/*! 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 61 A \(2 \mathrm{~kg}\) block is dr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 61

A \(2 \mathrm{~kg}\) block is dropped from a height of \(0.4 \mathrm{~m}\) on a spring of force constant \(K=1960 \mathrm{Nm}^{-1}\). The maximum compression of the spring is (a) \(0.1 m\) (b) \(0.2 m\) (c) \(0.3 m\) (d) \(0.4 m\)

Short Answer

Expert verified
The maximum compression of the spring is (a) \(0.1 \mathrm{m}\)

Step by step solution

01

Understand the Given Information

We are given that the mass \(m\) of the block is \(2 \mathrm{~kg}\), the height \(h\) from which it is dropped is \(0.4 \mathrm{~m}\), and the spring constant \(K\) is \(1960 \mathrm{Nm}^{-1}\). We have to find the maximum compression of the spring, which we'll denote as \(x\).
02

Formulate the Equation for Conservation of Energy

According to the principle of conservation of energy, the gravitational potential energy of the block at the height from which it is dropped will be equal to the potential energy of the spring when it is fully compressed. This gives us the equation \[mgh = \frac{1}{2}Kx^2\]
03

Substitute the Known Values into the Equation

Substitute the given values into the equation: \[(2 \mathrm{~kg}) \cdot (9.8 \mathrm{m/s}^2) \cdot (0.4 \mathrm{~m})= \frac{1}{2} \cdot (1960 \mathrm{Nm}^{-1}) \cdot x^2\] which simplifies to \[7.84 \mathrm{J}= 980x^2\]
04

Solve for x (the maximum compression)

Solve the equation for \(x\): \[x^2=\frac{7.84 \mathrm{~J}}{980}\] therefore, \[x = \sqrt{\frac{7.84}{980}} = 0.1\, \mathrm{m}\]

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 91影视!

Key Concepts

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

Conservation of Energy
The concept of energy conservation is pivotal in solving problems involving moving objects and springs. It states that the total energy in an isolated system remains constant, meaning energy is neither created nor destroyed, only transformed from one form to another. In the given exercise, when the block is released from a height, its gravitational potential energy is converted into the spring's potential energy as the block compresses it.
  • This ensures that the equation \(mgh = \frac{1}{2}Kx^2\) holds true.
  • Here, \(mgh\) is the gravitational potential energy, and \(\frac{1}{2}Kx^2\) is the elastic potential energy of the spring.
Since no external forces do work on the block and spring system, energy transformation is seamless, and the energy conservation principle is a powerful tool for finding unknown quantities, like the spring compression, within the system.
Potential Energy
Potential energy is the stored energy of an object due to its position or state. In our scenario, we deal with two types of potential energy:
  • Gravitational Potential Energy (GPE): When the block is at a height, it has gravitational potential energy calculated by \(mgh\), where \(m\) is mass, \(g\) is the gravitational acceleration \(9.8 \mathrm{m/s}^2\), and \(h\) is the height.
  • Elastic Potential Energy (EPE): When the spring is compressed by a length \(x\), it stores elastic potential energy given by \(\frac{1}{2}Kx^2\), where \(K\) is the spring constant.
The change from gravitational to elastic potential energy allows us to ascertain how much the spring compresses, revealing the dynamic interplay between different energy forms in physical interactions.
Spring Constant
The spring constant \(K\) is a core parameter describing the stiffness of a spring. The greater the value of \(K\), the stiffer the spring, and more force is required to compress it to the same extent.
  • It is expressed in newtons per meter (\(\mathrm{Nm}^{-1}\)) and captures the relationship between force applied and displacement in Hooke's Law: \(F = Kx\).
  • In the problem, the spring constant \(K\) is crucial as it determines how far the spring compresses under the impact of the block.
Knowing the spring constant allows the application of the energy conservation equation to solve for \(x\), the spring's maximum compression, as it tells us how resistive the spring is to the block's impact energy.

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

Consider the following statements Assertion ( \(A\) ): In an elastic collision of two billiard balls, the total kinetic energy is conserved during the short time of collision of the balls (i.e., when they are in contact) Reason \((R)\) : Energy spent against friction does not follow the law of conservation of energy of these statements (a) Both \(A\) and \(R\) are true and the \(R\) is a correct explanation of \(A\) (b) Both \(A\) and \(R\) are true but the \(R\) is not a correct explanation of the \(A\) (c) \(A\) is true but the \(R\) is false (d) Both \(A\) and \(R\) are false

Which one of the following statement does not hold good when two balls of masses \(m_{1}\) and \(m_{2}\) undergo elastic collision (a) When \(m_{1}m_{2}\) and \(m_{2}\) at rest, after collision the ball of mass \(m_{2}\) moves with four times the velocity of \(m_{1}\) (c) When \(m_{1}=m_{2}\) and \(m_{2}\) at rest, there will be maximum transfer of kinetic energy (d) When collision is oblique and \(m_{2}\) at rest with \(m_{1}=m_{2}\), after collision the balls move in opposite directions

A rubber ball is dropped from a height of \(5 m\) on a planet where the acceleration due to gravity is not known. On bouncing, it rises to \(1.8 m\). The ball loses its velocity on bouncing by a factor of \(\quad\) [CBSE PMT 1998] (a) \(16 / 25\) (b) \(2 / 5\) (c) \(3 / 5\) (d) \(9 / 25\)

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by \(1 \mathrm{~m} / \mathrm{s}\) so as to have same \(K . E\). as that of boy. The original speed of the man will be (a) \(\sqrt{2} \mathrm{~m} / \mathrm{s}\) (b) \((\sqrt{2}-1) \mathrm{m} / \mathrm{s}\) (c) \(\frac{1}{(\sqrt{2}-1)} \mathrm{m} / \mathrm{s}\) (d) \(\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)

A body of mass \(5 \mathrm{~kg}\) is placed at the origin, and can move only on the \(x\)-axis. A force of \(10 \mathrm{~N}\) is acting on it in a direction making an angle of \(60^{\circ}\) with the \(x\)-axis and displaces it along the \(x\) axis by 4 metres. The work done by the force is (a) \(2.5 J\) (b) \(7.25 J\) (c) \(40 J\) (d) \(20 J\)
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Magnetism

Read Explanation

Fluids

Read Explanation

Work Energy and Power

Read Explanation

Circular Motion and Gravitation

Read Explanation

Astrophysics

Read Explanation
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.