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Chapter 7: Problem 72

On an essentially frictionless horizontal ice-skating 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 the rough patch.

Short Answer

Expert verified
The rough patch is about 1.28 meters long.

Step by step solution

01

Identify Initial Speed and Reduction

The skater's initial speed is given as \(3.0 \text{ m/s}\). The friction reduces this speed by \(45\%\), so the final speed \(v_f\) is \(3.0 \times (1 - 0.45) = 1.65 \text{ m/s}\).
02

Understand Work-Energy Theorem

The work-energy theorem states that the work done by a force on an object is equal to the change in kinetic energy of that object. The formula is \( W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \), where \( v_i \) is the initial velocity, and \( v_f \) is the final velocity.
03

Calculate Change in Kinetic Energy

Substitute the initial and final speeds into the kinetic energy formula to find the change in kinetic energy, \( \Delta KE = \frac{1}{2} m (1.65)^2 - \frac{1}{2} m (3.0)^2 \).
04

Determine the Force of Friction

The friction force \( F_f \) is given as \(25\%\) of the skater's weight. If \( m \) is the mass of the skater, then the weight \( mg \) leads to a friction force \( F_f = 0.25mg \).
05

Relate Work Done to Friction and Distance

The work done by friction is \( W = F_f \cdot d \), where \( d \) is the distance of the rough patch. Substitute this into the work-energy theorem \( W = \Delta KE = 0.25mg \cdot d \).
06

Solve for Distance of the Rough Patch

Equating the work done by friction and the change in kinetic energy: \[ 0.25mg \cdot d = \frac{1}{2} m (1.65)^2 - \frac{1}{2} m (3.0)^2 \]Cancel \( m \) from both sides to simplify:\[ 0.25g \cdot d = \frac{1}{2} ((1.65)^2 - (3.0)^2) \]Solve for \( d \) by dividing both sides by \( 0.25g \):\[ d = \frac{\frac{1}{2} ((1.65)^2 - (3.0)^2)}{0.25g} \tilde \]
07

Final Calculation

Calculate the specific values:\[ d = \frac{\frac{1}{2} (2.7225 - 9)}{0.25 \times 9.81} \approx \frac{-3.13875}{2.4525} \approx 1.28 \text{ meters} \] The length of the rough patch is approximately \( 1.28 \text{ meters} \).

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Key Concepts

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

Friction Force
Friction force is a resistive force that occurs when two surfaces interact. In the context of physics, it is often seen as a hindrance to motion. However, friction is crucial for many everyday actions, like walking or driving. In physics problems, understanding the role friction plays can help solve various equations, especially when applying the work-energy theorem. In this exercise, the friction force acts as a decelerating factor for the skater. Given that the friction force accounts for 25% of the skater's weight, it significantly affects her movement. Calculating friction in physics involves weights and friction coefficients, which help determine how much motion is opposed.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. This concept is pivotal in the work-energy theorem, which evaluates how forces like friction alter kinetic energy. The basic formula to calculate the kinetic energy (KE) of an object is:
  • KE = \( \frac{1}{2} m v^2 \)
where \( m \) is the mass and \( v \) is the velocity. When a skater on the rink encounters a patch that reduces her speed, her kinetic energy decreases. This change is captured by subtracting the final kinetic energy from the initial one. Understanding these shifts in kinetic energy helps in determining the effects of external forces on motion.
Physics Problems
Solving physics problems often involves breaking down the task into manageable steps. This specific problem uses the work-energy theorem, integrating concepts like kinetic energy and friction to find an unknown value. By identifying the initial and reduced speeds and understanding the resistance caused by friction, you can apply mathematical formulas to quantify how forces affect motion. Students often learn to handle these problems by:
  • Identifying known values and the asks.
  • Applying appropriate formulas.
  • Performing algebraic manipulations.
Such exercises enhance problem-solving skills and deepen the understanding of core physics principles.
Ice-Skating Rink
Ice-skating rinks provide unique situations where friction is minimial because ice surfaces are relatively smooth. This makes such environments perfect for studying friction's effects in controlled physics problems. However, this slipperiness is interrupted by rough patches, increasing friction momentarily. When students analyze these scenarios, they observe how external conditions alter motion. Calculations in these contexts highlight real-world applications of friction and energy changes, allowing students to connect textbook solutions to tangible environments. Whether for recreational skating or competitions, understanding these dynamics is essential.
Velocity Reduction
Velocity reduction is a key concept when analyzing the effects of forces like friction. When the skater's velocity drops by 45% after hitting a rough patch, this directly relates to the friction's impact on her kinetic energy and motion. Calculating velocity reductions involves finding the difference between initial and final velocities, providing insight into how much the frictional force affects motion. Such calculations are commonplace in physics problems, serving as a basis for further analysis. They help elucidate the relation between force application and changes in speed, crucial for mastering dynamics and motion studies.

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Most popular questions from this chapter

A fisherman reels in \(12.0 \mathrm{~m}\) of line while landing a fish, using a constant forward pull of \(25.0 \mathrm{~N}\). How much work does the tension in the line do on the fish?

On flat ground, a \(70 \mathrm{~kg}\) person requires about \(300 \mathrm{~W}\) of metabolic power to walk at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}(1.4 \mathrm{~m} / \mathrm{s})\) Using the same metabolic power output, that person can bicycle over the same ground at \(15 \mathrm{~km} / \mathrm{h}\). A \(70 \mathrm{~kg}\) person walks at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}\) on a treadmill at a \(5.0 \%\) grade (that is, the vertical distance covered is \(5.0 \%\) of the horizontal distance covered). If we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required? A. \(300 \mathrm{~W}\) B. 315 W C. \(350 \mathrm{~W}\) D. \(370 \mathrm{~W}\)

A \(12.0 \mathrm{~N}\) package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is supported from below by a vertical spring of force constant \(325 \mathrm{~N} / \mathrm{m}\). If the pan has negligible weight, find the maximum distance the spring will be compressed if no energy is dissipated by friction.

The engine of a motorboat delivers \(30.0 \mathrm{~kW}\) to the propeller while the boat is moving at \(15.0 \mathrm{~m} / \mathrm{s}\). What would be the tension in the towline if the boat were being towed at the same speed?

II An \(8.00 \mathrm{~kg}\) package in a mail-sorting room slides \(2.00 \mathrm{~m}\) down a chute that is inclined at \(53.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the chute's surface is 0.40 . Calculate the work done on the package by (a) friction, (b) gravity, and (c) the normal force. (d) What is the net work done on the package?
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