/*! 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 81 A \(5.00-k g\) block is moving a... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(5.00-k g\) block is moving at \(v_{0}=6.00 \mathrm{m} / \mathrm{s}\) along a frictionless, horizontal surface toward a spring with force constant \(k=500 \mathrm{N} / \mathrm{m}\) that is attached to a wall (Fig. 6.36\()\) 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 \(0.150 \mathrm{m},\) what should be the maximum value of \(v_{0} ?\)

Short Answer

Expert verified
(a) Maximum compression is 0.60 m. (b) Maximum initial speed for 0.150 m compression is 1.50 m/s.

Step by step solution

01

Understand the Concept

In this problem, we need to use the principles of energy conservation. The kinetic energy of the block will be converted into the potential energy stored in the spring.
02

Write the Energy Conservation Equation

The initial kinetic energy of the block is given by \( KE = \frac{1}{2} m v_0^2 \). At maximum compression, all this energy is converted into the spring's potential energy \( PE = \frac{1}{2} k x^2 \), where \( x \) is the distance compressed. Thus, the equation becomes:\[\frac{1}{2} m v_0^2 = \frac{1}{2} k x^2\]
03

Solve for Maximum Compression (x)

Rearrange the energy conservation equation to solve for \( x \):\[ x^2 = \frac{m v_0^2}{k} \quad \Rightarrow \quad x = \sqrt{\frac{m v_0^2}{k}}\]Substitute \( m = 5.00 \text{ kg} \), \( v_0 = 6.00 \text{ m/s} \), and \( k = 500 \text{ N/m} \):\[x = \sqrt{\frac{5.00 \times 6.00^2}{500}} = \sqrt{0.36} = 0.60 \text{ m}\]
04

Understand the Limiting Condition

To find the maximum allowable initial speed \( v_0 \) for a compression not exceeding 0.150 m, use the same energy conservation approach with \( x = 0.150 \text{ m} \).
05

Arrange the Equation for Maximum Initial Speed

Reintroduce the energy equation with the new limit:\[\frac{1}{2} m v_0^2 = \frac{1}{2} k (0.150)^2\]Solve for \( v_0 \):\[v_0^2 = \frac{k (0.150)^2}{m} \quad \Rightarrow \quad v_0 = \sqrt{\frac{k (0.150)^2}{m}}\]
06

Calculate the Maximum Allowable Initial Speed

Substitute the values \( m = 5.00 \text{ kg} \), \( k = 500 \text{ N/m} \), and \( x = 0.150 \text{ m} \):\[v_0 = \sqrt{\frac{500 \times 0.150^2}{5.00}} = \sqrt{2.25} = 1.50 \text{ m/s}\]

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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. If you've ever seen a car speeding down a highway, that's a perfect example of kinetic energy in action. It's calculated using the formula: \[ KE = \frac{1}{2} m v^2 \]where:
  • KE is the kinetic energy.
  • m denotes the mass of the object.
  • v represents its velocity.
In our exercise, a 5 kg block moving at 6 m/s on a frictionless surface means it has kinetic energy. Its motion is what we're dealing with and trying to convert to another form. The kinetic energy concept teaches us that energy can change forms. This ability to change drives many natural and engineered processes. For example, the moving block can compress a spring, or you can feel the wind pushing against you as you ride a bike.
The faster an object moves and the heavier it is, the more kinetic energy it holds. That's why small bullets can pack a punch; they have high velocity, translating to high kinetic energy, even with small masses.
Potential Energy
Potential energy is all about the stored energy waiting to be transformed into movement or another form of energy. Think of a drawn bow, ready to shoot an arrow; it’s full of potential energy. The formula used for potential energy in springs is:\[ PE = \frac{1}{2} k x^2 \]where:
  • PE represents the potential energy stored in the spring.
  • k is the spring constant (how stiff the spring is).
  • x is the distance the spring is compressed or extended from its rest position.
In our scenario, the block's kinetic energy transforms into potential energy as it compresses the spring. As the block moves toward the spring, it slows down and transfers its kinetic energy to the spring until it completely stops. At this maximum compression point, all the block's kinetic energy is converted to potential energy stored in the spring. This potential energy can be released to push the block back or perform other work.
The fascinating part of potential energy is that it must always come from somewhere and go somewhere, showcasing the conservation of energy principle in action.
Spring Constant
The spring constant, denoted as k, is a measure of a spring’s stiffness. It shows how much force is needed to compress the spring by a specific distance. The larger the spring constant, the stiffer the spring. A common example is comparing a slinky and a coil spring from a car. The coil spring is much stiffer and thus has a higher spring constant.In essence, the spring constant is part of Hooke’s Law:\[ F = kx \]where:
  • F is the force exerted by the spring.
  • k is the spring constant.
  • x is the displacement of the spring from its equilibrium (or rest) position.
For the block and spring problem, understanding the spring constant is crucial to calculating how much energy the spring can store. It helps determine the point when the block’s kinetic energy fully turns into potential energy. In calculations, knowing k lets us find potential energy for different compression amounts and can be crucial in designing systems that rely on springs for energy conservation, like trampolines or car suspensions.
In our exercise, a spring constant of 500 N/m shows we have a relatively stiff spring capable of storing considerable energy through compression.

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Most popular questions from this chapter

A moving electron has kinetic energy \(K_{1}\) . After a net amount of work \(W\) has been done on it, the electron is moving one-quarter as fast in the opposite direction. (a) Find \(W\) in terms of \(K_{1}\) . (b) Does your answer depend on the final direction of the electron's motion?

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount \(x\) , a force along the \(x\) -axis with \(x\)-component \(F_{x}=k x-b x^{2}+c x^{3}\) must be applied to the free end. Here \(k=100 \mathrm{N} / \mathrm{m}, b=700 \mathrm{N} / \mathrm{m}^{2},\) and \(c=12,000 \mathrm{N} / \mathrm{m}^{3}\). Note that \(x>0\) when the spring is stretched and \(x<0\) when it is compressed. (a) How much work must be done to stretch this spring by 0.050 \(\mathrm{m}\) from its unstretched length? (b) How much work must be done to compress this spring by 0.050 \(\mathrm{m}\) from its unstretched length? (c) Is it easier to stretch or compress this spring? Explain why in terms of the dependence of \(F_{x}\) on \(x\) . (Many real springs behave qualitatively in the same way.)

You and your bicycle have combined mass 80.0 \(\mathrm{kg}\) . When you reach the base of a bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} .6 .35) .\) At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s}\) . You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

Pushing a Cat. Your cat "Ms" (mass 7.00 \(\mathrm{kg}\) ) is trying to make it to the top of a frictionless ramp 2.00 \(\mathrm{m}\) long and inclined upward at \(30.0^{\circ}\) above the horizontal. Since the poor cat can't get any traction on the ramp, you push her up the entire length of the ramp by excring a constant \(100-\mathrm{N}\) force parallel to the ramp. If Ms. takes a running start so that she is moving at 2.40 \(\mathrm{m} / \mathrm{s}\) at the bottom of the ramp, what is her speed when she reaches the top of the incline? Use the work-energy theorem.

Working Like a Horse. Your job is to lift \(30-k g\) crates a vertical distance of 0.90 \(\mathrm{m}\) from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)

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