Problem 32 - Thomas Calculus

Question

A ball bearing \(A\) 2-oz steel ball bearing is placed on a vertical spring whose force constant is \(k=18 \mathrm{lb} / \mathrm{ft}\) . The spring is com- pressed 2 in. and released. About how high does the ball bearing go?

Step-by-Step Solution

Answer

The ball bearing goes 1 ft high.

Step 1: Convert units

First, convert the units of mass and displacement into compatible units. The ball bearing is 2 oz, which is equivalent to \( \frac{2}{16} = 0.125 \) lbs. The spring is compressed 2 inches, which is \( \frac{2}{12} = \frac{1}{6} \) ft.

Step 2: Calculate potential energy in spring

The potential energy stored in the compressed spring is given by \( PE = \frac{1}{2} k x^2 \). Substituting the given values: \( k = 18 \text{ lb/ft} \) and \( x = \frac{1}{6} \text{ ft} \) gives \[ PE = \frac{1}{2} \times 18 \times \left( \frac{1}{6} \right)^2 = \frac{1}{2} \times 18 \times \frac{1}{36} = \frac{1}{4} \text{ lb-ft} \].

Step 3: Relate potential energy to gravitational potential energy

When the spring releases its potential energy, this energy is converted into gravitational potential energy at the highest point of the ball's trajectory. Hence, \( PE_{spring} = mgh \) where \( m = \frac{1}{32} \) for the weight of ball in slugs (1 lb = 32 slugs ft/s²), \( g = 32 \text{ ft/s}^2 \), and \( h \) is the height.

Step 4: Solve for height

Substitute the potential energy and other known values into the equation: \( \frac{1}{4} = \frac{1}{32} \times 32 \times h \). Solving for \( h \), we find \( h = 1 \text{ ft} \).

Key Concepts

Potential Energy

Potential energy is a type of energy that an object possesses due to its position or state. In the context of springs, it is the energy stored when the spring is compressed or stretched. The formula used to calculate the potential energy ( PE ) in a spring is:

• \( PE = \frac{1}{2} k x^2 \)

where \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
The stronger the spring (higher \( k \)) or the more you compress or stretch it (larger \( x \)), the greater the potential energy.
Once released, this stored energy can be converted to kinetic or other forms of energy, such as gravitational potential energy in this problem.

Springs

Springs are mechanical devices that can store and release energy. They are characterized by a spring constant ( k ), which measures the stiffness of the spring. The higher the spring constant, the stiffer the spring.

When a spring is compressed or stretched from its natural length, it exerts a force trying to return to its original position. This force is described by Hooke's Law:

• \( F = -kx \)

where \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement.

In many physics problems, like the one we are examining, springs are idealized to assume no energy loss.
This means all energy stored in the spring is converted to other forms of energy, such as potential energy, when the spring is released.

Unit Conversion

Unit conversion is a critical skill when solving physics problems to ensure consistency and accuracy.
In this problem, we primarily converted:

• Mass: from ounces to pounds, since there are 16 ounces in a pound.
  Thus, 2 ounces = \( \frac{2}{16} = 0.125 \) lbs.
• Displacement: from inches to feet, with 12 inches in a foot.
  So, 2 inches = \( \frac{2}{12} = \frac{1}{6} \) ft.
• Weight to mass in slugs: using the relationship that 1 lb is equivalent to 1/32 slugs under Earth's gravity.

Converting units correctly ensures that all calculations remain correct throughout the problem, especially when dealing with formulas involving constants such as the gravitational acceleration \( g = 32 \text{ ft/s}^2 \).

Projectile Motion

Projectile motion refers to the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. In this exercise, once the spring releases the ball bearing, the bearing follows a trajectory similar to a basic projectile motion.

Key factors in projectile motion include:

• Initial velocity, which in this instance, is provided by the released energy of the spring.
• Gravitational acceleration, which acts downwards to bring the ball back to the ground. Here it's \( 32 \text{ ft/s}^2 \).
• Maximum height, reached when all kinetic energy is converted to gravitational potential energy. This happens at the top of the ball bearing's path before it descends.

In our problem, this height was determined to be 1 foot by equating the spring's potential energy with the gravitational potential energy at this peak height. Understanding such transformations between energy types is essential in solving projectile motion problems.