Question

A block of mass \(0.773 \mathrm{~kg}\) on a spring with spring constant \(239.5 \mathrm{~N} / \mathrm{m}\) oscillates vertically with amplitude \(0.551 \mathrm{~m}\). What is the speed of this block at a distance of \(0.331 \mathrm{~m}\) from the equilibrium position? 6.49 A spring with \(k=10.0 \mathrm{~N} / \mathrm{cm}\) is initially stretched \(1.00 \mathrm{~cm}\) from its equilibrium length. a) How much more energy is needed to further stretch the spring to \(5.00 \mathrm{~cm}\) beyond its equilibrium length? b) From this new position, how much energy is needed to compress the spring to \(5.00 \mathrm{~cm}\) shorter than its equilibrium position?

Short answer

Question: Calculate the additional energy needed to stretch a spring from 1 cm to 5 cm beyond its equilibrium length, and the energy needed to compress the spring to 5 cm shorter than its equilibrium position from the 5 cm stretched position. The spring constant is 10 N/cm. Answer: The additional energy needed to stretch the spring to 5 cm beyond its equilibrium length is 1.2 J, and the energy needed to compress the spring to 5 cm shorter than its equilibrium position from the 5 cm stretched position is 3.75 J.

Step-by-step solution

Write down the given values

The spring constant k is 10 N/cm, and the initial stretched length is 1 cm.

Convert to SI units

Convert length values from centimeters to meters and spring constant from N/cm to N/m: 1 cm = 0.01 m, and 1 N/cm = 100 N/m. So, k = 1000 N/m, the initial stretch is 0.01 m, and the final stretch is 0.05 m.

Calculate the potential energy at the initial and final positions

We will use the formula for the potential energy stored in a spring: \(U = \frac{1}{2}kx^2\). Calculate the potential energy at the initial position (0.01 m) and the final position (0.05 m): \(U_{initial} = \frac{1}{2}(1000)(0.01)^2 = 0.05 \mathrm{~J}\) \(U_{final} = \frac{1}{2}(1000)(0.05)^2 = 1.25 \mathrm{~J}\)

Find the difference in potential energy

Subtract the initial potential energy from the final potential energy to find the additional energy needed to further stretch the spring to 5 cm beyond the equilibrium length: \(\Delta U = U_{final} - U_{initial} = 1.25 - 0.05 = 1.2 \mathrm{~J}\) Answer for part a: The additional energy needed to further stretch the spring to 5 cm beyond its equilibrium length is 1.2 J. ##Part b##

Determine the spring's compressed position

From the 5 cm stretched position, we need to compress the spring to 5 cm shorter than its equilibrium position, which is equal to compressing the spring 10 cm relative to its current position.

Calculate the potential energy at the compressed position

Convert the compressed length from centimeters to meters (10 cm = 0.1 m). Then, calculate the potential energy at the compressed position (0.1 m compressing): \(U_{compressed} = \frac{1}{2}(1000)(-0.1)^2 = 5 \mathrm{~J}\)

Find the difference in potential energy

Subtract the potential energy at the final stretched position (5 cm beyond the equilibrium) from the potential energy at the compressed position to find the energy needed to compress the spring to 5 cm shorter than its equilibrium position: \(\Delta U = U_{compressed} - U_{final} = 5 - 1.25 = 3.75 \mathrm{~J}\) Answer for part b: The energy needed to compress the spring to 5 cm shorter than its equilibrium position from the 5 cm stretched position is 3.75 J.

Key concepts

Hooke's Law

Understanding how springs work under force is essential in physics. Hooke's Law is a principle that helps us describe this exactly. It states that the force needed to extend or compress a spring by some distance (\( x \) ) is proportional to that distance. Mathematically, it's represented as \( F = -kx \) , where \( F \) is the force applied, \( k \) is the spring constant, indicating the stiffness of the spring, and \( x \) is the distance the spring is stretched or compressed from its original equilibrium position.

The negative sign indicates that the direction of the force is opposite to the direction of displacement, which is why the spring tries to return to its equilibrium position. The spring constant, \( k \) , can be determined experimentally and is different for each spring.

Hooke's Law is fundamental in solving problems involving spring potential energy. When dealing with oscillating blocks on springs or calculating the extra energy required to stretch or compress a spring, Hooke's Law provides the necessary starting point.

Elastic Potential Energy

When a spring is stretched or compressed, it stores elastic potential energy. This is the energy that has the potential to do work as the spring returns to its equilibrium state. The formula to calculate the elastic potential energy in a spring is given by \( U = \frac{1}{2}kx^2 \) , where \( U \) is the potential energy, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium.

In our exercise, we used this formula to determine the energy stored in a spring at different stretching points. The energy is purely elastic as long as we stay within the elastic limits of the spring. If stretched too far, springs can reach a plastic deformation range, where they won't return to their original shape, hence losing the 'elastic' in 'elastic potential energy'. For typical high-school physics problems, we assume the material remains elastic.

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 opposite direction of that displacement. It is a model for understanding various motions like the swing of a pendulum or the oscillation of a spring.

An object in SHM oscillates around an equilibrium position, and its motion can be described by sine or cosine functions. The most familiar example is a mass attached to a spring, where, if displaced and released, it will move back and forth around the equilibrium position. This motion is characterized by properties like amplitude, period, frequency, and energy conservation.

In our exercise, the block attached to a spring exhibits SHM, and using the principles of this type of motion, we can predict the speed of the block at any point in its oscillation cycle.

Mechanical Energy Conservation

The principle of mechanical energy conservation states that if only conservative forces, like gravitational and elastic forces, are acting on a system, then the total mechanical energy of the system remains constant. Mechanical energy is the sum of kinetic energy and potential energy. During SHM, the energy continuously transforms back and forth between kinetic and potential forms, but their sum remains unchanged.

For instance, as the block on the spring in our exercise moves from the equilibrium position, its kinetic energy decreases as potential energy increases, and vice versa. Knowing the total mechanical energy helps us find values like maximum speed or maximum displacement without enumerating all the forces at play. In our step-by-step solution, we use this conservation principle to find the energy needed to compress the spring from a stretched position without calculating the work done during the compression.