/*! 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 42 A 4.00-kg block of ice is placed... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 42

A 4.00-kg block of ice is placed against a horizontal spring that has force constant \(k\) = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. Ignore friction and the mass of the spring. (a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length. (b) What is the speed of the block after it leaves the spring?

Short Answer

Expert verified
The work done is 0.0625 J and the block's speed is 0.177 m/s.

Step by step solution

01

Determine the Potential Energy Stored in the Spring

The work done by the spring on the block is equal to the potential energy stored in the spring when it is compressed. The potential energy (or elastic potential energy) can be calculated using the formula:\[ PE = \frac{1}{2} k x^2 \]where \(k\) is the spring constant (200 N/m), and \(x\) is the compression distance (0.025 m). Substitute the values to find the energy:\[ PE = \frac{1}{2} \times 200 \times (0.025)^2 = 0.0625 \, \text{J} \]
02

Calculate the Work Done by the Spring

The work done on the block by the spring is equivalent to the potential energy initially stored in the spring because energy is conserved in this system (ignoring any friction). Therefore, the work done \(W\) on the block by the spring is:\[ W = 0.0625 \, \text{J} \]
03

Use Work-Energy Principle for Speed

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. The block starts from rest, so its initial kinetic energy is zero. The final kinetic energy \(KE\) is given by the work done, so:\[ KE = \frac{1}{2} mv^2 = W \]We can solve for the speed \(v\) of the block:\[ \frac{1}{2} \times 4.00 \times v^2 = 0.0625 \]
04

Solve for the Speed of the Block

Solve the equation from Step 3 for \(v\):\[ 2 \times 0.0625 = 4.00 \times v^2 \]\[ 0.125 = 4.00 \times v^2 \]\[ v^2 = \frac{0.125}{4.00} \]\[ v^2 = 0.03125 \]\[ v = \sqrt{0.03125} \]\[ v \approx 0.177 \, \text{m/s} \]
05

Conclusion: Results Summary

The work done on the block by the spring is 0.0625 Joules. After being released by the spring, the block achieves a speed of approximately 0.177 m/s.

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.

Potential Energy
Potential energy is a concept that helps us understand how energy can be stored in an object, ready to be released. It doesn't do any work while in its stored form, but it's all about potential for future action. In our example of a spring and a block, potential energy comes into play when the spring is compressed. Imagine you are squishing the spring. It's like you're filling an invisible energy tank inside it. This stored energy depends on how much you compress the spring, as well as the spring's stiffness, or force constant, denoted by the symbol \( k \). Potential energy (or more specifically, elastic potential energy in the case of springs) can be calculated using the formula:
  • \( PE = \frac{1}{2} k x^2 \)
where \( k \) is the spring constant (200 N/m in our exercise), and \( x \) is the compression distance (0.025 m). Substituting these values confirms that the spring stores 0.0625 Joules of potential energy. This energy, initially dormant when the spring is compressed, is what will soon propel our block.
Work-Energy Principle
The Work-Energy Principle is a key concept in physics. It links the work done on an object to the changes in that object's kinetic energy. In simple terms, work is the energy transferred to or from an object via the application of force along a displacement. Now, if we consider the spring and the block, the work done by the spring is actually the stored energy being transferred to the block. The principle tells us that:
  • The work done \( W \) equals the change in kinetic energy \( \Delta KE \) of the block.
Since the block begins its journey at rest, its initial kinetic energy is 0 Joules. When released, the spring's stored potential energy (0.0625 Joules) transforms into kinetic energy of the block. This conversion is why the block moves, highlighting the beauty of energy conservation. So, the work done by the spring is exactly 0.0625 Joules, matching the energy initially stored.
Kinetic Energy
Kinetic energy is the energy of motion. When the spring releases, the block moves, and the stored potential energy becomes kinetic energy. Kinetic energy \( KE \) is calculated using the formula:
  • \( KE = \frac{1}{2} mv^2 \)
where \( m \) is the mass of the block (4.00 kg) and \( v \) is the velocity of the block. Because the work done by the spring is equivalent to the change in kinetic energy, we set our equation \( \frac{1}{2} mv^2 = 0.0625 \). Solving this, we find that the velocity \( v \) of the block is approximately 0.177 meters per second. This transition demonstrates how the block gains speed as it travels away from the spring, illustrating kinetic energy in action. It's amazing how the stored energy in a compressed spring can be calculated and then seen as motion energy in the block, bringing theory to life.

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 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity. (b) Repeat part (a) for motion along the semicircle from the lowest to the highest point on the path.

A force in the \(+x\)-direction with magnitude \(F(x) = 18.0 \, \mathrm{N} - (0.530 \, \mathrm{N/m})x\) is applied to a 6.00-kg box that is sitting on the horizontal, frictionless surface of a frozen lake. \(F(x)\) is the only horizontal force on the box. If the box is initially at rest at \(x = 0\), what is its speed after it has traveled 14.0 m?

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1200-kg car moving at 0.65 m/s is to compress the spring no more than 0.090 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass.

A 5.00-kg block is moving at \(\upsilon_0\) \(=\) 6.00 m/s along a frictionless, horizontal surface toward a spring with force constant \(k\) = 500 N/m that is attached to a wall (\(\textbf{Fig. P6.79}\)). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than 0.150 m, what should be the maximum value of \(\upsilon_0\)?

When a car is hit from behind, its passengers undergo sudden forward acceleration, which can cause a severe neck injury known as \(whiplash\). During normal acceleration, the neck muscles play a large role in accelerating the head so that the bones are not injured. But during a very sudden acceleration, the muscles do not react immediately because they are flexible; most of the accelerating force is provided by the neck bones. Experiments have shown that these bones will fracture if they absorb more than 8.0 J of energy. (a) If a car waiting at a stoplight is rear-ended in a collision that lasts for 10.0 ms, what is the greatest speed this car and its driver can reach without breaking neck bones if the driver's head has a mass of 5.0 kg (which is about right for a 70-kg person)? Express your answer in m/s and in mi/h. (b) What is the acceleration of the passengers during the collision in part (a), and how large a force is acting to accelerate their heads? Express the acceleration in m/s\(^2\) and in \(g\)'s.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Kinematics Physics

Read Explanation

Linear Momentum

Read Explanation

Electric Charge Field and Potential

Read Explanation

Force

Read Explanation

Translational Dynamics

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.