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A 1.1\(\cdot \mathrm{kg}\) object is suspended from a vertical spring whose spring constant is 120 \(\mathrm{N} / \mathrm{m}\) . (a) Find the amount by which the spring is stretched from its unstrained length. (b) The object is pulled straight down by an additional distance of 0.20 \(\mathrm{m}\) and released from rest. Find the speed with which the object passes through its original position on the way up.

Short Answer

Expert verified
(a) The spring stretches 0.08983 m. (b) The object's speed is 1.738 m/s.

Step by step solution

01

Understanding the Problem – Part (a)

The object causes the spring to stretch until the force from the spring equals the gravitational force on the object. This is described by Hooke's Law: \( F_s = k \cdot x \) where \( k \) is the spring constant and \( x \) is the displacement. Also, \( F_g = m \cdot g \) where \( m \) is mass and \( g \) is gravitational acceleration \( (9.8 \, \text{m/s}^2) \). Equating forces gives: \( k \cdot x = m \cdot g \).
02

Solving for Stretch Amount – Part (a)

Rearrange the equation \( k \cdot x = m \cdot g \) to solve for \( x \): \[ x = \frac{m \cdot g}{k} \] Substituting values: \( m = 1.1 \, \text{kg} \), \( g = 9.8 \, \text{m/s}^2 \), \( k = 120 \, \text{N/m} \): \[ x = \frac{1.1 \cdot 9.8}{120} \approx 0.08983 \, \text{m} \]
03

Describing the Situation – Part (b)

The object is initially pulled an additional 0.20 m down from its stretched position and then released. It oscillates back to its original stretched position (equilibrium) where we need to find its speed. At the equilibrium, all the potential energy stored in spring converts into kinetic energy.
04

Applying Energy Conservation – Part (b)

Initially, the object has elastic potential energy after being displaced 0.20 m (plus initial stretch). Using the formula for elastic potential energy: \[ U = \frac{1}{2} k x^2 \] where \( x \) is the total displacement (0.08983 m + 0.20 m). Set this equal to the kinetic energy at the equilibrium point: \[ K = \frac{1}{2} mv^2 \].
05

Calculating Speed at Equilibrium – Part (b)

Equate the potential energy to kinetic energy: \[ \frac{1}{2} k x^2 = \frac{1}{2} mv^2 \] Cancel out the \( \frac{1}{2} \) and solve for \( v \): \[ k x^2 = mv^2 \] \[ v = \sqrt{\frac{k x^2}{m}} \] Substitute values \( k = 120 \, \text{N/m} \), \( x = 0.08983 + 0.20 \), \( m = 1.1 \, \text{kg} \). Calculate to find \( v \) as \( 1.738 \, \text{m/s} \).

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Key Concepts

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

Spring Constant
In the world of physics, the spring constant is a fundamental concept, especially when dealing with springs and Hooke's Law. The spring constant, denoted as \( k \), essentially tells us how stiff the spring is.

Hooke's Law forms the backbone of understanding the spring constant. It states that the force \( F_s \) required to stretch or compress a spring by a distance \( x \) is directly proportional to that distance. Mathematically, it's represented as:
  • \( F_s = k \cdot x \)
The larger the spring constant, the stiffer the spring. For example, a spring constant of 120 \( \text{N/m} \) means that it takes 120 Newtons of force to stretch the spring by one meter.

Understanding \( k \) helps us predict how a spring will react to different loads. Whether it's a tiny spring in a watch or a massive one in a suspension bridge, knowing the spring constant allows engineers and scientists to design and use materials appropriately.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as the result of their deformation. This type of potential energy is especially important in springs. When a spring is either compressed or stretched, it stores energy that can later be released.

The formula to calculate the elastic potential energy \( U \) of a spring is:
  • \( U = \frac{1}{2} k x^2 \)
Here, \( x \) is the displacement from the spring's original unstrained position. So, if a spring is stretched an additional 0.20 meters beyond its initial displacement, this extra distance also contributes to the stored energy.

As the spring returns to its equilibrium position, the energy converts into kinetic energy. This transformation is crucial in many mechanical systems, converting stored energy into motion.
Kinetic Energy
Kinetic energy is the energy of motion. Whenever an object moves, it possesses kinetic energy. It's an indispensable aspect of energy transformations in physics, especially in systems involving springs and motion.

The formula for calculating kinetic energy \( K \) is:
  • \( K = \frac{1}{2} mv^2 \)
where \( m \) is the mass of the object, and \( v \) is its velocity. In the case of the spring system, as the object is released and moves back toward the equilibrium position, its potential energy transforms into kinetic energy.

This converted kinetic energy helps us calculate the velocity of the object as it passes through the original point. By equating it to the initial elastic potential energy, we determine how fast the object moves, providing insights into its dynamic behavior.

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Most popular questions from this chapter

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of 12.0 \(\mathrm{Hz}\) . When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is 4.00 \(\mathrm{Hz}\) . Find the ratio \(m_{2} / m_{1}\) of the mases.

The fan blades on a jet engine make one thousand revolutions in a time of 50.0 \(\mathrm{ms}\) . Determine (a) the period (in seconds) and (b) the frequency (in \(\mathrm{Hz} )\) of the rotational motion. (c) What is the angular frequency of the blades?

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 1.90 s to complete one cycle. The height of each bounce above the equilibrium position is 45.0 \(\mathrm{cm}\) . Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 \(\mathrm{N} / \mathrm{m}\) . The block is shoved parallel to the spring axis and is given an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

A spiral staircase winds up to the top of a tower in an old castle. To measure the height of the tower, a rope is attached to the top of the tower and hung down the center of the staircase. However, nothing is available with which to measure the length of the rope. Therefore, at the bottom of the rope a small object is attached so as to form a simple pendulum that just clears the floor. The period of the pendulum is measured to be 9.2 s. What is the height of the tower?

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