/*! 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 30 A 10-kg block is attached to spr... [FREE SOLUTION] | 91Ó°ÊÓ

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A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs unstretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block.

Short Answer

Expert verified
(a) The velocity after the block has moved 50 mm is 2 m/s. (b) The maximum velocity is achieved at the equilibrium position.

Step by step solution

01

Identify and List the Given Information

We are given that there is a 10-kg block connected to two springs with spring constants of each spring as \( k = 2 \text{ kN/m} \). The block moves 50 mm down when released.
02

Set Up the Energy Balance Equation

To solve this problem, we consider the conservation of energy. Define the potential energy in the springs and the kinetic energy of the block. Initially, the potential energy is zero since both springs are unstretched, and the kinetic energy is also zero since the block starts from rest.
03

Write the Energy Conservation Equation

The potential energy stored in the springs after the block has moved 50 mm (0.05 m) will be converted into kinetic energy. We write:\[ \frac{1}{2} k x^2 + \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]where \( x = 0.05 \text{ m} \), \( m = 10 \text{ kg} \), and \( k = 2000 \text{ N/m} \).
04

Solve for the Velocity After 50 mm of Movement

Insert the given values into the equation:\[ 2 \times \frac{1}{2} \times 2000 \times (0.05)^2 = \frac{1}{2} \times 10 \times v^2 \]Solve for \( v \), the velocity: \[ v = \sqrt{2 \cdot 2000 \cdot 0.05^2 / 10} \]Calculate to find \( v = 2 \text{ m/s} \).
05

Determine the Maximum Velocity of the Block

The maximum velocity occurs when the potential energy in the springs is completely converted into kinetic energy and then back into potential energy. This is at the equilibrium position where both springs are neither fully stretched nor compressed. At this point, the potential energy is equal to zero again, as springs have equal force acting on them.
06

Use Energy Conservation to Find Maximum Velocity

The total energy initially stored in the springs when fully compressed or extended is at the same point where kinetic energy will be maximum. This can also happen when the block oscillates through its equilibrium position: \[ k x_{max}^2 = m v_{max}^2 \]For maximum velocity, potentials in springs are zero. Set initial elastic energy equals maximum kinetic energy: \[ 2 \times \frac{1}{2} \times k x^2 = \frac{1}{2} m v_{max}^2 \]Setting spring potential zero at equilibrium means solving \( v_{max} = \sqrt{4 k x^2 / m} \).

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Key Concepts

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

Spring Mechanics
In spring mechanics, we understand the behavior of springs subjected to forces. A spring's fundamental characteristic is its ability to store mechanical energy, exhibited as potential energy when the spring is either compressed or extended. The spring constant, denoted by \( k \), measures the stiffness of a spring. In our example, each spring attached to the 10-kg block has a spring constant of 2 kN/m, indicating how resistant it is to deformation.
Springs follow Hooke’s Law, which states that the force needed to extend or compress a spring by some distance \( x \) is proportional to that distance. This law is mathematically represented by \( F = kx \). When a spring returns to its original shape, it releases stored potential energy. This energy transformation is central to spring mechanics and is crucial when analyzing systems with multiple springs, like the one in the exercise.
Some key aspects of spring mechanics include:
  • Understanding spring constant \( k \): Determines energy storage capacity.
  • Hooke’s Law \( F = kx \): Describes the proportional relationship between force and displacement.
  • Potential energy stored \( \frac{1}{2} k x^2 \): Calculated when a spring is displaced by \( x \).
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When the 10-kg block moves after being released, it gains kinetic energy. The kinetic energy \( KE \) of the moving block is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the block.
In this problem, when the block is first released and moves 50 mm downward, the accumulated potential energy in the springs is converted into kinetic energy, making the block accelerate. Initially, when the block is at rest, its kinetic energy is zero. Once it starts moving, the equation \( \frac{1}{2} k x^2 + \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \) helps us find the velocity by calculating the amount of potential energy transformed into kinetic energy.
Some important points about kinetic energy:
  • Depends on mass (10 kg in this case) and velocity (\( v \)).
  • Calculated using \( \frac{1}{2} mv^2 \).
  • Essential in analyzing movement resulting from potential energy release.
Potential Energy
Potential energy is the stored energy of an object due to its position or configuration. In our exercise, potential energy is stored in the springs when they are compressed or extended. This energy can be calculated using the formula \( PE = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from its equilibrium position, which in this case is 0.05 m or 50 mm.
When the block is released, the potential energy stored in the springs is gradually converted into kinetic energy as the block moves. Initially, when both springs are unstressed, the potential energy is zero. As the springs stretch or compress, potential energy increases, peaking as they reach their maximum displacement before converting fully to kinetic energy.
Here's a summary of potential energy in spring systems:
  • Calculated using \( \frac{1}{2} k x^2 \) — significant in energy transformations.
  • Reflects the capability to convert into kinetic energy.
  • Peaks when springs are fully compressed or stretched, then transforms into kinetic energy.

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Most popular questions from this chapter

Rockfalls can cause major damage to roads and infrastructure. To design mitigation bridges and barriers, engineers use the coefficient of restitution to model the behavior of the rocks. Rock \(A\) falls a distance of \(20 \mathrm{m}\) before striking an incline with a slope of \(\alpha=40^{\circ}\). Knowing that the coefficient of restitution between rock \(A\) and the incline is \(0.2,\) determine the velocity of the rock after the impact.

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