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Chapter 14: Problem 39

A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; (b) at its lowest point; (c) at its equilibrium position.

Short Answer

Expert verified
At each point, the total energy is always 1.96 J.

Step by step solution

01

Identify Parameters

Mass of the cat, \( m = 4.00 \text{ kg} \). Amplitude of the oscillation, \( A = 0.050 \text{ m} \).At the highest point, the spring is unstretched, which implies at the highest point, the potential energy due to the spring is 0.
02

Calculate energies at the highest point (a)

At the highest point, the spring is unstretched, so the elastic potential energy \( U_s = 0 \).There is no kinetic energy as the velocity is zero, so \( K = 0 \).Gravitational potential energy \( U_g = mgh \) with \( h = A = 0.050 \text{ m} \):\[ U_g = 4.00 \times 9.8 \times 0.050 = 1.96 \text{ J} \].Therefore, the total energy \( E = U_s + K + U_g = 0 + 0 + 1.96 = 1.96 \text{ J} \).
03

Calculate energies at the lowest point (b)

At the lowest point, the spring is stretched by \( 2A \).Elastic potential energy \( U_s = \frac{1}{2} k x^2 \) where \( x = A + A = 0.100 \text{ m} \). Using total energy conservation from highest point, \( U_s = 1.96 \text{ J} \) and \( U_g = 0 \) because it is at the lowest point.Kinetic energy \( K = 0 \) because total mechanical energy remains constant.So, the total energy \( E = U_s + K + U_g = 1.96 + 0 + 0 = 1.96 \text{ J} \).
04

Calculate energies at the equilibrium position (c)

At the equilibrium position, the spring is stretched by \( A \).Elastic potential energy \( U_s = \frac{1}{2} k x^2 \) with \( x = A \), so \( U_s = 0.5 \times k \times (0.050)^2 = 0.5 \times \frac{1.96}{(0.100)^2} \times (0.050)^2 = 0.49 \text{ J} \).Kinetic energy \( K = E - U_s = 1.96 \text{ J} - 0.49 \text{ J} = 1.47 \text{ J} \).Gravitational potential energy \( U_g = mgh = m \times 9.8 \times 0 \), so \( U_g = 0 \).So the total energy \( E = U_s + K + U_g = 0.49 + 1.47 + 0 = 1.96 \text{ J} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Potential Energy
Elastic potential energy is the energy stored in a spring when it is either compressed or stretched from its natural length. In this scenario, a spring attached to a cat oscillates vertically, creating both compressed and stretched positions throughout its oscillation cycle. The formula used to calculate elastic potential energy is:\[U_s = \frac{1}{2} k x^2\]- **Variables:** - \( k \): Spring constant - \( x \): Displacement from the spring's natural length, also known as the stretching or compressing amount.When the spring is stretched or compressed, it stores energy. At the highest point, the spring returns to its natural, unstretched position, thus having zero elastic potential energy, as seen in part (a) of the solution. However, at the lowest point, the spring is stretched to twice the amplitude, so the energy is quite significant, calculated to be 1.96 J. At the equilibrium position, the cat's weight causes the spring to stretch only to the amplitude (0.050 m), leading to 0.49 J of stored energy.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its position relative to a lower point under the influence of gravity. This energy changes as the cat, attached to the spring, moves vertically during its oscillation.- **Formula:**\[U_g = mgh\]- **Variables:** - \( m \): Mass of the object (the cat, in this case) - \( g \): Acceleration due to gravity (approximately 9.8 m/s虏) - \( h \): Height above a chosen reference point (the lowest point in this problem)In this exercise, at the highest point, the gravitational potential energy is 1.96 J, because the cat is 0.050 m above the lowest point. However, at the lowest point, the energy is zero since we've defined it as our reference point. At the equilibrium position, the cat's potential energy is zero again as the vertical displacement from the lowest point is not considered.
Kinetic Energy
Kinetic energy is the energy that an object has due to its motion, and it is a crucial part of the understanding of Simple Harmonic Motion (SHM). As the cat moves, its kinetic energy becomes a significant part of the energy exchange in the system.The kinetic energy is given by:\[K = \frac{1}{2} mv^2\]- **Variables:** - \( m \): Mass of the object - \( v \): Velocity of the objectIn the context of SHM:- At the highest point, the velocity is zero since the spring is momentarily at rest, making the kinetic energy 0 J.- At the lowest point, again, the velocity reduces to zero due to change in direction.- At the equilibrium position, however, kinetic energy is maximized because the velocity is greatest here, calculated to be 1.47 J. This is when the speed of the cat is highest as it passes through the midpoint of its oscillation, maintaining the balance of total energy observed in SHM systems.

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Most popular questions from this chapter

A holiday ornament in the shape of a hollow sphere with mass \(M =\) 0.015 kg and radius \(R =\) 0.050 m is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (\(Hint\): Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

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