Problem 67 - University Physics with Modern Physics

Question

A 3.00-kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 900 N/m. The spring initially is neither stretched nor compressed. The fish is released from rest. (a) What is its speed after it has descended 0.0500 m from its initial position? (b) What is the maximum speed of the fish as it descends?

Step-by-Step Solution

Answer

(a) 0.48 m/s, (b) 0.990 m/s.

Step 1: Analyze the Energy Conservation Principle

The fish is initially released from rest, implying that its initial kinetic energy is zero, and it has potential energy due to its position. As it descends, it will exchange potential energy with kinetic energy while also storing some energy in the spring (elastic potential energy). According to the conservation of energy principle:\[E_i = E_f\]where the initial energy \(E_i\) is gravitational potential energy \(mgh\), and the final energy \(E_f\) consists of kinetic energy \(\frac{1}{2}mv^2\) and elastic potential energy \(\frac{1}{2}kx^2\).

Step 2: Set Up the Energy Conservation Equation

For the moment when the fish has descended 0.0500 m, the energies involved are:\[E_i = mgh = 3.00\, \text{kg} \times 9.81\, \text{m/s}^2 \times 0.0500\, \text{m}\]\[E_f = \frac{1}{2}mv^2 + \frac{1}{2}kx^2\]Setting initial and final energies equal gives:\[3.00 \times 9.81 \times 0.0500 = \frac{1}{2} \times 3.00 \times v^2 + \frac{1}{2} \times 900 \times (0.0500)^2\]

Step 3: Calculate the Speed After Descending 0.0500 m

First, calculate the gravitational potential energy and the elastic potential energy:\[E_g = 3.00 \times 9.81 \times 0.0500 = 1.4715\, \text{J}\]\[E_{elastic} = \frac{1}{2} \times 900 \times (0.0500)^2 = 1.125\, \text{J}\]Substituting back in the energy conservation equation:\[1.4715 = \frac{1}{2} \times 3.00 \times v^2 + 1.125\]Solve for \(v^2\):\[\frac{1}{2} \times 3.00 \times v^2 = 1.4715 - 1.125 = 0.3465\]\[v^2 = \frac{0.3465}{1.5}\]\[v = \sqrt{\frac{0.3465}{1.5}} = 0.48\, \text{m/s}\]

Step 4: Determine the Maximum Speed of the Fish

The maximum speed occurs when all the initial potential energy has been converted into kinetic energy right before it contributes further to elastic potential energy. Thus \(E_{g}\) converts solely into kinetic energy.\[E_i = \frac{1}{2}mv_{max}^2\]Substitute \(E_i = E_g = 1.4715\) from earlier:\[1.4715 = \frac{1}{2} \times 3.00 \times v_{max}^2\]Solve for \(v_{max}^2\):\[v_{max}^2 = \frac{1.4715}{1.5} = 0.981\]\[v_{max} = \sqrt{0.981} = 0.990\, \text{m/s}\]

Key Concepts

Energy Conservation

Energy conservation is a fundamental concept in physics that states the total energy in a closed system remains constant. In the case of the fish attached to a spring, the principle of energy conservation helps us understand how different forms of energy are transformed. When the fish is released from rest, all of its energy is initially stored as gravitational potential energy. As it descends, this energy is converted into two other forms:

• Kinetic energy, which is due to the fish's motion.
• Elastic potential energy, stored in the spring as it stretches.

By using the equation \(E_i = E_f\), where \(E_i\) is the initial energy and \(E_f\) is the final energy, we can analyze these transformations. This ensures that no energy is lost, just altered, which is the essence of energy conservation. It is crucial to consider each form of energy at different points as the fish moves, to solve for variables such as speed or displacement.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position relative to a gravitational source. For our exercise, GPE is determined by the formula \(mgh\), where:

• \(m\) is the mass of the object (3.00 kg in this problem).
• \(g\) is the acceleration due to gravity, approximately \(9.81 \text{ m/s}^2\).
• \(h\) is the height or distance the object has moved vertically (0.0500 meters in this case).

When the fish is released, it initially holds only GPE. As it descends and the spring stretches, part of this energy changes into elastic potential energy, while another part converts into kinetic energy when the fish moves. By calculating GPE, we determine the energy available to be transformed into these other types of energy.

Elastic Potential Energy

Elastic potential energy is the energy stored in an object when it is stretched or compressed. In this spring-mass system, the spring's elasticity allows it to store energy as the fish pulls it downward. The elastic potential energy in a spring is calculated using the formula \(\frac{1}{2}kx^2\), where:

• \(k\) is the spring constant (900 N/m here), a measure of the spring's stiffness.
• \(x\) is the displacement from the equilibrium position, which is 0.0500 meters in this exercise.

When the fish descends, the spring stretches, and elastic potential energy increases. At different points of the descent, some of the gravitational potential energy initially carried by the fish morphs into elastic potential energy. Understanding this transformation is crucial, as it contributes to determining parameters like the speed of the fish after a certain distance, and identifying when the speed is at its maximum as other energies diminish.