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A \(1.3-\mathrm{kg}\) block is pushed up against a stationary spring, compressing it a distance of \(4.2 \mathrm{~cm}\). When the block is released, the spring pushes it away across a frictionless, horizontal surface. What is the speed of the block, given that the spring constant of the spring is \(1400 \mathrm{~N} / \mathrm{m}\) ?

Short Answer

Expert verified
The speed of the block is approximately 1.378 m/s.

Step by step solution

01

Understand the Problem

We need to determine the speed of a block once it is released from a spring. The block compresses the spring, storing potential energy, which is then converted to kinetic energy as the spring returns to its equilibrium position.
02

Identify Known Values

The mass of the block is given as \( m = 1.3 \) kg. The spring constant is \( k = 1400 \) N/m. The compression distance of the spring is \( x = 4.2 \) cm, which is \( 0.042 \) m when converted to meters.
03

Apply the Conservation of Energy Principle

When the block is released, the potential energy stored in the spring (elastic potential energy) is converted into kinetic energy of the block. The equations for these energies are:- Elastic potential energy: \( U = \frac{1}{2} k x^2 \)- Kinetic energy: \( K = \frac{1}{2} m v^2 \)Since potential energy is converted entirely into kinetic energy, we equate:\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]
04

Solve for the Speed of the Block

From the equation \( \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \), we solve for \( v \):1. Eliminate \( \frac{1}{2} \) from both sides: \[ k x^2 = m v^2 \]2. Rearrange to solve for \( v \): \[ v^2 = \frac{k x^2}{m} \]3. Take the square root of both sides to find \( v \): \[ v = \sqrt{\frac{k x^2}{m}} \]
05

Substitute and Compute

Substitute the known values: \( k = 1400 \) N/m, \( x = 0.042 \) m, and \( m = 1.3 \) kg into the equation:\[ v = \sqrt{\frac{1400 \times (0.042)^2}{1.3}} \]Calculate:\[ v = \sqrt{\frac{1400 \times 0.001764}{1.3}} = \sqrt{1.8992} \approx 1.378 \text{ m/s} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. When a block moves, like the one in the exercise once it's released by the spring, it gains kinetic energy. The formula for kinetic energy (K) is given by:
  • \( K = \frac{1}{2} m v^2 \)
Here, \( m \) is the mass of the object, and \( v \) is its velocity. For kinetic energy to be maximized, the velocity has to be as large as possible. In our scenario, the spring releases all its stored energy to the block, converting it into kinetic energy. This transformation allows us to calculate the speed of the block using the conservation of energy principle. The more kinetic energy the block has, the faster it moves.
Elastic Potential Energy
Elastic potential energy is the energy stored in elastic materials as a result of their stretching or compressing. Springs are an ideal example. When a spring is compressed, it holds energy that can be released at a later time. The formula for elastic potential energy (U) in a spring is:
  • \( U = \frac{1}{2} k x^2 \)
Where:
  • \( k \) is the spring constant, a measure of the spring's stiffness.
  • \( x \) is the displacement from the spring's equilibrium position.
In this problem, the block compresses the spring by 4.2 cm (0.042 m), storing potential energy. When released, this energy converts entirely into kinetic energy. Understanding how energy shifts from potential to kinetic helps in realizing why the block can speed away from the spring so quickly once released.
Spring Constant
The spring constant (\( k \)) is a key factor in determining a spring's behavior. It represents the stiffness of the spring and is defined as the force per unit length required to compress or stretch the spring. The higher the spring constant, the stiffer the spring.
  • Measured in Newtons per meter (N/m).
In our exercise, the given spring constant is 1400 N/m. This value indicates a relatively stiff spring, capable of exerting a strong force. A stiffer spring stores more energy for a given amount of compression. Therefore, when the block compresses the spring to 0.042 meters, a substantial amount of elastic potential energy is accumulated. Consequently, when the spring releases, this high energy translates into considerable kinetic energy for the block, increasing its speed.

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Most popular questions from this chapter

Calculate A \(0.21-\mathrm{kg}\) apple falls from a tree to the ground, \(4.0 \mathrm{~m}\) below. Ignoring air resistance, determine the apple's kinetic energy, \(K E\), the gravitational potential energy of the system, \(P E_{\text {gravity, }}\) and the total mechanical energy of the system, \(E\), when the apple's height above the ground is \(3.0 \mathrm{~m}\).

Predict \(\&\) Explain You throw a ball upward and let it fall to the ground. Your friend drops an identical ball straight down to the ground from the same height. (a) Is the change in kinetic energy (from just after the ball is released until just before it hits the ground) of your ball greater than, less than, or equal to the change in kinetic energy of your friend's ball? (b) Choose the best explanation from among the following: C. The change in gravitational potential energy is the same for each ball, which means that the change in kinetic energy must also be the same. A. Your friend's ball converts all of its initial energy into kinetic energy. B. Your ball is in the air longer, which results in a greater change in kinetic energy.

A particle moves without friction. At point A the particle has a kinetic energy of \(12 \mathrm{~J}\); at point B the particle is momentarily at rest, and the potential energy of the system is \(25 \mathrm{~J}\); at point \(\mathrm{C}\) the potential energy of the system is \(5 \mathrm{~J}\). (a) What is the potential energy of the system when the particle is at point \(A\) ? (b) What is the kinetic energy of the particle at point \(C\) ?

Think \& Calculate A sled slides without friction down a small, ice-covered hill. If the sled starts from rest at the top of the hill, its speed at the bottom is \(7.50 \mathrm{~m} / \mathrm{s}\). (a) On a second run, the sled starts with a speed of \(1.50 \mathrm{~m} / \mathrm{s}\) at the top. When it reaches the bottom of the hill, is its speed \(9.00 \mathrm{~m} / \mathrm{s}\), more than \(9.00 \mathrm{~m} / \mathrm{s}\), or less than \(9.00 \mathrm{~m} / \mathrm{s}\) ? Explain. (b) Find the speed of the sled at the bottom of the hill after the second run.

Meteorite On October 9,1992, a \(12=\mathrm{kg}\) meteorite struck a car in Peekskill, New York, creating a dent about \(22 \mathrm{~cm}\) deep, as shown in Figure 6.16. If the initial speed of the meteorite was \(550 \mathrm{~m} / \mathrm{s}\), what was the average force exerted on the meteorite by the car?

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