/*! 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 17 A block of mass \(1 \mathrm{~kg}... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A block of mass \(1 \mathrm{~kg}\) is pressed against a spring of constant \(400 \mathrm{~N} / \mathrm{m}\). The spring is compressed by \(10 \mathrm{~cm}\) and block is released. Which of the following is a possible velocity of the block during subsequent motion? (1) \(2 \mathrm{~m} / \mathrm{s}\) (2) \(1 \mathrm{~m} / \mathrm{s}\) (3) \(3 \mathrm{~m} / \mathrm{s}\) (4) \(4 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The block can reach a velocity of \(2 \mathrm{~m/s}\).

Step by step solution

01

Identify Stored Energy in Spring

The potential energy stored in a compressed spring is given by the formula \( E_{s} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression distance. Substitute \( k = 400 \mathrm{~N/m} \) and \( x = 0.10 \mathrm{~m} \) (since 10 cm = 0.10 m) to find \( E_{s} = \frac{1}{2} \times 400 \times (0.10)^2 \).
02

Calculate Potential Energy

Calculate \( E_{s} = \frac{1}{2} \times 400 \times 0.01 = 2 \mathrm{~J} \). This is the energy stored in the spring when compressed.
03

Equate Stored Energy to Kinetic Energy

When the spring is fully released, all its stored potential energy converts to kinetic energy of the block. Therefore, set \( KE = E_s \). The kinetic energy formula is \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the block and \( v \) is the velocity. Substitute \( m = 1 \mathrm{~kg} \) and \( KE = 2 \mathrm{~J} \) to solve for \( v \).
04

Solve for Velocity

Using the equation \( 2 = \frac{1}{2} \times 1 \cdot v^2 \), solve for \( v \). Rearrange to find \( v^2 = 4 \), and then take the square root to find \( v = \sqrt{4} = 2 \mathrm{~m/s} \).
05

Compare Possible Velocities

Compare the calculated velocity \( 2 \mathrm{~m/s} \) with the options provided: 1) \(2 \mathrm{~m/s} \), 2) \(1 \mathrm{~m/s} \), 3) \(3 \mathrm{~m/s} \), 4) \(4 \mathrm{~m/s} \). The calculated velocity corresponds to option 1.

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.

Potential Energy
In a spring-mass system, potential energy plays a key role when an external force compresses a spring. The spring stores this energy as 'potential energy'. The potential energy in an elastic spring is determined using the formula:
  • \( E_{s} = \frac{1}{2} k x^2 \)
Here, \( k \) is the spring constant, which indicates the stiffness of the spring, and \( x \) is the compression distance. Substituting the values from the problem, we have \( k = 400 \mathrm{~N/m} \) and \( x = 0.10 \mathrm{~m} \) (since 10 cm = 0.10 m). Thus, your calculation would look like this:
  • \( E_{s} = \frac{1}{2} \times 400 \times (0.10)^2 = 2 \mathrm{~J} \)
This means that the spring stores 2 joules of potential energy when it is fully compressed. Each element within the equation reflects how much the spring is able to store based on its physical properties.
Understanding potential energy in springs allows us to predict how energy stored can later contribute to the motion of the mass once it is released.
Kinetic Energy
Kinetic energy arises when the potential energy stored in the spring is converted into motion energy once the object is released. The kinetic energy of an object depends on its mass and velocity, and it is described by the equation:
  • \( KE = \frac{1}{2} mv^2 \)
For the given problem, where the block has a mass \( m \) of \( 1 \mathrm{~kg} \) and the potential energy of the spring is converted to kinetic energy at 2 joules, we can use the formula to solve for velocity \( v \):
  • Plug \( KE = 2 \mathrm{~J} \) and \( m = 1 \mathrm{~kg} \) into the equation: \( 2 = \frac{1}{2} \times 1 \times v^2 \)
  • Rearrange the equation to solve for \( v^2 \): \( v^2 = 4 \)
  • Take the square root: \( v = \sqrt{4} = 2 \mathrm{~m/s} \)
This tells us that when all the potential energy is converted to kinetic energy, the velocity of the block reaches \( 2 \mathrm{~m/s} \). It illustrates how energy conversions within the system determine the object's motion.
Energy Conservation
The principle of energy conservation is fundamental when analyzing spring-mass systems. It states that energy cannot be created or destroyed, only transformed from one form to another. In the context of our problem:
  • The potential energy stored in the compressed spring is converted entirely into kinetic energy of the block.
  • Throughout this process, the total mechanical energy within the system remains constant.
By calculating the potential energy, we knew that 2 joules were stored in the spring. Upon release, this 2 joules became kinetic energy, allowing us to determine the velocity of the block using energy conservation. The transformation looks like:
  • Potential Energy (Spring) \( \rightarrow \) Kinetic Energy (Block)
If there were no external forces (like friction), the energy would retain its magnitude throughout the motion. This demonstrates a basic understanding of energy conservation, highlighting how shifts in energy states govern the dynamics of motion in spring-mass systems.

