Problem 7 - Applied Mathematics

Question

Prove that the work done in stretching a light elastic string from its natural length \(a\) to a length \((a+x)\) is proportional to \(x^{2}\). One end of this string is fastened to a fixed point \(A\), and at the other end a particle of mass \(m\) is attached. The particle is released from rest at \(A\), and first . comes to rest when it has fallen a distance \(3 a\). Show that at the lowest point of its path the acceleration of the particle is \(2 g\) upwards. Find in terms of \(g\) and \(a\) the speed of the particle at the instants when the magnitude of its acceleration is \(\frac{1}{2} g\). (U of L)

Step-by-Step Solution

Answer

The work done in stretching the string is proportional to \(x^2\). The acceleration at the lowest point is \2g\ upwards, and the speed of the particle is \sqrt{3ga}\.

Step 1: Determine work done in stretching the string

The work done in stretching an elastic string can be found utilizing Hooke's law. For a light elastic string with natural length \(a\), the work \(W\) done in stretching it to length \((a+x)\) can be expressed as \(\frac{1}{2} k x^2\), where \(k\) is the elastic constant.

Step 2: Prove work is proportional to \(x^2\)

Since the work \(W = \frac{1}{2} k x^2\), it clearly shows that the work done is proportional to \(x^2\).

Step 3: Calculate potential energy change due to particle's fall

When the particle falls a distance of \(3a\), the gravitational potential energy lost is \(mg \times 3a = 3mga\).

Step 4: Calculate extension of the string and its potential energy

At the lowest point, the string is extended by \(2a\). The elastic potential energy stored in the string is \(\frac{1}{2} k (2a)^2 = 2ka^2\). From energy conservation, \(3mga=2ka^2\). Hence, \(k=\frac{3mg}{2a}\).

Step 5: Determine acceleration at the lowest point

At the lowest point, the forces acting are the weight of the particle, \(mg\), and the elastic force, \(k \times 2a = 3mg\) upwards, resulting in a net force of \(3mg - mg = 2mg\) upwards. Thus, the acceleration is \(2g\) upwards.

Step 6: Find speed when acceleration is \(\frac{1}{2}g\)

Determine the point where the net force on the particle results in an acceleration of \(\frac{1}{2}g\). If the string's extension is \(x\), then the elastic force is \(kx\) and the net force equation is \(kx - mg = \frac{m}{2}g\). Therefore, \(\frac{3mg}{2a} x - mg = \frac{mg}{2}\). Solving this gives \(x = a\). Using energy conservation, the potential energy is converted into kinetic energy \(\frac{1}{2} k a^2 = \frac{1}{2} mv^2 \), leading to \(\frac{1}{2} k a^2 = \frac{1}{2} mv^2\). Substituting \(k = \frac{3mg}{2a}\), simplifies to \(v^2 = 3ga\). Thus \(v = \frac{2\frac{3a}{3}\frac{g}{g}}{2} v = \frac{2ga^{\frac{3}{2}}\frac{g}{2} = -}\), bringing conclusion as \(v = \frac{\frac{2a^{\frac{3}{2}}g \).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle used in physics to describe the behavior of elastic materials. When you stretch a light elastic string, the force needed to extend it is proportional to the amount of stretch, or displacement, from its original length. Mathematically, this is expressed as: \[F = kx\] where:

• F is the force applied to the string.
• k is the elastic constant (a measure of the string's stiffness).
• x is the extension of the string from its natural length.

Hooke's Law is useful because it helps us calculate how much force is needed to stretch or compress an elastic object. In our problem, a light elastic string follows this law, making the calculations predictable and straightforward.

Work Done in Stretching an Elastic String

When you stretch an elastic string, the work done is stored as elastic potential energy in the string. The work done, \(W\), to stretch the string from its natural length, \(a\), to \(a+x\), can be calculated using: \[ W = \frac{1}{2} k x^2 \] This formula tells us that the work required to stretch the string is proportional to the square of the extension, \(x^2\). This proportionality is key because it shows that as you increase the stretch, the work done increases quadratically.

• If you double the length of the stretch, the work done increases four times.
• This relationship helps us understand the amount of energy stored in the string at different extensions.

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In our problem, the total mechanical energy (potential and kinetic) is conserved. When the particle attached to the string is released, the gravitational potential energy is converted into the elastic potential energy of the stretched string and kinetic energy. Let's denote the key energies involved:

• Gravitational Potential Energy: The energy due to the height of the particle, calculated as \(mgh\), where \(g\) is the acceleration due to gravity and \(h\) is the height.
• Elastic Potential Energy: Stored in the string when stretched, given by \(\frac{1}{2}kx^2\).
• Kinetic Energy: The energy of motion, expressed as \(\frac{1}{2}mv^2\).

As the particle falls, its gravitational potential energy converts into elastic potential energy and any remaining energy becomes kinetic.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field, typically relative to a reference point, such as the ground. The formula for GPE is: \[ GPE = mgh \]where:

• m is the mass of the object.
• g is the acceleration due to gravity.
• h is the height above the reference point.

In our problem, when the particle falls a distance of \(3a\), its gravitational potential energy decreases by:

• \( GPE = mg(3a) = 3mga \).

This lost GPE is converted to elastic potential energy in the string and serves to illustrate the principle of energy conservation.

Acceleration and Forces

The forces acting on the particle include the gravitational force pulling it downward and the elastic force of the stretched string pulling it upward. At different points in its path, the net force acting on the particle changes, leading to different accelerations.

• At the lowest point, the forces include downward gravitational force \(mg\) and upward elastic force \(k \times 2a\).
• The net upward force results in acceleration: \(2mg\), leading to an upward acceleration of \(2g\).
• To find the velocity when the acceleration equals \( \frac{1}{2}g\), we solve for the extension \(x\) using: \[ kx - mg = \frac{mg}{2} \]

By balancing the forces, we derive that the extension needed for this is. Using energy principles, we can then determine the velocity at this specific point.