/*! 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 85 On an essentially frictionless, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 85

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 \(\mathrm{m} / \mathrm{s}\) encounters a rough patch that reduces her speed by 45\(\%\) due to a friction force that is 25\(\%\) of her weight. Use the work- energy theorem to find the length of this rough patch.

Short Answer

Expert verified
The skater travels 8.1 meters across the rough patch.

Step by step solution

01

Establish Initial and Final Speeds

The skater initially moves at a speed of 3.0 m/s. We are told that the rough patch decreases her speed by 45%, so the speed after the patch will be \( v_f = 3.0 \times (1 - 0.45) = 1.65 \, \mathrm{m/s} \).
02

Identify the Forces and Work Done

The work-energy theorem states that the work done on the skater equals the change in kinetic energy. The friction force, which is given as 25% of her weight \( F_f = 0.25 mg \), does the work. The work done, \( W = F_f \times d \), where \( d \) is the distance over the rough patch, is what slows the skater down.
03

Calculate the Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) is equal to the final kinetic energy minus the initial kinetic energy. The initial kinetic energy is \( KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (3.0)^2 \), and the final kinetic energy is \( KE_f = \frac{1}{2} m (1.65)^2 \). Therefore, \( \Delta KE = KE_f - KE_i \).
04

Apply the Work-Energy Theorem

According to the work-energy theorem, the work done by friction \( -F_f \times d \) is equal to the change in kinetic energy \( \Delta KE \). Thus, we set up the equation: \(-0.25 mg \times d = \Delta KE \).
05

Solve for the Distance \(d\)

Re-arrange the equation to solve for \( d \): \( d = -\frac{\Delta KE}{0.25 mg} \). Plug in the expression for \( \Delta KE \) from Step 3, and simplify: \( \Delta KE = \frac{1}{2} m [(1.65)^2 - (3.0)^2] \). Substitute this into the equation for \( d \) to find \( d \). Note that in this equation, the mass \( m \) cancels out.

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 91影视!

Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In physics, we represent it using the formula: \[ KE = \frac{1}{2} m v^2 \]Here, \( KE \) stands for kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity. Kinetic energy is always a positive value, since both mass and the square of velocity are positive.
In our exercise, we calculate both initial and final kinetic energies to understand how the skater's energy changes as she passes over the rough patch. Initially, she is moving at 3.0 m/s, providing her with an initial kinetic energy of \( \frac{1}{2} m (3.0^2) \). After slowing down due to friction, her speed reduces to 1.65 m/s, resulting in a new kinetic energy of \( \frac{1}{2} m (1.65^2) \).
The difference in these values, referred to as the change in kinetic energy, helps us determine how much work the friction force has done.
Friction Force
Friction force is a resistive force that acts parallel to the surfaces in contact. It always opposes the motion of the object. In our scenario, the skater experiences a friction force when encountering the rough patch. This force is described as being 25% of the skater's weight.
To calculate it, we use the formula:\[ F_f = 0.25 mg \]where \( mg \) represents the weight of the skater. The friction force here performs negative work, which means it extracts energy from the skater, slowing her down.
Incorporating this friction force into the work-energy theorem allows us to relate the force to a specific distance鈥攖he length of the rough patch that caused the skater to decelerate. The work done by friction is calculated using the expression \( -F_f \times d \), highlighting that it is equal and opposite to the energy change experienced by the skater.
Physics Problem Solving
Physics problem solving often requires a step-by-step approach. Let's break down the process used in our exercise:
  • Identify given information and what needs to be found鈥攊n this case, the skater's speed reduction and friction force.
  • Utilize appropriate physics principles, here the work-energy theorem, which relates the work done by friction to changes in kinetic energy.
  • Calculate initial and final kinetic energies to understand how the skater's velocity affected her energy levels.
  • Use known forces, like the friction force, to find work done over a distance.
  • Set up and solve equations systematically to isolate the desired variable, such as distance traveled over the rough patch.
  • Ensure all calculations account for units and dimensional consistency for accuracy.
This structured approach aids in breaking down complex problems into manageable parts, ensuring clarity and precision in finding solutions. Good problem-solving practices enhance our understanding of physics and strengthen analytical skills.

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

Stopping Distance. A car is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{x}\) between the tires and the road. (b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

On a winter's day in Maine, a warehouse worker is shoving boxes up a rough plank inclined at an angle \(\alpha\) above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with the distance \(x\) along the plank: \(\mu=A x\), where \(A\) is a positive constant and the bottom of the plank is at \(x=0 .\) (For this plank the coefficients of kinetic and static friction are equal: \(\mu_{k}=\mu_{s}=\mu\) . The worker shoves a box up the plank so that it leaves the bottom of the plank moving at speed \(v_{0}\) . Show that when the box first comes to rest, it will remain at rest if $$v_{0}^{2} \geq \frac{3 g \sin ^{2} \alpha}{A \cos \alpha}$$

Half of a Spring. (a) Suppose you cut a massless ideal spring in half. If the full spring had a force constant \(k\) , what is the force constant of each half, in terms of \(k ?\) (Hint: Think of the original spring as two equal halves, each producing the same force as the entire spring. Do you see why the forces must be equal? (b) If you cut the spring into three equal segments instead, what is the force constant of each one, in terms of \(k ?\)

Rotating Bar. A thin, uniform \(12.0-\mathrm{kg}\) bar that is 2.00 \(\mathrm{m}\) long rotates uniformly about a pivot at one end, making 5.00 complete revolutions every 3.00 seconds. What is the kinetic energy of this bar? (Hint: Different points in the bar have different speeds. Break the bar up into infinitesimal segments of mass \(d m\) and integrate to add up the kinetic energy of all these segments.)

The gravitational pull of the earth on an object is inversely proportional to the square of the distance of the object from the center of the earth. At the earth's surface this force is equal to the object's normal weight \(m g,\) where \(g=9.8 \mathrm{m} / \mathrm{s}^{2},\) and at large distances, the force is zero. If a \(20,000-k g\) asteroid falls to earth from a very great distance away, what will be its minimum speed as it strikes the earth's surface, and how much kinetic energy will it impart to our planet? You can ignore the effects of the earth's atmosphere.
See all solutions

Recommended explanations on Physics Textbooks

Medical Physics

Read Explanation

Nuclear Physics

Read Explanation

Rotational Dynamics

Read Explanation

Electricity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Physical Quantities And Units

Read Explanation
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.