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 chain of length \(l\) and mass \(m\) lies on the surface of a smooth sphere of radius \(R>l\) with one end tied to the top of the sphere. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {Gravitational potential energy w.r.t. centre of the sphere} & \text { a. } \text{\(\frac{\operatorname{Rg}}{l}\left[1-\cos \left(\frac{l}{R}\right)\right]\)} \\ \hline \text { ii. } \text{The chain is released and slides down, its \(\mathrm{KE}\) when it has slid by \(\theta\)} & \text { b. } \text{\(\frac{2 R g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \text { iii. } \text{The initial tangential acceleration} & \text { c. } \text{\(\frac{M R^{2} g}{l} \sin \left(\frac{l}{R}\right)\)}\\\ \hline \text { iv. } \text{The radial acceleration \(a_{r}\)} & \text { d. } \text{\(\frac{M R^{2} g}{l}\left[\sin \left(\frac{l}{R}\right)+\sin \theta-\sin \left(\theta+\frac{l}{R}\right)\right]\)} \\ \hline \end{array} $$

One end of an unstretched vertical spring is attached to the ceiling and an object attached to the other end is slowly lowered to its equilibrium position. If \(S\) is the gain in spring energy and \(G\) is the loss in gravitational potential energy in the process, then (1) \(S=G\) (2) \(S=2 G\) (3) \(G=2 S\) (4) None of these

Two blocks \(A\) and \(B\) of mass \(m\) and \(2 m\) respectively are connected by a massless spring of spring constant \(K\). This system lies over a smooth horizontal surface. At \(t=0\) the block. \(A\) has velocity \(u\) towards right as shown while the speed of block \(B\) is zero, and the length of spring is equal its natural length at that instant. $$ \begin{array}{|c|c|} \hline {\begin{array}{c} \text { Column I } \\ \end{array}} & {\begin{array}{c} \text { Column II } \\ \end{array}} \\ \hline \text { i. } \text {The velocity of block A} & \text { a. } \text{can never be zero} \\ \hline \text { ii. } \text{The velocity of block B} & \text { b. } \text{may be zero at certain instants of time} \\ \hline \text { iii. } \text{The kinetic energy of system of two blocks} & \text { c. } \text{is maximum at maximum compression of spring}\\\ \hline \text { iv. } \text{The potential energy of spring} & \text { d. } \text{is maximum at maximum extension of spring} \\ \hline \end{array} $$ Now match the given columns and select the correct option from the codes given below. Codes: $$ \begin{array}{lllll} & \text { i. } & \text { ii. } & \text { iii. } & \text { iv. } \\ \text { (1) } &\text { b }& \text { b} & \text { a, c} & \text { b, d}\\\ \text { (2) } &\text { b }& \text { a, c } & \text { a } & \text { d}\\\ \text { (3) } &\text { d }& \text { a, c } & \text { b } & \text { b, c }\\\ \text { (4) } &\text { a, d }& \text { c } & \text { b } & \text { d } \end{array} $$

A particle of mass \(m\) moves with a variable velocity \(v\), which changes with distance covered \(x\) along a straight line as \(v=k \sqrt{x}\), where \(k\) is a positive constant. The work done by all the forces acting on the particle, during the first \(t\) seconds is (l) \(\frac{m k^{4}}{t^{2}}\) (2) \(\frac{m k^{4} t^{2}}{4}\) (3) \(\frac{m k^{4} t^{2}}{8}\) (4) \(\frac{m k^{4} t^{2}}{16}\)

A single conservative force \(F(x)\) acts on a \(1.0-\mathrm{kg}\) particle that moves along the \(x\)-axis. The potential energy \(U(x)\) is given by \(U(x)=20+(x-2)^{2}\) where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\), the particle has a kinetic energy of \(20 \mathrm{~J}\). Determine the equation of \(F(x)\) as a function of \(x\). (1) \(F=2+x\) (2) \(F=2+3 x\) (3) \(F=2(2-x)\) (4) \(F=3+2 x\)

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.