Problem 54 - College Physics

Question

\(\bullet\) \(\bullet\) A spring of negligible mass has force constant \(k=\) 1600 \(\mathrm{N} / \mathrm{m}\) (a) How far must the spring be compressed for 3.20 \(\mathrm{J}\) of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then drop a \(1.20-\) -kg book onto it from a height of 0.80 m above the top of the spring. Find the maximum distance the spring will be compressed.

Step-by-Step Solution

Answer

(a) 0.0632 m; (b) 0.108 m.

Step 1: Understanding the Formula for Potential Energy in a Spring

The potential energy stored in a compressed or stretched spring can be calculated using the formula: \[ U = \frac{1}{2} k x^2 \]where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the compression distance from its equilibrium position.

Step 2: Calculate Compression for Stored Energy

To find the distance \( x \) for which the spring stores 3.20 J of energy, set \( U = 3.20 \) J and \( k = 1600 \) N/m in the potential energy formula: \[ 3.20 = \frac{1}{2} \times 1600 \times x^2 \]Solve for \( x \): \[ x^2 = \frac{3.20 \times 2}{1600} = \frac{6.40}{1600} \]\[ x^2 = 0.004 \]\[ x = \sqrt{0.004} \approx 0.0632 \text{ m} \]

Step 3: Calculate Initial Mechanical Energy for Dropped Book

The potential energy when the book is at height 0.80 m is given by:\[ PE = mgh = 1.20 \times 9.8 \times 0.80 \]\[ PE = 9.408 \text{ J} \]This is the potential energy due to gravity before the book hits the spring.

Step 4: Considering Energy Conservation

When the book compresses the spring to its maximum point, all gravitational potential energy is converted into spring potential energy: \[ \frac{1}{2} k x^2 = mgh \]From step 3, \( mgh = 9.408 \text{ J} \). Already known \( k = 1600 \text{ N/m} \). Solve for \( x \): \[ 9.408 = \frac{1}{2} \times 1600 \times x^2 \]\[ 9.408 = 800x^2 \]\[ x^2 = \frac{9.408}{800} \]\[ x = \sqrt{0.01176} \approx 0.108 \text{ m} \]

Step 5: Conclusion for Both Part A and Part B

For part (a), the spring must be compressed about 0.0632 m to store 3.20 J of energy. For part (b), the maximum compression distance of the spring due to the drop of the book is approximately 0.108 m.

Key Concepts

Force Constant

In the study of springs, the term 'force constant', or spring constant, denoted by \( k \), is pivotal. This constant represents the stiffness of a spring. A higher \( k \) value implies a stiffer spring that requires more force to compress or stretch it by a certain length. The unit of \( k \) is Newton per meter (N/m). An understanding of the force constant is crucial when calculating how much potential energy a spring can store when compressed or stretched.

Consider the formula for potential energy in a spring:

• \( U = \frac{1}{2} k x^2 \)

Here, \( U \) stands for potential energy, while \( x \) is the displacement of the spring from its equilibrium position. The equation shows that energy stored in the spring depends on both the force constant and the square of the displacement. In practical applications, this means knowing \( k \) allows you to predict how much energy will be stored for a given compression or stretch, which is crucial in designing systems that use spring mechanics efficiently.

Energy Conservation

Energy conservation is a fundamental principle in physics which states that energy cannot be created or destroyed but only transformed from one form to another. In this exercise, energy conservation becomes evident when a book is dropped onto a spring. Initially, the book possesses gravitational potential energy due to its height above the ground.

As the book falls, this gravitational potential energy is converted into spring potential energy at the point of maximum compression. The equation that reflects this conversion is:

• \( mgh = \frac{1}{2} k x^2 \)

Here, \( mgh \) represents the initial gravitational potential energy, while \( \frac{1}{2} k x^2 \) symbolizes the energy stored in the spring. By setting these two equations equal, you can solve for \( x \), finding the spring's compression distance that matches the book's potential energy loss. This principle is applicable in various practical situations, ensuring efficient design, from amusement park rides to engineering shock absorption systems.

Mechanical Energy

Mechanical energy comprises both potential and kinetic energy in a system. In the context of the spring problem, we focus primarily on potential energy. Initially, the book's position contributes gravitational potential energy, calculated as \( mgh \). Upon impact with the spring, this energy is transformed into another form of potential energy stored within the spring.

The conservation of mechanical energy involves maintaining the total energy of the system constant as long as no external forces, like friction, interfere. This is why understanding mechanical energy components is so vital.

• Potential Energy in Gravity: \( PE = mgh \)
• Potential Energy in a Spring: \( PE = \frac{1}{2} k x^2 \)

These formulas highlight how mechanical energy changes form but remains constant in quantity, providing a predictable and calculable way to assess energy transfers in systems. This principle helps engineers and scientists design systems where energy transitions smoothly, yielding functionality such as energy-efficient springs or controlled motion mechanics.