/*! 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 89 A box slides on a frictionless s... [FREE SOLUTION] | 91Ó°ÊÓ

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A box slides on a frictionless surface with a total energy of \(50 \mathrm{J}\). It hits a spring and compresses the spring a distance of \(25 \mathrm{cm}\) from equilibrium. If the same box with the same initial energy slides on a rough surface, it only compresses the spring a distance of \(15 \mathrm{cm},\) how much energy must have been lost by sliding on the rough surface?

Short Answer

Expert verified
The energy lost due to friction on the rough surface is \(32\mathrm{J}\).

Step by step solution

01

Calculate the spring constant k using the frictionless case

The total energy of the box on the frictionless surface is equal to the potential energy stored in the spring when it's compressed from equilibrium. This potential energy can be calculated using the formula: \(U = \frac{1}{2}kx_{1}^2\) where, U = total energy k = spring constant \(x_{1}\) = compression distance in frictionless case Plugging in the given values: \(50\mathrm{J} = \frac{1}{2}k(0.25\mathrm{m})^2\) Solving for k: \(k = \frac{2 \cdot 50\mathrm{J}}{(0.25\mathrm{m})^2} = 1600\mathrm{N/m}\)
02

Calculate the potential energy in the spring on the rough surface

Now that we have the spring constant, we can calculate the potential energy stored in the spring when the box compresses it by \(15\mathrm{cm}\) on the rough surface. We will still use the formula: \(U_{2} = \frac{1}{2}kx_{2}^2\) where, \(U_{2}\) = potential energy on the rough surface \(x_{2}\) = compression distance on the rough surface Plugging in the values: \(U_{2} = \frac{1}{2}(1600\mathrm{N/m})(0.15\mathrm{m})^2 = 18\mathrm{J}\)
03

Calculate the energy lost on the rough surface

Now, we can find the energy lost on the rough surface by subtracting the final potential energy on the rough surface (\(U_{2}\)) from the initial total energy. Energy lost = Initial energy - Final energy Energy lost = \(50\mathrm{J} - 18\mathrm{J} = 32\mathrm{J}\) Therefore, \(32\mathrm{J}\) of energy is lost due to friction on the rough surface.

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Key Concepts

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

Potential Energy
Potential energy represents stored energy associated with an object's position or condition. Imagine holding a ball at a certain height; the ball has potential energy due to gravity, which could be converted into kinetic energy when released. Similarly, in the context of springs, potential energy is stored when a spring is compressed or stretched from its natural, 'equilibrium' position.

The formula for potential energy in a spring is given by \( U = \frac{1}{2}kx^2 \), where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the distance the spring has been displaced from its equilibrium position. Remember, this potential energy can be harnessed to do work, for example, propelling the box forward when the spring is released.
Spring Constant Calculation
The spring constant, symbolized as \( k \), is a parameter that describes the stiffness of a spring. The stiffer the spring, the larger the spring constant.

To calculate the spring constant, we use the potential energy formula mentioned earlier. By rearranging the equation to solve for \( k \), you get \( k = \frac{2U}{x^2} \). In the provided problem, by knowing the energy stored (\( 50 \mathrm{J} \) in the frictionless case) and the displacement (\( 0.25 \mathrm{m} \) or \( 25 \mathrm{cm} \)), you can solve for \( k \) and find that the spring able to store that amount of energy at that displacement must have a spring constant of \( 1600 \mathrm{N/m} \). This calculation allows you to understand how springs will respond to different forces applied to them.
Work-Energy Principle
The work-energy principle links the work done on an object to its energy change. The principle states that work done by forces acting on an object results in a change in kinetic energy. In situations involving conservative forces, like gravity or ideal spring forces, the work done can also change the potential energy.

In the context of our exercise, the box sliding on a rough surface loses some of its initial energy to friction before compressing the spring. The work done by the force of friction is equal to the energy lost by the box. By applying the work-energy principle, we calculate this loss by subtracting the spring's potential energy after being compressed on the rough surface (\( 18 \mathrm{J} \) in this case) from the initial total energy (\( 50 \mathrm{J} \) before encountering the spring). So, the work done by friction, and thus the energy lost, is \( 32 \mathrm{J} \), an important insight into the effects of non-conservative forces in real-life situations.

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Most popular questions from this chapter

A dropped ball bounces to one-half its original height. Discuss the energy transformations that take place.

A mysterious force acts on all particles along a particular line and always points towards a particular point \(P\) on the line. The magnitude of the force on a particle increases as the cube of the distance from that point; that is \(F \infty r^{3},\) if the distance from \(P\) to the position of the particle is \(r .\) Let \(b\) be the proportionality constant, and write the magnitude of the force as \(F=b r^{3} .\) Find the potential energy of a particle subjected to this force when the particle is at a distance \(D\) from \(P\), assuming the potential energy to be zero when the particle is at \(P\).

\(\begin{array}{lllll}\text { In the } & \text { Hunger } & \text { Games } & \text { movie }\end{array}\) (https:llopenstaxcollege.org/II21HungGamesclip) Katniss Everdeen fires a \(0.0200-\mathrm{kg}\) arrow from ground level to pierce an apple up on a stage. The spring constant of the bow is \(330 \mathrm{~N} / \mathrm{m}\) and she pulls the arrow back a distance of \(0.55 \mathrm{~m}\). The apple on the stage is \(5.00 \mathrm{~m}\) higher than the launching point of the arrow. At what speed does the arrow (a) leave the bow? (b) strike the apple?

In the Back to the Future movies (https:I/openstaxcollege.org/l/21bactofutclip) DeLorean car of mass \(1230 \mathrm{kg}\) travels at 88 miles per hour to venture back to the future. (a) What is the kinetic energy of the DeLorean? (b) What spring constant would be needed to stop this DeLorean in a distance of \(0.1 \mathrm{m}\) ?

In the Hunger Games movie (https://openstaxcollege.org/I/21HungGamesclip) Katniss Everdeen fires a \(0.0200-\mathrm{kg}\) arrow from ground level to pierce an apple up on a stage. The spring constant of the bow is \(330 \mathrm{N} / \mathrm{m}\) and she pulls the arrow back a distance of \(0.55 \mathrm{m}\). The apple on the stage is \(5.00 \mathrm{m}\) higher than the launching point of the arrow. At what speed does the arrow (a) leave the bow? (b) strike the apple?

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