/*! 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 26 A 3-kg block rests on top of a 2... [FREE SOLUTION] | 91Ó°ÊÓ

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A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block.

Short Answer

Expert verified
Maximum speed is approximately 5.477 m/s, and maximum height is 1.53 m.

Step by step solution

01

Understanding the Problem

We have a 3-kg block on top of a 2-kg block which is supported by a spring with a constant of 40 N/m. When the upper block is removed, the spring will naturally accelerate the 2-kg block upward. We need to find the maximum speed and height of the 2-kg block.
02

Finding Potential Energy of the Spring

Initially, the spring is compressed by weight of both blocks, so potential energy is stored in it. The force exerted by both blocks on the spring is \( (3 \, \text{kg} + 2 \, \text{kg}) \cdot 9.8 \, \text{m/s}^2 = 49 \, \text{N} \). The spring is compressed by \( x = \frac{49}{40} = 1.225 \, \text{m} \). Hence, potential energy \( E_p = \frac{1}{2}kx^2 = \frac{1}{2} \cdot 40 \cdot (1.225)^2 = 30.0125 \, \text{J} \).
03

Applying Energy Conservation for Maximum Speed

When the upper block is removed, the energy stored in the spring is converted to kinetic energy of the 2-kg block. We set the potential energy equal to its kinetic energy. Thus, \( \frac{1}{2}mv^2 = 30.0125 \, \text{J} \). Solving for \( v \) gives \( v = \sqrt{\frac{2 \cdot 30.0125}{2}} = \sqrt{30.0125} \approx 5.477 \, \text{m/s} \).
04

Finding Maximum Height Using Energy Conservation

When the 2-kg block has reached its maximum height, all kinetic energy is converted into gravitational potential energy. Setting \( mgh = \frac{1}{2}mv^2 \), where \( h \) is maximum height and solving, \( 2 \cdot 9.8 \cdot h = 30.0125 \implies h = \frac{30.0125}{19.6} \approx 1.53 \, \text{m} \).

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Key Concepts

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

Potential Energy
Potential energy is the stored energy an object possesses due to its position or state. In this exercise, a spring holds potential energy when compressed by the blocks resting on it. The formula for potential energy stored in a spring is given by:
  • \( E_p = \frac{1}{2} k x^2 \)
where:
  • \( k \) is the spring constant (40 N/m in this case).
  • \( x \) is the compression distance (1.225 m here).
The potential energy initially stored (30.0125 J) is due to the weight of the blocks compressing the spring. This energy becomes available to do work when the 3-kg block is suddenly removed.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. As the spring releases, this stored potential energy converts into kinetic energy in the 2-kg block.The formula to calculate kinetic energy is:
  • \( E_k = \frac{1}{2}mv^2 \)
where:
  • \( m \) is the mass of the object (2 kg in this situation).
  • \( v \) is the velocity of the object.
When the spring releases, the velocity of the block at its maximum speed can be calculated using the stored energy, resulting in a potential maximum speed of approximately 5.477 m/s.
Energy Conservation
Energy conservation is a fundamental principle stating that energy cannot be created or destroyed; it can only change forms. In this problem, the principle of energy conservation is pivotal.When the spring is compressed, it stores energy as potential energy. Once the block is released and the spring returns to its equilibrium position, this energy transforms into kinetic energy. For the maximum height, the kinetic energy at its peak is then completely converted into gravitational potential energy. Using the formula:
  • \( mgh = \frac{1}{2}mv^2 \)
we determine the height reached. Gravitational force (9.8 m/s²) allows us to calculate the maximum height reached by the 2-kg block, approximately 1.53 m here. No energy is lost; it simply transitions between forms.
Mass and Force
In dynamics, understanding the relationship between mass and force is essential. Here, mass affects how force interacts with an object. The 3-kg block on top of the 2-kg block, along with gravitational acceleration (9.8 m/s²), creates a force of 49 N compressing the spring. Force is calculated by the formula:
  • Force = Mass × Acceleration
Upon removal of the 3-kg block, the spring no longer supports its weight, meaning the system re-evaluates how force impacts the 2-kg block alone. Therefore, the 2-kg block accelerates upward, influenced by the force of the spring's release. Each block’s mass determines how it interacts with the forces applied or released, controlling the motion behavior observed during this interaction. Understanding these dynamics is key to predicting motion outcomes correctly.

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Most popular questions from this chapter

A 5 -kg sphere is dropped from a height of \(y=2 \mathrm{m}\) to test newly designed spring floors used in gymnastics. The mass of the floor section is \(10 \mathrm{kg}\), and the effective stiffness of the floor is \(k=120 \mathrm{kN} / \mathrm{m}\). Knowing that the coefficient of restitution between the ball and the platform is \(0.6,\) detrmine \((a)\) the height \(h\) reached by the sphere after rebound, \((b)\) the maximum force in the springs.

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.

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A ballistic pendulum is used to measure the speed of high-speed projectiles. A \(6-\mathrm{g}\) bullet \(A\) is fired into a \(1-\mathrm{kg}\) wood block \(B\) suspended by a cord with a length of \(l=2.2 \mathrm{m}\). The block then swings through a maximum angle of \(\theta=60^{\circ} .\) Determine \((a)\) the initial speed of the bullet \(v_{0},(b)\) the impulse imparted by the bullet on the block, \((c)\) the force on the cord immediately after the impact.

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