Problem 1
Question
(1) A spring has a spring constant \(k\) of 82.0 \(\mathrm{N} / \mathrm{m} .\) How much must this spring be compressed to store 35.0 \(\mathrm{J}\) of potential energy?
Step-by-Step Solution
Verified Answer
The spring must be compressed approximately 0.924 meters.
1Step 1: Understand the Formula for Potential Energy in a Spring
The potential energy (PE) stored in a compressed or stretched spring can be calculated using the formula: \[ PE = \frac{1}{2} k x^2 \]where \(k\) is the spring constant, and \(x\) is the displacement or compression of the spring. In this exercise, the potential energy \(PE\) is given as 35.0 \(\mathrm{J}\), and the spring constant \(k\) is 82.0 \(\mathrm{N/m}\). We need to find \(x\).
2Step 2: Rearrange the Formula to Solve for Compression
To find \(x\), we need to rearrange the potential energy formula:\[ 35.0 = \frac{1}{2} \times 82.0 \times x^2 \]To isolate \(x^2\), we first multiply both sides by 2 to get rid of the fraction:\[ 70.0 = 82.0 \times x^2 \]Then, divide both sides by 82.0 to solve for \(x^2\):\[ x^2 = \frac{70.0}{82.0} \]
3Step 3: Calculate the Compression Distance
Now calculate \(x^2\):\[ x^2 = \frac{70.0}{82.0} \approx 0.85366 \]Next, take the square root of both sides to find \(x\):\[ x = \sqrt{0.85366} \approx 0.924 \]Thus, the spring must be compressed approximately 0.924 meters.
Key Concepts
Hooke's LawSpring ConstantMechanical Energy
Hooke's Law
Hooke's Law is fundamental to understanding how springs behave under forces. It states that the force required to compress or extend a spring is directly proportional to the amount of displacement it experiences. This is expressed as:
Understanding and applying this law is crucial when dealing with problems involving potential energy in springs. It helps us identify how much force a spring can handle before it becomes permanently deformed.
- \( F = kx \)
Understanding and applying this law is crucial when dealing with problems involving potential energy in springs. It helps us identify how much force a spring can handle before it becomes permanently deformed.
Spring Constant
The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to extend or compress the spring by a unit length. The spring constant is either given in scenarios like our problem or can be calculated through experimentation.Think of it this way:
Knowing the spring constant is essential for calculating potential energy and dealing with mechanical energy in systems involving springs.
- A high spring constant means the spring is stiff, requiring more force to change its length.
- A low spring constant indicates a more flexible spring.
Knowing the spring constant is essential for calculating potential energy and dealing with mechanical energy in systems involving springs.
Mechanical Energy
Mechanical energy is the sum of kinetic and potential energy in any given system. When dealing with springs, we often focus on potential energy, especially when they are compressed or stretched. In the spring potential energy formula:
This energy can be converted from potential to kinetic when the spring is released, propelling objects attached to it or returning to its original form.
Understanding this energy transformation is crucial. It allows us to grasp how mechanical energy can be conserved and manipulated in systems. It also helps us solve real-world problems, such as those requiring calculations of spring compression or energy storage.
- \( PE = \frac{1}{2} k x^2 \)
This energy can be converted from potential to kinetic when the spring is released, propelling objects attached to it or returning to its original form.
Understanding this energy transformation is crucial. It allows us to grasp how mechanical energy can be conserved and manipulated in systems. It also helps us solve real-world problems, such as those requiring calculations of spring compression or energy storage.
Other exercises in this chapter
Problem 1
(I) A spring has a spring constant \(k\) of \(82.0 \mathrm{~N} / \mathrm{m} .\) How much must this spring be compressed to store \(35.0 \mathrm{~J}\) of potenti
View solution Problem 2
(I) A 6.0-kg monkey swings from one branch to another \(1.3 \mathrm{~m}\) higher. What is the change in gravitational potential energy?
View solution Problem 3
(II) A spring with \(k=63 \mathrm{~N} / \mathrm{m}\) hangs vertically next to a ruler. The end of the spring is next to the \(15-\mathrm{cm}\) mark on the ruler
View solution Problem 4
(II) A \(56.5-\mathrm{kg}\) hiker starts at an elevation of \(1270 \mathrm{~m}\) and climbs to the top of a 2660-m peak. (a) What is the hiker's change in poten
View solution