Question

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

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

Step-by-step solution

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.

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.

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:\[ \frac{1}{2} k x^2 = \frac{1}{2} m v^2 \]

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 \( v \): \[ v = \sqrt{\frac{k x^2}{m}} \]

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:\[ v = \sqrt{\frac{1400 \times 0.001764}{1.3}} = \sqrt{1.8992} \approx 1.378 \text{ m/s} \]

Key concepts

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.