Problem 58 - College Physics

Question

A vertical spring with a force constant of \(300 \mathrm{~N} / \mathrm{m}\) is compressed \(6.0 \mathrm{~cm}\) and a \(0.25-\mathrm{kg}\) ball placed on top. The spring is released and the ball flies vertically upward. How high does the ball go?

Step-by-Step Solution

Answer

The ball reaches a height of approximately 22.0 cm.

Step 1: Convert Units

First, we convert the units of the spring compression from centimeters to meters. Since there are 100 cm in a meter, the spring compression is \(6.0 \text{ cm} = 0.06 \text{ m}\).

Step 2: Calculate the Spring Potential Energy

The potential energy stored in a compressed spring is given by \(U = \frac{1}{2} k x^2\). Here, \(k = 300 \text{ N/m}\) and \(x = 0.06 \text{ m}\). Substitute into the formula to find \(U\): \[U = \frac{1}{2} \times 300 \times (0.06)^2 = 0.54 \text{ J}.\]

Step 3: Apply the Energy Conservation Principle

When the spring releases the ball, the spring potential energy is converted into gravitational potential energy at the ball's highest point. Thus, we have \(U = mgh\), where \(m = 0.25 \text{ kg}\), \(g = 9.81 \text{ m/s}^2\) (acceleration due to gravity), and \(h\) is the height the ball reaches. Set \(U = mgh\):\[0.54 = 0.25 \times 9.81 \times h.\]

Step 4: Solve for Height

Rearrange the equation from Step 3 to solve for \(h\):\[h = \frac{0.54}{0.25 \times 9.81}.\]Calculate to find the height:\[h \approx 0.220 \text{ m} \text{ or } 22.0 \text{ cm}.\]

Key Concepts

Spring Potential Energy

Spring potential energy is a fascinating concept. It describes the energy stored in a spring when it is either compressed or stretched. Imagine a spring being squeezed; it is eager to return to its original position. This "eagerness" is the potential energy stored in the spring.
To figure out this energy, we use the formula: \[ U = \frac{1}{2} k x^2 \]

• \( k \) stands for the spring constant; it tells us how stiff the spring is. In our exercise, this is given as \(300 \text{ N/m}\).
• \( x \) represents the displacement from the spring's natural length. For example, our spring is compressed by \(0.06 \text{ m}\).

By plugging these values into the formula, we can find that the spring stores roughly \(0.54 \text{ J}\) of energy.
This stored energy has the potential to do work, such as launching a ball upward.

Gravitational Potential Energy

Gravitational potential energy explains the energy an object gains as it moves higher in the air, particularly when opposed by gravity. It can be thought of as the capacity for an object to fall back down under the influence of gravity.
In our exercise, the ball gains gravitational potential energy as it ascends in the air due to the energy imparted from the spring. The energy relationship here is direct, with the spring potential energy converting into gravitational potential energy as the ball reaches its peak height.
The key formula used for gravitational potential energy is:\[ U = mgh \]

• \( m \) stands for the object's mass, which is given as \(0.25 \text{ kg}\).
• \( g \) is the gravitational acceleration constant, approximately \(9.81 \text{ m/s}^2\) on Earth.
• \( h \) represents the height, or how far the ball ascends.

By equating the spring potential energy and gravitational potential energy, we solve for \( h \) to determine how high the ball flies upwards. In our example, that height turns out to be around \(0.220 \text{ m}\), or \(22.0 \text{ cm}\).

Unit Conversion

Unit conversion is a vital skill in solving physics problems. It involves transforming measurements from one unit to another. This step is crucial for ensuring all calculations use compatible measurements.
In our exercise, we initially have the compression of the spring measured in centimeters (\(6.0 \text{ cm}\)). To use it in our energy formulas, we need to convert it to meters, the standard SI unit for length.
Here’s a simple conversion rule we use frequently:

• 100 cm = 1 m

Applying this rule, \(6.0 \text{ cm} = 0.06 \text{ m}\).
Using compatible units like these makes math easier and prevents mistakes, especially in energy calculations where the units need to match SI standards.