/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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) Speed: 8.75 m/s, (b) Kinetic Energy: 2617.34 J, (c) Average Power: 1636.71 W.

Step by step solution

01

Calculate Acceleration

Given that the sprinter's initial velocity \(u = 0\), distance \(s = 7.0\, \mathrm{m}\), and time \(t = 1.6\, \mathrm{s}\). We use the equation of motion: \(s = ut + \frac{1}{2} a t^2\). Substituting the given values: \[7.0 = 0 + \frac{1}{2} a (1.6)^2\] Solve for \(a\): \[a = \frac{2 \times 7.0}{(1.6)^2} = \frac{14}{2.56} = 5.47 \, \mathrm{m/s^2}\].
02

Calculate Speed at the End

Use the equation \(v = u + at\), where \(u = 0\), \(a = 5.47 \mathrm{~m/s^2}\), and \(t = 1.6\mathrm{~s}\). Calculate \(v\): \[v = 0 + 5.47 \times 1.6 = 8.75 \mathrm{~m/s}\].
03

Calculate Kinetic Energy

The kinetic energy \(KE\) can be calculated using the formula \(KE = \frac{1}{2} m v^2\). First, find the mass from the weight \(W = 670 \mathrm{~N}\) using \(W = mg\), where \(g = 9.81 \mathrm{~m/s^2}\). Hence, \(m = \frac{670}{9.81} \approx 68.34 \mathrm{~kg}\). Now, find \(KE\): \[KE = \frac{1}{2} \times 68.34 \times (8.75)^2 \approx 2617.34 \mathrm{~J}\].
04

Calculate Average Power

Power is the work done over time, and the work done is the change in kinetic energy. Since the initial kinetic energy is zero (starting from rest), \(\Delta KE = 2617.34 \mathrm{~J}\). Power \(P = \frac{\Delta KE}{t}\): \[P = \frac{2617.34}{1.6} \approx 1636.71 \mathrm{~W}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Acceleration
Acceleration is a fundamental concept in physics that describes how quickly an object's velocity changes over time. It is given by the formula: \( a = \frac{\Delta v}{\Delta t} \), where \( \Delta v \) is the change in velocity and \( \Delta t \) is the change in time. In our problem, we calculated the sprinter's uniform acceleration using the equation of motion:
  • \( s = ut + \frac{1}{2} a t^2 \) , where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time.
Starting from rest means the initial velocity \( u = 0 \). By rearranging the equation, we can solve for \( a \):
  • For the given problem: \( a = \frac{2s}{t^2} \).
This approach provides the sprinter's acceleration, helping us understand how fast the sprinter was speeding up over the 1.6 seconds. A uniform acceleration like this suggests a steady increase in speed.
Gaining Insight into Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is calculated using the formula: \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity. It's important to grasp that kinetic energy depends on both mass and the square of the velocity, meaning that it increases rapidly with higher speeds. In this scenario, we first determined the sprinter's mass from the given weight \( W = mg \), with \( g \approx 9.81 \, \mathrm{m/s^2} \), to find:
  • \( m = \frac{W}{g} \).
Once we had the speed \( v \) from the previous calculation steps, substituting these values into the kinetic energy formula allowed us to find the energy the sprinter had at the end of the 1.6 seconds. Kinetic energy tells us how much work the sprinter can perform due to the motion.
Breaking Down Power Calculation
Power in physics refers to the rate at which work is done or energy is transferred over time. The power generated by the sprinter can be calculated using the formula: \( P = \frac{\Delta KE}{t} \), where \( \Delta KE \) is the change in kinetic energy and \( t \) is the time taken. Power is measured in watts, with one watt equaling one joule per second. In this exercise, since the sprinter starts from rest, the initial kinetic energy is zero, and consequently, all the change in kinetic energy equals the final kinetic energy. By dividing this energy change by the time interval (1.6 seconds), we can find the average power generated by the sprinter during this interval. High power values indicate that the sprinter was able to exert a large amount of energy quickly, showcasing their explosive performance capability.
The Equations of Motion Explained
Equations of motion are crucial for analyzing and predicting the behavior of moving objects, especially when they are undergoing constant acceleration. These equations connect the core aspects of motion: displacement, initial velocity, final velocity, acceleration, and time. The primary equations include:
  • \( v = u + at \) - relates velocity, initial velocity, acceleration, and time.
  • \( s = ut + \frac{1}{2} a t^2 \) - connects displacement, initial velocity, acceleration, and time.
  • \( v^2 = u^2 + 2as \) - links final velocity, initial velocity, acceleration, and displacement.
These equations allow for a comprehensive analysis of motion when an object accelerates at a constant rate. In our scenario, they were utilized to compute both the acceleration and final speed of the sprinter. It's important to ensure that all values are in coherent units, like meters, seconds, and meters per second, for consistent and accurate calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(2.50 \mathrm{~kg}\) beverage can is thrown directly downward from a height of \(4.00 \mathrm{~m}\), with an initial speed of \(3.00 \mathrm{~m} / \mathrm{s}\). The air drag on the can is negligible. What is the kinetic energy of the can (a) as it reaches the ground at the end of its fall and (b) when it is halfway to the ground? What are (c) the kinetic energy of the can and (d) the gravitational potential energy of the can-Earth system \(0.200 \mathrm{~s}\) before the can reaches the ground? For the latter, take the reference point \(y=0\) to be at the ground.

A small block of mass \(m=0.032 \mathrm{~kg}\) can slide along the frictionless loop-the-loop, with loop radius \(R=12 \mathrm{~cm}\). The block is released from rest at point \(P\), at height \(h=5.0 R\) above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point \(P\) to (a) point \(Q\) and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point \(P\), (d) at point \(Q\), and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?

A block is sent sliding down a frictionless ramp. Its speeds at points \(A\) and \(B\) are \(2.00 \mathrm{~m} / \mathrm{s}\) and \(2.60 \mathrm{~m} / \mathrm{s}\), respectively. Next, it is again sent sliding down the ramp, but this time its speed at point \(A\) is \(4.00 \mathrm{~m} / \mathrm{s}\). What then is its speed at point \(B ?\)

A conservative force \(\vec{F}=(6.0 x-12) \hat{\mathrm{i}} \mathrm{N}\) where \(x\) is in meters, acts on a particle moving along an \(x\) axis. The potential energy \(U\) associated with this force is assigned a value of \(27 \mathrm{~J}\) at \(x=0\). (a) Write an expression for \(U\) as a function of \(x\), with \(U\) in joules and \(x\) in meters. (b) What is the maximum positive potential energy? At what (c) negative value and (d) positive value of \(x\) is the potential energy equal to zero?

The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$U=\frac{A}{r^{12}}-\frac{B}{r^{6}}$$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation - that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.