vertical spring (ignore its mass), whose spring constant is \(875 \mathrm{~N} / \mathrm{m},\) is attached to a table and is compressed down by \(0.160 \mathrm{~m} .\) ( \(a\) ) What upward speed can it give to a \(0.380-\mathrm{kg}\) ball when released? \((b)\) How high above its original position (spring compressed) will the ball fly?

(II) A 72 -kg trampoline artist jumps vertically upward from the top of a platform with a \(\begin{array}{llll}\text { speed of } & 4.5 \mathrm{~m} / \mathrm{s} . & (a) & \text { How }\end{array}\) fast is he going as he lands on the trampoline, \(2.0 \mathrm{~m}\) below (Fig. \(8-31\) )? (b) If the trampoline behaves like a spring of spring constant \(5.8 \times 10^{4} \mathrm{~N} / \mathrm{m}\), how far does he depress it?

(II) A \(180-\mathrm{g}\) wood block is firmly attached to a very light horizontal spring, Fig. \(8-35 .\) The block can slide along a table where the coefficient of friction is \(0.30 .\) A force of \(25 \mathrm{~N}\) compresses the spring \(18 \mathrm{~cm}\). If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first cycle?

(II) A particle is constrained to move in one dimension along the \(x\) axis and is acted upon by a force given by \(\overrightarrow{\mathbf{F}}(x)=-\frac{k}{x^{3}} \hat{\mathbf{i}}\) where \(k\) is a constant with units appropriate to the SI system. Find the potential energy function \(U(x)\), if \(U\) is arbitrarily defined to be zero at \(x=2.0 \mathrm{~m},\) so that \(U(2.0 \mathrm{~m})=0 .\)

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