/*! 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 31 A block with mass \(m=2.00 \math... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 31

A block with mass \(m=2.00 \mathrm{~kg}\) is placed against a spring on a frictionless incline with angle \(\theta=30.0^{\circ}\) (Fig. 8-42). (The block is not attached to the spring.) The spring, with spring constant \(k=19.6 \mathrm{~N} / \mathrm{cm}\), is compressed \(20.0 \mathrm{~cm}\) and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block- Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

Short Answer

Expert verified
(a) 39.2 J, (b) 39.2 J, (c) 4.00 m

Step by step solution

01

Calculate Elastic Potential Energy

The elastic potential energy stored in a compressed or stretched spring is given by the formula: \( U = \frac{1}{2} k x^2 \), where \( U \) is the elastic potential energy, \( k \) is the spring constant, and \( x \) is the compression distance of the spring. Convert the spring constant \( k \) from \( \mathrm{N/cm} \) to \( \mathrm{N/m} \): \( k = 19.6 \mathrm{~N/cm} = 1960 \mathrm{~N/m} \), and the compression distance \( x = 20.0 \mathrm{~cm} = 0.20 \mathrm{~m} \). Substitute these values into the equation: \( U = \frac{1}{2} (1960) (0.20)^2 \). Calculate \( U \): \( U = 39.2 \mathrm{~J} \).
02

Consider Energy Conservation

As the block moves up the incline, the mechanical energy is conserved because there is no friction. Therefore, the elastic potential energy initially stored in the spring will be completely converted into gravitational potential energy of the block at its highest point.
03

Calculate Gravitational Potential Energy Change

The change in gravitational potential energy \( \Delta U_g \) as the block moves to its highest point is equal to the initial elastic potential energy: \( \Delta U_g = U = 39.2 \mathrm{~J} \).
04

Determine Height and Incline Distance

The change in gravitational potential energy is also described as \( \Delta U_g = mgh \), where \( h \) is the vertical height the block ascends, \( m = 2.00 \mathrm{~kg} \), \( g = 9.8 \mathrm{~m/s}^2 \). Substitute the known values to find \( h \): \( 39.2 = (2.00)(9.8) h \), leading to \( h = 2.00 \mathrm{~m} \). Finally, to find the distance \( d \) along the incline, use \( h = d \sin(\theta) \), thus, \( 2.00 = d \sin(30.0^{\circ}) \). Solving for \( d \) gives \( d = 4.00 \mathrm{~m} \).

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Key Concepts

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

Elastic Potential Energy
Elastic potential energy is stored in an object when it is compressed or stretched. In the context of springs, this type of energy is crucial. When you compress or stretch a spring, you're adding energy to it. This energy is stored and can be released to do work later.

For a spring, elastic potential energy can be calculated using the formula: \[ U = \frac{1}{2} k x^2 \]where:
  • \( U \) is the elastic potential energy,
  • \( k \) is the spring constant,
  • \( x \) is the distance the spring is compressed or stretched from its natural length.
The spring constant \( k \) tells us how stiff the spring is. A higher value means more force is needed to compress or stretch the spring. In this exercise, the spring was compressed by 20 centimeters, resulting in an elastic potential energy of 39.2 Joules.
Gravitational Potential Energy
Gravitational potential energy relates to an object's position relative to Earth. Generally, it depends on the height of the object and the gravitational force acting on it. As an object moves upwards, energy is required to overcome the gravitational pull, hence increasing its potential energy.

This energy is represented by the formula:\[ \Delta U_g = mgh \]where:
  • \( \Delta U_g \) is the change in gravitational potential energy,
  • \( m \) is the mass of the object,
  • \( g \) is the acceleration due to gravity,
  • \( h \) is the increase in height.
In the exercise, as the block moves to the highest point on the incline, its gravitational potential energy increases by 39.2 Joules, which matches the elastic potential energy initially stored in the spring.
Energy Conservation
Energy conservation is a fundamental concept in physics, stating that energy in a closed system is constant. It cannot be created or destroyed; it only changes forms. This law is essential for solving mechanics problems like the one involving the inclined plane.

In this scenario, the block-spring system is a closed system with no external forces like friction. Hence, as the spring releases its stored elastic potential energy, it converts completely into gravitational potential energy as the block reaches its highest point. During the ascent up the incline, no energy is lost, only transformed.
Inclined Plane
An inclined plane is a simple machine that helps raise objects by distributing weight along the plane itself, thus requiring less force over a greater distance. In mechanics problems, slopes or inclines often come into play to study gravitational effects and energy transformations.

Inclined planes make it easier to calculate the distance an object travels vertically versus along the surface of the plane. For this exercise, calculating the highest point reached by the block involves understanding the angle of the incline and using trigonometric relationships. The height and distance along the incline are connected through the sine function: \[ h = d \sin(\theta) \]
Spring Constant
The spring constant, denoted by \( k \), measures a spring's stiffness. It's a fundamental attribute in Hooke's law and is crucial in calculating elastic potential energy. The larger the spring constant, the stiffer the spring, requiring more force for compression or extension.

For our exercise, converting the spring constant from \( N/cm \) to \( N/m \) allowed us to use consistent units in calculations. Here, the spring constant was 1960 N/m, indicating a fairly stiff spring. This value helps determine how much energy the spring can store when compressed over a certain distance.

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