Problem 28 - Physics

Question

A vertical spring with a spring constant of \(450 \mathrm{~N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{~cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

Step-by-Step Solution

Answer

The block was dropped from a height of approximately 4.79 cm above the compressed spring.

Step 1: Understand the Problem

We need to find the height from which a block was dropped onto a spring. The spring compresses, and we are given the spring constant, block mass, and compressed distance. This is an energy conservation problem, where potential energy is converted into elastic potential energy in the spring.

Step 2: Write the Energy Conservation Equation

Initially, the block has gravitational potential energy at height \( h \). When the block momentarily stops, all initial energy is stored as elastic potential energy in the spring. Thus, \( mgh = \frac{1}{2}kx^2 \).

Step 3: Identify Variables and Values

The mass of the block \( m = 0.30 \, ext{kg} \), gravitational acceleration \( g = 9.8 \, ext{m/s}^2 \), spring constant \( k = 450 \, ext{N/m} \), and compression distance \( x = 0.025 \, ext{m} \) since it is given in cm and needs to be converted to meters.

Step 4: Substitute Values into Energy Equation

Substitute the known values into the energy conservation equation: \( (0.30)(9.8)h = \frac{1}{2}(450)(0.025)^2 \).

Step 5: Simplify and Solve for Height \( h \)

Calculate the right side: \( \frac{1}{2}(450)(0.025)^2 = 0.140625 \, ext{J} \). Solve for \( h \): \((0.30)(9.8)h = 0.140625 \Rightarrow h = \frac{0.140625}{2.94} \approx 0.04785 \, ext{m} \).

Step 6: Convert Height to Centimeters

Convert \( h = 0.04785 \, ext{m} \) to centimeters: \( h = 0.04785 \, ext{m} \times 100 = 4.785 \, ext{cm} \).

Key Concepts

Spring Constant

The spring constant is a central concept in understanding springs and their behavior. It is denoted by the symbol \( k \) and measured in Newtons per meter (N/m). The spring constant indicates how much force is needed to compress or extend a spring by a unit length. Essentially, it tells us about the stiffness of the spring.

• A higher spring constant means a stiffer spring, requiring more force to compress or extend.
• A lower spring constant indicates a less stiff spring, which compresses or extends more easily.

In this exercise, the spring constant is given as 450 N/m. This value helps us calculate how much energy is stored in the spring when compressed by the block. It's crucial for understanding how the spring "pushes back" against the block when it's compressed.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy that an object possesses due to its position in a gravitational field. It is determined by the formula \( U_g = mgh \), where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height above a reference point.
When an object is elevated, it has more potential energy because it can fall, converting that potential energy into kinetic energy.

• The higher the object, the greater its gravitational potential energy.
• The more massive the object, the more GPE it has.

In this scenario, the block starts with gravitational potential energy at a height \( h \), which it then converts into elastic potential energy as it compresses the spring.

Elastic Potential Energy

Elastic potential energy is the energy stored in elastic materials as they are stretched or compressed. For springs, it is calculated using the formula \( U_e = \frac{1}{2}kx^2 \), where \( k \) is the spring constant, and \( x \) is the compression or extension from the equilibrium position.
This form of energy is potential because it can be converted back into kinetic energy as the spring returns to its original shape.

• The further the spring is compressed or stretched, the more elastic potential energy it stores.
• Elastic potential energy is equal to the work done on the spring to compress or extend it.

In this exercise, the spring compresses 2.5 cm under the block's weight, converting gravitational potential energy into elastic potential energy. This conversion stops the block momentarily at the spring's maximum compression.

Energy Conversion in Physics

In physics, energy conversion refers to the process of transforming one type of energy into another. This concept is fundamental to understanding how different systems and machines operate.
For the block and spring scenario, energy conversion is all about transforming the gravitational potential energy of the block into the elastic potential energy of the spring.

• Initially, the block has a certain amount of gravitational potential energy due to its height above the spring.
• As the block falls, it loses gravitational potential energy and gains kinetic energy.
• When the block compresses the spring, the kinetic energy is then converted into elastic potential energy within the spring.
• At the moment the spring is fully compressed, all the initial energy is stored as elastic potential energy.

This exemplifies how energy is never lost but transformed from one form to another through the process of compression and motion.