/*! 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 13 A \(5.0 \mathrm{~g}\) marble is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(5.0 \mathrm{~g}\) marble is fired vertically upward using a spring gun. The spring must be compressed \(8.0 \mathrm{~cm}\) if the marble is to just reach a target \(20 \mathrm{~m}\) above the marble's position on the compressed spring. (a) What is the change \(\Delta U_{g}\) in the gravitational potential energy of the marble-Earth system during the \(20 \mathrm{~m}\) ascent? (b) What is the change \(\Delta U_{s}\) in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

Short Answer

Expert verified
(a) 0.98 J; (b) 0.98 J; (c) 306.25 N/m.

Step by step solution

01

Identify the Change in Gravitational Potential Energy

To find the change in gravitational potential energy (\( \Delta U_g \)) as the marble ascends 20 meters, apply the formula: \( \Delta U_g = mgh \), where \( m = 5.0 \mathrm{~g} = 0.005 \mathrm{~kg} \) (converted to kilograms), \( g = 9.8 \mathrm{~m/s^2} \), and \( h = 20 \mathrm{~m} \). Compute \( \Delta U_g \): \[ \Delta U_g = 0.005 \times 9.8 \times 20 = 0.98 \mathrm{~J} \].
02

Calculate Change in Elastic Potential Energy

At the start, the spring is compressed by 8.0 cm (which is 0.08 meters). The entire elastic potential energy (\( \Delta U_s \)) stored in the spring is used to lift the marble up. Therefore, \( \Delta U_s = \Delta U_g = 0.98 \mathrm{~J} \).
03

Determine the Spring Constant

The change in elastic potential energy of the spring (\( \Delta U_s \)) can be expressed as \( \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x = 0.08 \mathrm{~m} \) is the compression. Solve for \( k \): \[ \Delta U_s = \frac{1}{2} k x^2 \] \[ 0.98 = \frac{1}{2} k (0.08)^2 \] \[ k = \frac{2 \times 0.98}{0.08^2} \approx 306.25 \mathrm{~N/m} \].

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Key Concepts

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

Gravitational Potential Energy
Gravitational potential energy is a form of energy that an object possesses due to its position in a gravitational field. It is directly related to the object's mass, the gravitational force, and the height from which it is elevated.
This type of energy can be calculated using the formula \( \Delta U_g = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, typically taken as \( 9.8 \ \mathrm{m/s^2} \), and \( h \) is the height.
By understanding gravitational potential energy, students can grasp how energy conservation plays out in vertical movements. For example, as a marble fired from a spring gun ascends, its potential energy increases while converting from its initial kinetic and elastic potential energy.
This concept highlights the transformation and conservation of energy, emphasizing that in an isolated system, energy is never lost but merely changes forms. When the marble reaches its maximum height, all kinetic energy initially imparted by the spring has been fully converted to potential energy.
Elastic Potential Energy
Elastic potential energy is stored in elastic materials as a result of their deformation, such as stretching or compressing a spring. In this context, the energy is stored when the spring is compressed to launch the marble.
The elastic potential energy is calculated using the formula \( \Delta U_s = \frac{1}{2} k x^2\), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position (compression or stretching distance).
This type of potential energy is crucial because it illustrates how energy can be stored in objects temporarily and released to perform work, like propelling the marble upward. Understanding elastic potential energy helps in comprehending energy storage in everyday objects, such as rubber bands, trampolines, and even car suspensions.
In our exercise, the elastic potential energy of the compressed spring becomes kinetic energy that propels the marble, eventually turning into gravitational potential energy at the peak of ascent.
Spring Constant
The spring constant, often denoted by \( k \), is a measure of a spring's stiffness. It represents the force required to compress or stretch the spring by a unit length.
A higher spring constant value means a stiffer spring, which requires more force to compress or stretch. Conversely, a lower value indicates a more flexible spring.
In the context of our exercise, the spring constant \( k \) can be derived from the elastic potential energy formula. Here, the marble's kinetic energy (from the spring) and its gravitational potential energy tell us how the spring's energy is stored and used.
Calculating the spring constant is crucial for understanding the amount of potential energy stored and predicting the spring's behavior under different conditions. When a spring gun shoots a marble, the stiffness directly influences how much energy can be transferred from the spring to the marble, ultimately defining how far or high the object can be propelled.

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Most popular questions from this chapter

A \(70 \mathrm{~kg}\) firefighter slides, from rest, \(4.3 \mathrm{~m}\) down a vertical pole. (a) If the firefighter holds onto the pole lightly, so that the frictional force of the pole on her is negligible, what is her speed just before reaching the ground floor? (b) If the firefighter grasps the pole more firmly as she slides, so that the average frictional force of the pole on her is \(500 \mathrm{~N}\) upward, what is her speed just before reaching the ground floor?

A \(1500 \mathrm{~kg}\) car starts from rest on a horizontal road and gains a speed of \(72 \mathrm{~km} / \mathrm{h}\) in \(30 \mathrm{~s}\). (a) What is its kinetic energy at the end of the \(30 \mathrm{~s}\) ? (b) What is the average power required of the car during the 30 s interval? (c) What is the instantaneous power at the end of the 30 s interval, assuming that the acceleration is constant?

A \(3.2 \mathrm{~kg}\) sloth hangs \(3.0 \mathrm{~m}\) above the ground. (a) What is the gravitational potential energy of the sloth-Earth system if we take the reference point \(y=0\) to be at the ground? If the sloth drops to the ground and air drag on it is assumed to be negligible, what are the (b) kinetic energy and (c) speed of the sloth just before it reaches the ground?

A constant horizontal force moves a \(50 \mathrm{~kg}\) trunk \(6.0 \mathrm{~m}\) up a \(30^{\circ}\) incline at constant speed. The coefficient of kinetic friction is \(0.20\). What are (a) the work done by the applied force and (b) the increase in the thermal energy of the trunk and incline?

The surface of the continental United States has an area of about \(8 \times 10^{6} \mathrm{~km}^{2}\) and an average elevation of about \(500 \mathrm{~m}\) (above sea level). The average yearly rainfall is \(75 \mathrm{~cm}\). The fraction of this rainwater that returns to the atmosphere by evaporation is \(\frac{2}{3}\); the rest eventually flows into the ocean. If the decrease in gravitational potential energy of the water-Earth system associated with that flow could be fully converted to electrical energy, what would be the average power? (The mass of \(1 \mathrm{~m}^{3}\) of water is \(1000 \mathrm{~kg}\).)

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