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Refer to Multiple-Concept Example 10 for help in solving problems like this one. An ice skater is gliding horizontally across the ice with an initial velocity of \(+6.3 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between the ice and the skate blades is \(0.081,\) and air resistance is negligible. How much time elapses before her velocity is reduced to \(+2.8 \mathrm{~m} /\) s?

Short Answer

Expert verified
The skater takes approximately 4.41 seconds to slow to 2.8 m/s.

Step by step solution

01

Identify Given Values

The initial velocity of the ice skater is given as \( u = 6.3 \text{ m/s} \), the final velocity is \( v = 2.8 \text{ m/s} \), and the coefficient of kinetic friction is \( \mu_k = 0.081 \). Also, note that air resistance is negligible, so it doesn't need to be included in the calculations.
02

Understand the Forces Involved

The force acting on the skater due to friction is given by the formula \( f_k = \mu_k \cdot m \cdot g \). However, since the actual mass \( m \) is not provided, we can use Newton's second law, \( F = m \cdot a \), to express the deceleration as \( a = -\mu_k \cdot g \) where \( g = 9.8 \text{ m/s}^2 \). The negative sign indicates deceleration.
03

Calculate the Deceleration

Using the formula from the previous step, calculate the deceleration: \( a = -\mu_k \cdot g = -0.081 \times 9.8 = -0.7938 \text{ m/s}^2 \).
04

Use Kinematic Equation to Find Time

Use the equation \( v = u + at \) to solve for the time \( t \). Rearrange the equation to \( t = \frac{v - u}{a} \). Substitute the known values: \( t = \frac{2.8 - 6.3}{-0.7938} \).
05

Calculate the Time

Substitute the values into the formula: \( t = \frac{-3.5}{-0.7938} \approx 4.41 \text{ seconds} \).

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Key Concepts

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

Kinetic Friction: How Skaters Glide Smoothly
When an ice skater glides over the ice, she's experiencing kinetic friction. This is the force that acts against her motion when she's moving. Unlike static friction, which keeps objects from starting to move, kinetic friction works on objects already in motion. This friction is determined by two main factors: the roughness of the surfaces in contact and the force pressing them together. In our ice skater's case, the surface of the ice and the blades of her skates create this friction. The formula we use is:
  • \( f_k = \mu_k \cdot m \cdot g \)
This signifies that the force of kinetic friction \( f_k \) is the product of the coefficient of kinetic friction \( \mu_k \), mass \( m \), and gravitational acceleration \( g \). In this scenario, the coefficient is 0.081, which is quite low. This signifies that the ice is very smooth, which helps the skater to glide with minimal resistance. The small value of kinetic friction on ice allows skaters to move swiftly.Understanding kinetic friction helps us appreciate why sports like ice skating and skiing are possible, as these depend on low friction between surfaces for smooth motion.
Kinematic Equations: Describing Motion
Kinematic equations help us describe the motion of objects. They relate velocity, acceleration, time, and displacement. These equations are fundamental in physics to predict the future motion of moving objects, like our skater. In this case, we used the following kinematic equation:
  • \( v = u + at \)
  • Where:
  • \( v \) is the final velocity
  • \( u \) is the initial velocity
  • \( a \) is the acceleration (or deceleration)
  • \( t \) is the time
By rearranging this equation to solve for time - \( t = \frac{v - u}{a} \) - we can figure out how long it takes for the skater to reduce her velocity from 6.3 m/s to 2.8 m/s. Kinematic equations are powerful tools for calculating the different aspects of motion and are used widely in physics and engineering.
Deceleration: Slowing Down with Control
Deceleration is simply negative acceleration. While acceleration means an increase in velocity, deceleration means a decrease. It occurs when a moving object slows down due to external forces like friction or air resistance.In the case of our ice skater, deceleration is caused by kinetic friction between the ice and the skate blades. We calculate it using:
  • \( a = -\mu_k \cdot g \)
The coefficient of kinetic friction \( \mu_k \) and gravitational acceleration \( g \) combine to give us the deceleration \( a \). Here, the result is \( -0.7938 \text{ m/s}^2 \). The negative sign reflects that the skater is slowing down. Understanding deceleration is crucial for safely controlling motion, whether you are a driver applying brakes or a skater coming to a graceful stop.
Initial and Final Velocity: The Start and End Points of Motion
Velocity is an essential aspect of motion, representing the speed and direction of an object's travel. Initial velocity refers to how fast and in what direction an object moves at the beginning of its observation, while final velocity is how fast it's moving at the end. For our skater, her initial velocity is given: 6.3 m/s. This is the speed at which she starts gliding over the ice. As time passes, and due to the friction acting upon her, her speed will reduce until she reaches a final velocity - in this instance, 2.8 m/s. Using the initial and final velocities, and knowing the deceleration, we can determine how long the transition from the initial to final velocity takes. These concepts of initial and final velocity are key to problem-solving in physics, allowing us to calculate time intervals, and are critical in understanding how objects speed up or slow down.

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Most popular questions from this chapter

A car is driving up a hill. (a) Is the magnitude of the normal force exerted on the car equal to the magnitude of its weight? Why or why not? (b) If the car drives up a steeper hill, does the normal force increase, decrease, or remain the same? Justify your answer. (c) Does the magnitude of the normal force depend on whether the car is traveling up the hill or down the hill? Give your reasoning. Problem A car is traveling up a hill that is inclined at an angle of \(\theta\) above the horizontal. Determine the ratio of the magnitude of the normal force to the weight of the car when (a) \(\theta=15^{\circ}\) and \((\mathrm{b}) \theta=35^{\circ} .\) Check to see that your answers are consistent with your answers to the Concept Questions.

Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes on the man with a certain force that is parallel to the ground. (a) Must the man accelerate under the action of this force? If so, what three factors determine the magnitude and direction of his acceleration? (b) Is there a corresponding force exerted on the woman? If so, where does it originate? Is this force related to the magnitude and direction of the force the woman exerts on the man? If so, how? The mass of the man is \(82 \mathrm{~kg}\) and that of the woman is \(48 \mathrm{~kg}\). The woman pushes on the man with a force of \(45 \mathrm{~N}\) due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.

A \(325-\mathrm{kg}\) boat is sailing \(15.0^{\circ}\) north of east at a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Thirty seconds later, it is sailing \(35.0^{\circ}\) north of east at a speed of \(4.00 \mathrm{~m} / \mathrm{s}\). During this time, three forces act on the boat: a \(31.0-\mathrm{N}\) force directed \(15.0^{\circ}\) north of east (due to an auxiliary engine), a 23.0-N force directed \(15.0^{\circ}\) south of west (resistance due to the water), and \(\vec{F}_{W}\) (due to the wind). Find the magnitude and direction of the force \(\overrightarrow{\mathrm{F}}_{\mathrm{W}} .\) Express the direction as an angle with respect to due east.

A \(1.40-\mathrm{kg}\) bottle of vintage wine is lying horizontally in the rack shown in the drawing. The two surfaces on which the bottle rests are \(90.0^{\circ}\) apart, and the right surface makes an angle of \(45.0^{\circ}\) with respect to the ground. Each surface exerts a force on the bottle that is perpendicular to the surface. What is the magnitude of each of these forces?

A supertanker (mass \(\left.=1.70 \times 10^{8} \mathrm{~kg}\right)\) is moving with a constant velocity. Its engines generate a forward thrust of \(7.40 \times 10^{5} \mathrm{~N}\). Determine (a) the magnitude of the resistive force exerted on the tanker by the water and (b) the magnitude of the upward buoyant force exerted on the tanker by the water.

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