Question

A vertical spring with spring constant \(23.73 \mathrm{~N} / \mathrm{m}\) is hanging from a ceiling. A small object with mass \(1.103 \mathrm{~kg}\) is attached to the lower end of the spring, and the spring stretches to its equilibrium length. The object is then pulled down and released. The speed of the object when it is \(4.985 \mathrm{~cm}\) from the equilibrium position is \(0.4585 \mathrm{~m} / \mathrm{s}\). How far down was the object pulled?

Short answer

Based on the given information and using conservation of mechanical energy, we have determined that the object was pulled down 1.2 cm from its equilibrium position before being released.

Step-by-step solution

Find the equilibrium length of the spring

To find the equilibrium length, we need to balance the gravitational force acting on the object and the force exerted by the spring. The forces acting on the object are gravitational force \(F_g = mg\) and the spring force \(F_s = kx\), where \(k\) is the spring constant and \(x\) is the displacement of the object from its equilibrium position. At equilibrium, \(F_s = F_g\), so \(kx = mg\). Rearrange the equation to solve for the displacement \(x\): \(x=\frac{mg}{k}\) Use the given values: \(m=1.103\mathrm{~kg}\), \(g=9.81 \mathrm{~m/s^2}\), and \(k=23.73 \mathrm{~N/m}\): \(x=\frac{(1.103)(9.81)}{23.73}=0.457\mathrm{~m}\)

Calculate the initial potential energy in the spring

Now, we can calculate the potential energy in the spring when it is at its equilibrium length using the formula: \(U_s=\frac{1}{2}kx^2\) Plugging in the values we found in step 1: \(U_s=\frac{1}{2}(23.73)(0.457)^2 = 2.472\mathrm{~J}\)

Calculate the total mechanical energy in the system

Initially, when the object was pulled down, there was only potential energy in the spring. When the object is at a distance of \(4.985\mathrm{~cm}\) from the equilibrium position, it has a velocity of \(0.4585\mathrm{~m/s}\). Let's find the total mechanical energy in the system when the object moves: \(E_{total}= K + U_s + U_g\) where \(K=\frac{1}{2}mv^2\) is the kinetic energy of the object, \(U_s\) is the potential energy in the spring and \(U_g=mgh\) is the gravitational potential energy of the object. Plugging in the values: \(E_{total} = (\frac{1}{2}(1.103)(0.4585)^2) + (\frac{1}{2}(23.73)(0.457)^2) + (1.103)(9.81)(h)\) \(E_{total} = 1.640\mathrm{~J}\)

Use conservation of mechanical energy to find the distance the object was pulled down

Since the mechanical energy is conserved in this problem, we can set the initial total mechanical energy equal to the final total mechanical energy: \(E_{initial}=E_{total}\) Initially, there was only potential energy stored in the spring: \(\frac{1}{2}kx_{initial}^2=1.640\mathrm{~J}\) We can solve for the initial displacement \(x_{initial}\): \(x_{initial}=\sqrt{\frac{2(1.640)}{23.73}}=0.469\mathrm{~m}\) Finally, we find the difference in initial and equilibrium displacement: \(Δx=x_{initial}-x=0.469-0.457=0.012\mathrm{~m}=1.2\mathrm{~cm}\) The object was pulled down 1.2 cm from its equilibrium position.

Key concepts

Understanding Hooke's Law

Hooke's Law is a principle of physics that describes the behavior of springs when they are subjected to external forces. It states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it's expressed as:
\( F = -kx \)
Here, \( F \) represents the force applied to the spring, \( x \) is the displacement of the spring from its equilibrium position, and \( k \) is the spring constant, a measure of the stiffness of the spring. The negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement.
In our problem, when the mass causes the spring to stretch to its equilibrium length, the spring force balances with the gravitational force on the mass, illustrating Hooke's Law in action. This relationship allows us to calculate the equilibrium position and analyze the forces involved in stretching or compressing the spring. When dealing with spring constant physics problems, understanding Hooke's Law is essential to predicting how a spring will behave under the influence of a given force.

Mechanical Energy Conservation

The principle of mechanical energy conservation is a fundamental concept that asserts that in a closed system with no external forces, the total mechanical energy is constant. The total mechanical energy is the sum of kinetic energy, which is energy due to motion, and potential energy, which is energy stored due to position. In the context of springs and masses, the potential energy can be the elastic potential energy of the spring and the gravitational potential energy of the mass.
The formula for the conservation of mechanical energy is:
\( E_{total} = K + U_{s} + U_{g} \)
In our spring constant physics problem, we use this principle to calculate the total energy when the mass is a certain distance from the equilibrium and use that to determine how far the mass was initially pulled down. By observing that the total mechanical energy at the starting point (where the mass is pulled down and still) is equal to the total energy at any other point in the motion, students can solve for unknown variables such as displacement.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that displacement. An object in SHM moves back and forth about an equilibrium position. Some familiar examples of SHM include the motion of a pendulum, the vibrations of a guitar string, and the oscillation of a mass attached to a spring.
The motion of the mass in our exercise is an example of simple harmonic motion, where the mass oscillates about the equilibrium position of the spring. The speed of the mass changes as it moves, being greatest at the equilibrium point and reducing to zero at the points of maximum displacement. The mechanical energy conservation equation we use is part of analyzing SHM, since it involves potential and kinetic energy that transform into each other as the mass oscillates. By fully understanding SHM, students can predict the motion characteristics such as amplitude, period, frequency, and phase of the object undergoing this type of motion.