Question

A spring is compressed by \(1 \mathrm{~cm}\) by a force of \(4 \mathrm{~N}\). Find the potential energy of the spring when it is compressed by \(10 \mathrm{~cm}\) (A) \(2 \mathrm{~J}\) (B) \(0.2 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(200 \mathrm{~J}\)

Short answer

The potential energy of the spring when it is compressed by 10 cm is \(2 \mathrm{~J}\). The correct answer is (A) \(2 \mathrm{~J}\).

Step-by-step solution

Apply Hooke's Law to find the spring constant

According to Hooke's Law, the force needed to compress a spring is proportional to its displacement (compression). Mathematically, it is represented as: \(F = kx\) where \(F\) is the force needed to compress the spring, \(k\) is the spring constant, and \(x\) is the displacement (the amount the spring is compressed). We are given that when the spring is compressed by 1 cm (0.01 m), the force required is 4 N. We can plug these values into Hooke's Law to find the spring constant, \(k\). \(4 \mathrm{~N} = k(0.01 \mathrm{~m})\)

Calculate the spring constant

To find the spring constant \(k\), we can simply divide the force by the displacement: \(k = \dfrac{4 \mathrm{~N}}{0.01 \mathrm{~m}}\) \(k = 400 \mathrm{~N/m}\) Now we have the spring constant, which is 400 N/m.

Apply the Potential Energy formula to find the potential energy when the spring is compressed by 10 cm

The potential energy stored in a compressed spring is given by the formula: PE = 0.5 × k × x^2 where PE is the potential energy, k is the spring constant, and x is the displacement (the amount the spring is compressed). We are given that the spring is compressed by 10 cm (0.1 m), and we have already found the spring constant, so now we can plug these values into the formula to find the potential energy. PE = 0.5 × 400 N/m × (0.1 m)^2

Calculate the potential energy

Now, we just need to compute the potential energy: PE = 0.5 × 400 N/m × (0.1 m)^2 PE = 0.5 × 400 N/m × 0.01 m^2 PE = 200 N/m × 0.01 m^2 PE = 2 J The potential energy of the spring when it is compressed by 10 cm is 2 J. The correct answer is (A) \(2 \mathrm{~J}\).

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes the behavior of springs in reaction to applied forces. It states that the force required to compress or stretch a spring is directly proportional to the distance it is deformed. This can be mathematically expressed as:\[ F = kx \]Where:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant, which indicates the stiffness of the spring.
• \( x \) is the displacement, the change from the spring's natural length.

This principle allows us to calculate the spring constant, as shown in the problem, by rearranging the equation to solve for \( k \). When using correct units, we get precise insights into how a spring behaves under varying forces.

Spring Constant

The spring constant, symbolized as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to compress or extend the spring by a unit of length. It's crucial in determining how a spring will react under load.To calculate the spring constant, you can use the formula derived from Hooke's Law:\[ k = \frac{F}{x} \]Where:

• \( F \) is the force applied.
• \( x \) is the amount of displacement from the equilibrium position, typically in meters.

In the exercise, by knowing that a force of 4 N compresses the spring by 0.01 m, we calculated the spring constant as 400 N/m. This information is critical for further calculations involving the spring, including determining its stored potential energy.

Elastic Potential Energy

Elastic potential energy is the energy stored in objects as a result of mechanical deformation, such as stretching or compressing, within the limits of elasticity (meaning the material returns to its original shape). For springs, this energy's formula is given by:\[ PE = \frac{1}{2} k x^2 \]Where:

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

This particular formula is quadratic with respect to \( x \), which means the energy increases with the square of the displacement. Hence, doubling the compression or extension quadruples the stored energy. In the exercise, calculating with \( x = 0.1 \) m for a spring constant of 400 N/m gives a potential energy of 2 J when compressed.

Compression Displacement

Compression displacement refers to the change in position that occurs when force is applied to a spring. It is the difference between the spring's natural length and its length when compressed. In our exercise, understanding compression displacement is crucial since all calculations depend on the accurate measurement of this parameter. The greater the displacement, the more energy is stored in the spring. It's measured in meters in the SI unit system, but can also be given in centimeters or other units depending on context. For precise calculations, always convert small units like centimeters to meters by dividing by 100. This ensures that the spring constant and potential energy calculations align properly within standard measurement units.