Problem 27 - College Physics

Question

\(\bullet\) \(\bullet\) A spring of force constant 300.0 \(\mathrm{N} / \mathrm{m}\) and unstretched length 0.240 \(\mathrm{m}\) is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to 15.0 \(\mathrm{N} .\) How long will the spring now be, and how much work was required to stretch it that distance?

Step-by-Step Solution

Answer

The spring's new length is 0.290 m and 0.375 J of work was required.

Step 1: Understand Hooke's Law

Hooke's Law describes the relationship between the force applied to a spring and the amount of stretch (or compression) it experiences. The formula is given by \[ F = kx \]where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the deformation (stretch or compression) from its original length.

Step 2: Calculate the stretch of the spring

Given that the total force applied is 15.0 N and the force constant \( k = 300.0 \, \mathrm{N/m} \), we can find the stretch using the formula derived from Hooke's Law:- Solving for \( x \), we have\[ x = \frac{F}{k} = \frac{15.0}{300.0} = 0.050 \, \mathrm{m} \]

Step 3: Determine the new length of the spring

The original length of the spring is given as 0.240 m. The spring stretches an additional 0.050 m. Therefore, the new length is:\[ \text{New Length} = 0.240 + 0.050 = 0.290 \, \mathrm{m} \]

Step 4: Calculate the work done in stretching the spring

The work done \( W \) on the spring is given by the formula:\[ W = \frac{1}{2} k x^2 \] Substituting the values into the equation:\[ W = \frac{1}{2} (300.0) (0.050)^2 = \frac{1}{2} \times 300.0 \times 0.0025 = 0.375 \, \mathrm{J} \]

Key Concepts

The Spring Constant

When discussing springs, the spring constant is crucial. It’s represented as \( k \) in physics formulas and measures a spring's stiffness. The higher the constant, the stiffer the spring. This constant is unique to each spring and is expressed in Newtons per meter (\( \mathrm{N/m} \)).
To imagine this, consider a slinky and a tight metal spring. The metal spring will have a much higher spring constant compared to the slinky because it resists deformation more. In our exercise, the spring had a constant of 300.0 \( \mathrm{N/m} \), meaning it requires 300 Newtons of force to stretch it by one meter.
Understanding the spring constant is essential because it helps predict how much force is needed to stretch or compress the spring by a certain length. This helps in various applications like designing suspension systems in cars or measuring forces in mechanical testing.

Work Done on a Spring

The concept of work done on a spring is tied closely to stretching or compressing it. To calculate the work, one uses the formula \( W = \frac{1}{2} k x^2 \), where \( W \) is the work done, \( k \) is the spring constant, and \( x \) denotes the deformation or stretch from its original position.
In simple terms, work is the energy transferred to the spring, enabling it to stretch. It's calculated by incorporating both the force applied and the distance the spring moved (or changed) from its rest state.

• In the context of our exercise, applying a force to stretch the spring by \( 0.050 \) m resulted in work done amounting to \( 0.375 \) Joules.
• This reflects not only the magnitude of force but also how much the spring constant impacts energy requirements.

Knowing how to find the work done on springs is crucial for engineering tasks where energy efficiency and system performance matter.

Force and Deformation

Force and deformation are central to understanding Hooke's Law. This law provides a straightforward correlation: more force results in more deformation, but only to the spring's limit:

• Mathematically, \( F = kx \)], where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the deformation.

When force is applied, the spring stretches from its original length. This deformation is directly proportional to the applied force, as long as the material's elastic limit isn't exceeded.
For example, if two springs of identical lengths have different spring constants, a greater force will deform the weaker spring more compared to the stiffer one.
In our example with the spring, a force of 15 N led to an additional stretch of \( 0.050 \) m from the original length. This practicality is what allows engineers and designers to create systems that can reliably resist, absorb, or release forces.

Elastic Potential Energy

Elastic potential energy is stored energy in an elastically deformed system like a stretched spring. It's 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 extent of deformation.
This potential energy is all about the ability of the spring to do work when released. When we stretch or compress a spring, we store energy, which can be used to do work when the spring returns to its natural length.

• In our scenario, the elastic potential energy stored in the spring as it stretched by \( 0.050 \) meters was exactly \( 0.375 \) Joules.

Understanding elastic potential energy is vital for many practical applications, including designing toys, watches, and various mechanical devices. It’s a clean, efficient form of energy that’s easy to harness in systems requiring precise mechanical movements.