/*! 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 16 A \(700 \mathrm{~g}\) block is r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=450 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?\) (d) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?

Short Answer

Expert verified
(a) 8.115 J, (b) -8.115 J, (c) 1.187 m, (d) 26.8 cm

Step by step solution

01

Find Work Done by Block on Spring

The work done by the block on the spring is equal to the potential energy stored in the spring at maximum compression. Use the formula for elastic potential energy: \[ W = \frac{1}{2} k x^2 \] where \( x = 0.19 \text{ m} \) (compression) and \( k = 450 \text{ N/m} \). Substitute the values to get: \[ W = \frac{1}{2} \times 450 \times (0.19)^2 = 8.115 \text{ J} \] Thus, the work done by the block on the spring is \( 8.115 \text{ J} \).
02

Work Done by Spring on Block

According to the work-energy principle, the work done by the spring on the block is equal and opposite to the work done by the block on the spring. Hence, the work done by the spring on the block is also \(-8.115 \text{ J}\).
03

Find Initial Height Using Energy Conservation

Using energy conservation, the initial potential energy of the block is converted to potential energy of the spring. The gravitational potential energy at height \( h_0 \) is given by: \[ mgh_0 = \frac{1}{2} k x^2 \] where \( m = 0.7 \text{ kg} \) and \( g = 9.81 \text{ m/s}^2 \). Substitute known values: \[ \times 9.81h_0 = 8.115 \] Solve for \( h_0 \): \[ h_0 = \frac{8.115}{0.7 \times 9.81} \approx 1.187 \text{ m} \] Therefore, \( h_0 \approx 1.187 \text{ m} \).
04

Determine Maximum Spring Compression for Double Height

If the block is released from height \( 2h_0 \), the gravitational potential energy is double: \[ mgh = \frac{1}{2} k x^2 \] Let initial height be \( 2h_0 = 2 \times 1.187 = 2.374 \text{ m} \), then: \[ 0.7 \times 9.81 \times 2.374 = \frac{1}{2} \times 450 \times x^2 \] Solve for \( x \): \[ x^2 = \frac{16.23}{225} \rightarrow x = \sqrt{0.0721} \approx 0.268 \text{ m} \] The maximum compression of the spring is approximately \( 26.8 \text{ cm} \).

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 91Ó°ÊÓ!

Key Concepts

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

Elastic Potential Energy
Elastic potential energy refers to the energy stored in elastic materials, like springs, when they are compressed or stretched. Whenever you press a spring or pull a rubber band, energy is being stored until you release it, allowing it to return to its original shape. This energy can be calculated using the formula:
  • \( E = \frac{1}{2} k x^2 \)
This formula calculates the potential energy (E) stored in the spring, where \( k \) is the spring constant and \( x \) is the distance the spring is compressed or stretched from its equilibrium position. In the original exercise, when the block compresses the spring by 19 cm, the energy stored in the spring as elastic potential energy is the work done by the block. The greater the compression or stretch, the more energy is stored.
The spring constant \( k \) is crucial as it shows how stiff the spring is. A higher spring constant means a stiffer spring, requiring more force to compress it by a certain amount.
Gravitational Potential Energy
Gravitational potential energy is the energy stored in an object due to its position in a gravitational field. It is heavily tied to the object's height above the ground. This type of energy is calculated using:
  • \( E_{p} = mgh \)
Here, \( m \) is mass, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point. In our exercise, the block starts with gravitational potential energy at height \( h_0 \). When it's released, this energy turns into kinetic energy as it falls and is then transferred into elastic potential energy when the block compresses the spring.
This shows the conservation of energy principle where at any state, the energy present must equal the total energy at another state in the system. In this case, the gravitational energy at the block's initial height converts to elastic energy in the spring when compressed.
Hooke's Law
Hooke's Law is fundamental in understanding how springs and other elastic materials compress and extend. It states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed and is calculated with:
  • \( F = -kx \)
In this expression, \( F \) is force, \( k \) is the spring constant, and \( x \) is the displacement from its equilibrium position. The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement, essentially a restoring force trying to bring the spring back to its original length.
Understanding Hooke's Law allows us to predict how a spring behaves when forces are applied, leading to potential energy changes within the system. In the exercise example, when the block compresses the spring by 19 cm, Hooke’s Law helps explain the force the spring exerts upward on the block. This force becomes key in understanding both the elastic potential energy and the work done by and on the spring.

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 \(5.0 \mathrm{~g}\) 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?

A cookie jar is moving up a \(40^{\circ}\) incline. At a point \(45 \mathrm{~cm}\) from the bottom of the incline (measured along the incline), the jar has a speed of \(1.4 \mathrm{~m} / \mathrm{s}\). The coefficient of kinetic friction between jar and incline is \(0.15\). (a) How much farther up the incline will the jar move? (b) How fast will it be going when it has slid back to the bottom of the incline? (c) Do the answers to parts (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction (but do not change the given speed or location)?

A worker pushed a \(23 \mathrm{~kg}\) block \(8.4 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block - floor system?

The total mechanical energy of a \(2.00 \mathrm{~kg}\) particle moving along an \(x\) axis is \(5.00\) J. The potential energy is given as \(U(x)=\left(x^{4}-\right.\) \(\left.2.00 x^{2}\right) \mathrm{J}\), with \(x\) in meters. Find the maximum velocity.

A \(60 \mathrm{~kg}\) skier leaves the end of a ski-jump ramp with a velocity of \(27 \mathrm{~m} / \mathrm{s}\) directed \(25^{\circ}\) above the horizontal. Suppose that as a result of air drag the skier returns to the ground with a speed of 22 \(\mathrm{m} / \mathrm{s}\), landing \(14 \mathrm{~m}\) vertically below the end of the ramp. From the launch to the return to the ground, by how much is the mechanical energy of the skier-Earth system reduced because of air drag?

See all solutions

Recommended explanations on Physics Textbooks

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.