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A force of 520 N keeps a certain spring stretched a distance of 0.200 m. (a) What is the potential energy of the spring when it is stretched 0.200 m? (b) What is its potential energy when it is compressed 5.00 cm?

Short Answer

Expert verified
(a) 52 J for 0.200 m stretch; (b) 3.25 J for 5 cm compression.

Step by step solution

01

Understand Hooke's Law

Hooke's Law states that the force needed to stretch or compress a spring is proportional to the displacement from its equilibrium position. This can be expressed as \( F = kx \), where \( F \) is the force applied, \( x \) is the displacement, and \( k \) is the spring constant.
02

Calculate the Spring Constant

Using Hooke's Law, we can find the spring constant \( k \). Given that \( F = 520 \text{ N} \) and \( x = 0.200 \text{ m} \), we use the formula \( k = \frac{F}{x} \). Substituting the values, we get \( k = \frac{520}{0.200} = 2600 \, \text{N/m}. \)
03

Calculate Potential Energy for Stretched Spring

The potential energy stored in a spring is given by the formula \( PE = \frac{1}{2}kx^2 \). For a stretch of \( 0.200 \text{ m} \), we use \( k = 2600 \text{ N/m} \) and \( x = 0.200 \text{ m} \). Thus, \( PE = \frac{1}{2} \times 2600 \times (0.200)^2 = 52 \text{ J}. \)
04

Calculate Potential Energy for Compressed Spring

For compression of \( 0.050 \text{ m} \) (since 5.00 cm = 0.050 m), using the same spring constant \( k = 2600 \text{ N/m} \), the potential energy is \( PE = \frac{1}{2} \times 2600 \times (0.050)^2 = 3.25 \text{ J}. \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
When dealing with springs, one of the most important values to understand is the spring constant, often denoted as \( k \). This constant acts as a measure of a spring's stiffness.
This value tells us how much force is needed to stretch or compress the spring by a certain distance.
Here's what you need to know about the spring constant:
  • Measured in Newtons per meter (N/m).
  • A high spring constant means a stiff spring, needing more force to stretch it.
  • A low spring constant indicates a less stiff spring, needing less force for the same amount of stretch.
By using Hooke's Law, expressed as \( F = kx \), where \( F \) is the force applied and \( x \) is the displacement from the equilibrium position, we can determine the value of \( k \).
In our example, with a force \( F = 520\text{ N} \) and a displacement \( x = 0.200 \text{ m} \), simply divide the force by the displacement: \( k = \frac{520}{0.200} = 2600 \text{ N/m} \).
This means the spring is quite stiff.
Potential Energy
When a spring is stretched or compressed, it stores energy which is known as potential energy. This energy is related to the work done to change the spring's displacement from its equilibrium position.
Potential energy in the spring can be found through the formula:\[PE = \frac{1}{2}kx^2\]Where:
  • \( PE \) is the potential energy.
  • \( k \) is the spring constant.
  • \( x \) is the displacement from the equilibrium.
In our example:
  • For a displacement of \( 0.200 \text{ m} \), using \( k = 2600 \text{ N/m} \), the potential energy calculated is \( 52 \text{ J} \).
  • For a compression displacement of \( 0.050 \text{ m} \), the potential energy is \( 3.25 \text{ J} \).
It shows that the potential energy depends strongly on both the displacement and the spring's stiffness.
Equilibrium Position
The equilibrium position is the point where the spring is neither stretched nor compressed, effectively in a state of rest. It is from this neutral point that any displacement is considered for calculating force and potential energy.
Understanding equilibrium helps in determining how much a spring stretches or compresses under a specific force.
In the context of Hooke's Law, the displacement \( x \) is measured from this position.
Key points about equilibrium position:
  • It is the default, unstressed length of the spring.
  • Any deviation from this point requires a force to either stretch or compress the spring.
  • Only when the displacement from the equilibrium position is positive, do we need to apply Hooke's Law or calculate potential energy.
Force Applied
Force is a critical component when discussing springs, as it's the initial action that causes a change in displacement from the equilibrium position.
The force applied to a spring can be calculated based on how much the spring stretches or compresses according to Hooke's Law.
Here's how force plays a role:
  • Expressed mathematically as \( F = kx \), where \( k \) is the spring constant and \( x \) is the displacement.
  • A larger force results in a greater stretch or compression, assuming a constant spring stiffness.
  • In practical terms, the applied force lets us determine the potential energy stored in the spring.
With our example, knowing the force applied was \( 520\text{ N} \), we could find the displacement details along with the spring constant to calculate potential energy.

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Most popular questions from this chapter

A 1.20-kg piece of cheese is placed on a vertical spring of negligible mass and force constant \(k =\) 1800 N/m that is compressed 15.0 cm. When the spring is released, how high does the cheese rise from this initial position? (The cheese and the springare \(not\) attached.)

A force parallel to the \(x\)-axis acts on a particle moving along the x-axis. This force produces potential energy \(U(x)\) given by \(U(x) = \alpha x^4\), where \(\alpha =\) 0.630 J/m\(^4\). What is the force (magnitude and direction) when the particle is at \(x = -0.800\) m?

A 75-kg roofer climbs a vertical 7.0-m ladder to the flat roof of a house. He then walks 12 m on the roof, climbs down another vertical 7.0-m ladder, and finally walks on the ground back to his starting point. How much work is done on him by gravity (a) as he climbs up; (b) as he climbs down; (c) as he walks on the roof and on the ground? (d) What is the total work done on him by gravity during this round trip? (e) On the basis of your answer to part (d), would you say that gravity is a conservative or nonconservative force? Explain.

A spring stores potential energy \(U_0\) when it is compressed a distance \(x_0\) from its uncompressed length. (a) In terms of \(U_0\), how much energy does the spring store when it is compressed (i) twice as much and (ii) half as much? (b) In terms of \(x_0\), how much must the spring be compressed from its uncompressed length to store (i) twice as much energy and (ii) half as much energy?

A spring of negligible mass has force constant \(k =\) 1600 N/m. (a) How far must the spring be compressed for 3.20 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 0.800 m above the top of the spring. Find the maximum distance the spring will be compressed.

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