/*! 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 24 Block \(A\) in Fig. E8.24 has ma... [FREE SOLUTION] | 91Ó°ÊÓ

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Block \(A\) in Fig. E8.24 has mass \(1.00 \mathrm{kg},\) and block \(B\) has mass 3.00 \(\mathrm{kg}\) . The blocks are forced together, compressing a spring \(S\) between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block \(B\) acquires a speed of 1.20 \(\mathrm{m} / \mathrm{s}\) . (a) What is the final speed of block \(A\) ? (b) How much potential energy was stored in the compressed spring?

Short Answer

Expert verified
(a) 3.6 m/s, (b) 6.48 J.

Step by step solution

01

Analyze the problem

We have two masses, Block A and Block B, with a spring between them. The spring is released, propelling the blocks in opposite directions. We'll use conservation of momentum and energy principles to solve the questions.
02

Apply conservation of momentum

According to the conservation of momentum, the total momentum before and after the release must be equal. Before release, the system is at rest, so initial momentum is zero. Thus, \( m_A \, v_A + m_B \, v_B = 0 \). Given \( m_A = 1.00 \, \text{kg} \), \( m_B = 3.00 \, \text{kg} \), and \( v_B = 1.20 \, \text{m/s} \), we find \( v_A \).
03

Solve for velocity of Block A

Using the equation \( m_A \, v_A + m_B \, v_B = 0 \), substitute the known values: \( 1.00 \, \text{kg} \cdot v_A + 3.00 \, \text{kg} \cdot 1.20 \, \text{m/s} = 0 \). Solving for \( v_A \) yields: \( v_A = -3.6 \, \text{m/s} \). The negative sign indicates Block A moves in the opposite direction to Block B.
04

Calculate the potential energy in the spring

Since the system's kinetic energy came from the potential energy stored in the spring, use energy conservation: \( \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = E_{\text{spring}} \). Substitute \( v_A = -3.6 \, \text{m/s} \) and \( v_B = 1.20 \, \text{m/s} \) to find the spring energy.
05

Compute spring potential energy

Substituting values, \( E_{\text{spring}} = \frac{1}{2} \cdot 1.00 \, \text{kg} \cdot (-3.6 \, \text{m/s})^2 + \frac{1}{2} \cdot 3.00 \, \text{kg} \cdot (1.20 \, \text{m/s})^2 \). This results in \( E_{\text{spring}} = 6.48 \, \text{J} \).

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Key Concepts

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

Conservation of Energy
The principle of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another. In the context of our exercise, the system initially holds potential energy in the compressed spring. When released, this energy converts into kinetic energy, propelling the blocks in opposite directions. Understanding this transformation is crucial for solving problems involving mechanical systems.
  • Potential energy in the spring is converted into kinetic energy of the blocks.
  • The total energy in the system remains constant, assuming no energy loss due to external factors like friction.
In our example, the total potential energy stored in the spring equals the total kinetic energy of both blocks post-release. Ensuring you account for all energy types at different stages is key to accurately analyzing such systems.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. In the exercise, both Block A and Block B acquired kinetic energy once the spring was released. The change from being at rest (zero kinetic energy) to moving is directly related to the energy released by the spring.
  • Block B, with a mass of 3 kg and a velocity of 1.20 m/s, has a significant kinetic energy component.
  • Block A, despite having less mass (1 kg), moves faster (3.6 m/s, found from calculations) and also contributes to the system's total kinetic energy.
Each block's kinetic energy can be calculated individually and added together to verify it matches the potential energy initially stored in the spring. Paying attention to direction and mass distribution is fundamental when assessing kinetic energy distribution.
Potential Energy
Potential energy represents the stored energy in a system due to its position or arrangement. In our exercise, the spring stores potential energy while compressed. Once the system is released, this energy becomes kinetic energy that moves the blocks. The formula for elastic potential energy (like that in a spring) is:\[ PE_{spring} = \frac{1}{2}kx^2 \]where \( k \) is the spring constant, and \( x \) is the displacement from its natural length. Although not directly given in this problem, understanding this relationship helps comprehend how potential energy works in similar contexts.
  • The energy stored depends on how much the spring is compressed.
  • Upon release, all stored energy transitions into kinetic energy.
This transition highlights how fields and forces, like that of a spring, can store and later release energy, driving motion and change in mechanical systems.

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Most popular questions from this chapter

A \(68.5-k g\) astronaut is doing a repair in space on the orbit ing space station. She throws a 2.25 -kg tool away from her at 3.20 \(\mathrm{m} / \mathrm{s}\) relative to the space station. With what speed and in what direction will she begin to move?

BIO Biomechanics. The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 \(\mathrm{ms}\) . (This number can also vary, depending on the racket and swing.) We shall assume a 30.0 -ms contact time for this exercise. The fastest-known served tennis ball was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be 73.14 \(\mathrm{m} / \mathrm{s}\) . (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of \(55 \mathrm{m} / \mathrm{s},\) what force and what impulse did he exert on the ball, assuming only horizontal motion?

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) (Fig. P8.95). You can ignore friction between the cart and the floor. A 15.0 -kg package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

A 1200 -kg station wagon is moving along a straight highway at 12.0 \(\mathrm{m} / \mathrm{s}\) . Another car, with mass 1800 \(\mathrm{kg}\) and speed \(20.0 \mathrm{m} / \mathrm{s},\) has its center of mass 40.0 \(\mathrm{m}\) ahead of the center of mass of the station wagon (Fig. E8.54). (a) Find the position of the center of mass of the system consisting of the two automobiles. (b) Find the magnitude of the total momentum of the system from the given data. (c) Find the speed of the center of mass of the system. (d) Find the total momentum of the system, using the speed of the center of mass. Compare your result with that of part (b).

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