/*! 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 10 A hockey puck with mass 0.160 \(... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hockey puck with mass 0.160 \(\mathrm{kg}\) is at rest on the horizontal, frictionless surface of a rink. A player applies a force of 0.250 \(\mathrm{N}\) to the puck, parallel to the surface of the ice, and continues to apply this force for 2.00 \(\mathrm{s}\) . What are the position and speed of the puck at the end of that time?

Short Answer

Expert verified
The puck's position is 3.125 meters, and its speed is 3.125 m/s after 2 seconds.

Step by step solution

01

Identify Given Values

We will start by identifying the given values in the problem. A mass of the puck: \( m = 0.160 \, \text{kg} \), a force applied: \( F = 0.250 \, \text{N} \), and the duration of force application: \( t = 2.00 \, \text{s} \). The initial velocity \( v_0 \) is 0 since the puck starts at rest.
02

Calculate the Acceleration

Use Newton's second law to find acceleration. According to the law, \( F = ma \). Therefore, acceleration \( a = \frac{F}{m} = \frac{0.250 \, \text{N}}{0.160 \, \text{kg}} = 1.5625 \, \text{m/s}^2 \).
03

Determine the Final Velocity

Use the equation of motion \( v = v_0 + at \) to calculate the final velocity. Since \( v_0 = 0 \), \( v = 0 + (1.5625 \, \text{m/s}^2)(2.00 \, \text{s}) = 3.125 \, \text{m/s} \).
04

Calculate the Final Position

Use the equation \( s = v_0 t + \frac{1}{2} at^2 \) to find the final position. Substituting the known values, \( s = 0 \times 2 + \frac{1}{2} \times 1.5625 \times 2^2 = 3.125 \, \text{m} \).
05

Summarize the Results

The final position of the puck is 3.125 meters, and the final speed is 3.125 meters per second after 2.00 seconds.

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.

Acceleration Calculation
When understanding the movement of objects, calculating acceleration is crucial. Acceleration represents how an object's velocity changes over time. To find it, we use Newton's Second Law which connects force, mass, and acceleration with the formula: \[ F = ma \] Where:- \( F \) is the force applied, - \( m \) is the mass of the object, - \( a \) is the acceleration.In the case of the hockey puck, we know the force applied is \( 0.250 \, \text{N} \), and the puck's mass is \( 0.160 \, \text{kg} \). By rearranging the formula to solve for acceleration, we get\[ a = \frac{F}{m} \] Substituting the given values gives us:\[ a = \frac{0.250}{0.160} \approx 1.5625 \, \text{m/s}^2 \]This tells us how fast the puck's velocity is changing, or accelerating, as it is pushed along the ice.
Equations of Motion
Equations of motion allow us to predict the future movement of an object given its initial conditions and the forces acting on it. These equations incorporate information about time, velocity, acceleration, and displacement. For the hockey puck, we can use these equations to determine both the speed and position of the puck after the force is applied.
  • The formula for final velocity \( v \) is based on initial velocity \( v_0 \), acceleration \( a \), and time \( t \):\[ v = v_0 + at \]Since the puck starts from rest, \( v_0 = 0 \), so:\[ v = 1.5625 \, \text{m/s}^2 \times 2 \, \text{s} = 3.125 \, \text{m/s} \]
  • The equation to find the position \( s \) is:\[ s = v_0 t + \frac{1}{2} a t^2 \]Plugging in the values:\\[ s = 0 \times 2 + \frac{1}{2} \times 1.5625 \times 2^2 = 3.125 \, \text{m} \]
Each of these calculations helps us understand the progression of the puck's movement over the course of the time it was subject to the force.
Force and Mass Relationship
The interplay between force, mass, and acceleration is foundational to Newton's Second Law. This relationship dictates how an object's velocity will change when a force is applied. Understanding this concept helps us predict how objects respond to different forces based on their mass.
  • **Force**: Measured in newtons (N), it represents the push or pull exerted on an object.
  • **Mass**: A scalar quantity that indicates how much matter is in an object, measured in kilograms (kg). It tells us how resistant an object is to acceleration when a force is applied.
  • **Acceleration**: The rate of change of velocity that occurs when a force acts on mass. It's measured in meters per second squared (\( \text{m/s}^2 \)).
These elements are intertwined in the formula \( F = ma \). A larger force results in greater acceleration, but only if the mass remains the same. Likewise, if the same force is applied to objects of differing mass, the lighter object accelerates more. In our example, this relationship enabled us to precisely calculate the puck's acceleration and subsequent motion after the advent of force.

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

Jumping to the ground. A 75.0 \(\mathrm{kg}\) man steps off a platform 3.10 \(\mathrm{m}\) above the ground. He keeps his legs straight as he falls, but at the moment his feet touch the ground his knees begin to bend, and, treated as a particle, he moves an additional 0.60 \(\mathrm{m}\) before coming to rest. (a) What is his speed at the instant his feet touch the ground? (b) Treating him as a particle, what are the magnitude and direction of his acceleration as he slows down if the acceleration is constant? (c) Draw a free-body diagram of this man as he is slowing down. (d) Use Newton's laws and the results of part (b) to calculate the force the ground exerts on him while he is slowing down. Express this force in newtons and also as a multiple of the man's weight. (e) What are the magnitude and direction of the reaction force to the force you found in part (c)?

A uniform 25.0 \(\mathrm{kg}\) chain 2.00 \(\mathrm{m}\) long supports a 50.0 \(\mathrm{kg}\) chandelier in a large public building. Find the tension in (a) the bottom link of the chain, (b) the top link of the chain, and (c) the middle link of the chain.

A tennis ball traveling horizontally at 22 \(\mathrm{m} / \mathrm{s}\) suddenly hits a vertical brick wall and bounces back with a horizontal velocity of 18 \(\mathrm{m} / \mathrm{s}\) . Make a free-body diagram of this ball (a) just before it hits the wall, (b) just after it has bounced free of the wall, and (c) while it is in contact with the wall.

Extraterrestrial physics. You have landed on an unknown planet, Newtonia, and want to know what objects will weigh there. You find that when a certain tool is pushed on a frictionless horizontal surface by a 12.0 \(\mathrm{N}\) force, it moves 16.0 \(\mathrm{m}\) in the first 2.00 \(\mathrm{s}\) starting from rest. You next observe that if you release this tool from rest at 10.0 \(\mathrm{m}\) above the ground, it takes 2.58 s to reach the ground. What does the tool weigh on Newtonia and what would it weigh on Earth?

An acrobat is hanging by his feet from a trapeze, while supporting with his hands a second acrobat who hangs below him. Draw separate free-body diagrams for the two acrobats.

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.