/*! 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 103 A sprinter who weighs \(670 \mat... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 103

A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s}\) ? (c) What average power does the sprinter generate during the \(1.6 \mathrm{~s}\) interval?

Short Answer

Expert verified
(a) 8.75 m/s; (b) 2619.97 J; (c) 1637.48 W

Step by step solution

01

Convert Weight to Mass

The sprinter's weight is given as 670 N. Using the relationship between weight and mass, we can find the mass:\[ W = mg \]where \( W = 670 \text{ N} \) and \( g = 9.8 \text{ m/s}^2 \). Solving for \( m \):\[ m = \frac{670}{9.8} \approx 68.37 \text{ kg} \]
02

Calculate Acceleration

The sprinter runs 7.0 m in 1.6 s starting from rest, with uniform acceleration. We can use the formula:\[ s = ut + \frac{1}{2}at^2 \]where \( s = 7.0 \text{ m} \), \( u = 0 \text{ m/s} \), \( t = 1.6 \text{ s} \). Rearrange to find \( a \):\[ 7.0 = 0 + \frac{1}{2}a(1.6)^2 \]\[ 7.0 = 1.28a \]\[ a = \frac{7.0}{1.28} \approx 5.47 \text{ m/s}^2 \]
03

Calculate Speed at 1.6 Seconds

Using the acceleration found, apply it to find the final speed using the equation:\[ v = u + at \]where \( u = 0 \text{ m/s} \) and \( a = 5.47 \text{ m/s}^2 \):\[ v = 0 + 5.47 \times 1.6 \]\[ v = 8.75 \text{ m/s} \]
04

Calculate Kinetic Energy

The kinetic energy at the end of 1.6 seconds can be calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]where \( m = 68.37 \text{ kg} \) and \( v = 8.75 \text{ m/s} \):\[ KE = \frac{1}{2} \times 68.37 \times (8.75)^2 \]\[ KE \approx 2619.97 \text{ J} \]
05

Calculate Average Power

Average power is the work done divided by the time interval.The work done is equal to the change in kinetic energy, so:\[ P_{avg} = \frac{\Delta KE}{\Delta t} \]where \( \Delta KE \approx 2619.97 \text{ J} \) and \( \Delta t = 1.6 \text{ s} \):\[ P_{avg} = \frac{2619.97}{1.6} \approx 1637.48 \text{ W} \]

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Key Concepts

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

Uniform Acceleration
Uniform acceleration refers to a condition where an object experiences a constant acceleration, meaning its velocity changes at a steady rate over time. This is a common scenario in physics, especially when analyzing the motion of objects without any resisting forces like friction. In the given exercise, the sprinter starts from rest and accelerates uniformly, which allows us to use certain kinematic equations to determine his motion parameters.

To calculate uniforms acceleration, we apply the formula:
  • \[ s = ut + \frac{1}{2}at^2 \]
Here, \(s\) represents the displacement, \(u\) is the initial velocity, which is zero, as the sprinter starts from rest, and \(t\) is the time interval. Rearranging the equation allows us to solve for \(a\) (acceleration).

Understanding this concept is crucial because it directly links the covered distance with time and acceleration, giving insights into how objects speed up when uniformly accelerating. This foundational principle helps in dissecting complex motions in physics into understandable segments.
Kinetic Energy Calculation
Kinetic energy is the energy possessed by an object due to its motion. When an object moves, it has kinetic energy, which can be calculated using the formula:

  • \[ KE = \frac{1}{2}mv^2 \]
where \(KE\) denotes kinetic energy; \(m\) is mass, and \(v\) is velocity. This equation reveals how kinetic energy depends on an object's mass and the square of its velocity. For the sprinter's case, once his speed at \(1.6\) seconds is known, we can calculate his kinetic energy using the previously computed mass and velocity.

Kinetic energy is a vital concept in physics because it helps quantify the motion energy of objects, crucial in scenarios involving moving systems. Comprehending how to calculate kinetic energy not only develops a better understanding of motion dynamics but also underlines conservation laws, where energy transitions between different forms in isolated systems.
Power in Physics
In physics, power measures the rate at which work is done or energy is transferred. It indicates how much energy is used in a given time interval. The average power generated during an event can be calculated using the formula:

  • \[ P_{avg} = \frac{\Delta KE}{\Delta t} \]
where \(P_{avg}\) is the average power, \(\Delta KE\) is the change in kinetic energy, and \(\Delta t\) is the time over which the change occurs. For the sprinter, the change in kinetic energy is from rest to his kinetic energy after \(1.6\) seconds, computed earlier, providing a convenient way to calculate the average power output during that interval.

Understanding power is important because it relates energy use and efficiency in various systems and applications. It's a concept that transcends across many areas of physics, illustrating how quickly energy is utilized in processes like the one with the sprinter accelerating during the race.

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Most popular questions from this chapter

A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(24 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 22 \(\mathrm{m} / \mathrm{s}\), landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

A particle can move along only an \(x\) axis, where conservative forces act on it (Fig. 8-66 and the following table). The particle is released at \(x=5.00 \mathrm{~m}\) with a kinetic energy of \(K=14.0 \mathrm{~J}\) and \(\mathrm{a}\) potential energy of \(U=0\). If its motion is in the negative direction of the \(x\) axis, what are its (a) \(K\) and (b) \(U\) at \(x=2.00 \mathrm{~m}\) and its (c) \(K\) and \((\) d \() U\) at \(x=0\) ? If its motion is in the positive direction of the \(x\) axis, what are its (e) \(K\) and (f) \(U\) at \(x=11.0 \mathrm{~m}\), its \((\mathrm{g}) \mathrm{K}\) and (h) \(U\) at \(x=12.0 \mathrm{~m}\), and its (i) \(K\) and \((\mathrm{j}) U\) at \(x=13.0 \mathrm{~m} ?(\mathrm{k})\) Plot \(U(x)\) versus \(x\) for the range \(x=0\) to \(x=13.0 \mathrm{~m}\). Next, the particle is released from rest at \(x=0\). What are (l) its kinetic energy at \(x=5.0 \mathrm{~m}\) and \((\mathrm{m})\) the maximum positive position \(x_{\max }\) it reaches? (n) What does the particle do after it reaches \(x_{\max } ?\) $$ \begin{array}{lc} \hline {\text { Range }} & {\text { Force }} \\ \hline 0 \text { to } 2.00 \mathrm{~m} & \vec{F}_{1}=+(3.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 2.00 \mathrm{~m} \text { to } 3.00 \mathrm{~m} & \vec{F}_{2}=+(5.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 3.00 \mathrm{~m} \text { to } 8.00 \mathrm{~m} & F=0 \\ 8.00 \mathrm{~m} \text { to } 11.0 \mathrm{~m} & \vec{F}_{3}=-(4.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 11.0 \mathrm{~m} \text { to } 12.0 \mathrm{~m} & \vec{F}_{4}=-(1.00 \mathrm{~N}) \hat{\mathrm{i}} \\ 12.0 \mathrm{~m} \text { to } 15.0 \mathrm{~m} & F=0 \\ \hline \end{array} $$

A child whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is \(0.10\). (a) How much energy is transferred to thermal energy? (b) If she starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s}\), what is her speed at the bottom?

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20\). What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

A \(70.0 \mathrm{~kg}\) man jumping from a window lands in an elevated fire rescue net \(11.0 \mathrm{~m}\) below the window. He momentarily stops when he has stretched the net by \(1.50 \mathrm{~m}\). Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by \(1.50 \mathrm{~m}\).
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