Question

When an elastic spring is given a displacement of \(10 \mathrm{~mm}\), it gains an potential energy equal to \(\mathrm{U}\). If this spring is given an additional displacement of \(10 \mathrm{~mm}\), then its potential energy will be.............. (A) \(\mathrm{U}\) (B) \(2 \mathrm{U}\) (C) \(4 \mathrm{U}\) (D) \(\mathrm{U} / 4\).

Short answer

The potential energy when the spring is further displaced by 10 mm is 4 times the initial potential energy. So, the correct answer is (C) \(4 \mathrm{U}\).

Step-by-step solution

Write down Hooke's Law formula for potential energy stored in a spring.

Hooke's Law states that the potential energy (PE) stored in a spring is proportional to the square of its elongation (displacement) from the equilibrium position. The formula for the potential energy is: \[ PE = \frac{1}{2} k x^2 \] Where: PE = potential energy stored in the spring, k = spring constant (a factor specific to the spring), x = displacement of the spring from its equilibrium position.

Write the formula for the initial potential energy U.

Using Hooke's Law formula, we'll write down the potential energy (U) when the spring is displaced by 10 mm (0.01 m). \[ U = \frac{1}{2} k(0.01)^2 \]

Calculate the potential energy when the spring is further displaced by 10 mm.

Now, we need to find the potential energy when the spring is further displaced by another 10 mm (0.01 m). The total displacement now is 20 mm (0.02 m). Let's call the new potential energy as V. \[ V = \frac{1}{2} k(0.02)^2 \]

Simplify and compare the new potential energy V with the initial potential energy U.

Let's rewrite and simplify the equations for U and V: \[ U = \frac{1}{2} k(0.01)^2 = \frac{1}{2} k \times 0.0001 \] \[ V = \frac{1}{2} k(0.02)^2 = \frac{1}{2} k \times 0.0004 \] Now, we can see that V is 4 times greater than U: \[ V = 4U \]

Choose the correct answer option.

From the analysis, we have found that the potential energy when the spring is further displaced by 10 mm is 4 times the initial potential energy. So, the correct answer is: (C) \(4 \mathrm{U}\)

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes the behavior of springs and elastic materials. When you apply a force to stretch or compress a spring, Hooke's Law tells us that the force required is directly proportional to the amount of displacement from the spring's original position. This relationship can be expressed with the formula:\[F = -kx\]Here,

• \( F \) represents the restoring force exerted by the spring.
• \( k \) is the spring constant, which measures how stiff the spring is.
• \( x \) is the displacement from the spring's equilibrium position.

Note that the negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement. This ensures the spring always returns to its equilibrium position, explaining why it bounces back when stretched or compressed.

Spring Constant

The spring constant, denoted as \( k \), is a crucial value that determines the stiffness of a spring. It is a measure of how difficult it is to compress or stretch the spring. A high spring constant means the spring is stiff, and requires more force for the same amount of displacement compared to a spring with a lower constant.The units for the spring constant \( k \) are Newtons per meter (N/m). To find the spring constant in a practical situation, you can rearrange Hooke's Law:\[k = \frac{F}{x}\]This calculation allows you to determine \( k \) if you know the force exerted on the spring and the displacement it caused. Understanding the spring constant is vital for designing systems where springs play a crucial role, such as in vehicle suspension systems or measuring forces in instruments.

Elastic Potential Energy

Elastic potential energy is the energy stored in an elastic object, particularly in springs, when it is deformed. Whether you stretch or compress a spring, it accumulates energy that can be released when the object returns to its original shape. This energy is described by the formula:\[PE = \frac{1}{2}kx^2\]Where:

• \( PE \) is the potential energy stored in the spring.
• \( k \) is the spring constant.
• \( x \) is the displacement from the equilibrium position.

The formula highlights that potential energy is directly proportional to the square of the displacement, which means if the displacement doubles, the potential energy increases by four times. This quadratic relationship is key to predicting how the energy stored in a spring changes with varying displacements.

Displacement in Springs

Displacement in springs refers to how much a spring is stretched or compressed from its equilibrium position. When you move a spring from its original shape, you're essentially changing its potential energy. Displacement can be a result of applying a force, and it's measured in meters (m).Understanding displacement is essential when calculating the potential energy stored. When a change is made in displacement, it dramatically affects elastic potential energy, as shown by the formula \( PE = \frac{1}{2}kx^2 \). A small change in displacement can lead to a significant change in energy stored, due to its squared relationship in the formula.When solving problems involving springs, always ensure that you measure displacement consistently, converting units if necessary to maintain accuracy. For instance, if the displacement is given in millimeters, convert it to meters before performing energy calculations to keep everything in compatible units.