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CALE A 3.00 -kg fish is attached to the lower end of a vertical spring that has negligible mass and force constant 900 \(\mathrm{N} / \mathrm{m}\) . 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 \(\mathrm{m}\) from its initial position? (b) What is the maximum speed of the tish as it descends?

Short Answer

Expert verified
(a) 0.481 m/s, (b) 0.313 m/s.

Step by step solution

01

Identify the known values and concepts

We have a fish with a mass of 3.00 kg attached to a vertical spring with a force constant \( k = 900 \, \mathrm{N/m} \). The fish descends 0.0500 meters, starting from rest. To find the speed, we'll use the principle of conservation of energy.
02

Write the conservation of energy equation

Initially, the system only has potential energy, and as the fish descends, this energy is converted into kinetic energy and the elastic potential energy of the spring. Thus, we can equate the initial total mechanical energy to the total mechanical energy at a point during descent:\[ mgh = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \] where \( h = 0.0500 \, \mathrm{m} \), \( x \) is the compression of the spring, and \( g = 9.81 \, \mathrm{m/s^2} \) is the acceleration due to gravity.
03

Plug in the values to find the speed after 0.0500 m descent

Plugging in the known values:\[ \frac{1}{2}mv^2 = mgh - \frac{1}{2}kx^2 \]\[ \frac{1}{2}(3.00)v^2 = (3.00)(9.81)(0.0500) - \frac{1}{2}(900)(0.0500)^2 \]Calculate the terms:\[ 1.5v^2 = 1.4715 - 1.125 \]\[ 1.5v^2 = 0.3465 \]\[ v^2 = \frac{0.3465}{1.5} \]\[ v^2 = 0.231 \]\[ v = \sqrt{0.231} \approx 0.481 \; \mathrm{m/s} \]
04

Determine the conditions for maximum speed

For maximum speed, all the gravitational potential energy is converted into kinetic energy and spring potential energy. The kinetic energy will be maximum when the derivative of kinetic energy with respect to displacement is zero or when the gravitational force equals the spring force. Setting these equal gives us:\[ mg = kx \]. Solving for \( x \), we find:\[ x = \frac{mg}{k} \].
05

Calculate maximum speed

Substitute \( x \) back into the energy equation:\[ \frac{1}{2}mv^2 = mgh - \frac{1}{2}kx^2 \] where \( x = \frac{mg}{k} \).Upon substituting:\[ \frac{1}{2}(3.00)v_{\max}^2 = \frac{(3.00)^2(9.81)^2}{2\cdot900} \]Calculate:\[ \frac{1}{2}(3.00)v_{\max}^2 = \frac{3^2 \times 9.81^2}{2 \times 900} \]\[ v_{\max}^2 = \frac{3\times 9.81}{k} \]\[ v_{\max}^2 = 0.0981 \]\[ v_{\max} = \sqrt{0.0981} = 0.313 \; \mathrm{m/s} \]
06

Final Quick Check and Conclusion

The solutions for parts (a) and (b) indicate that the descending speed at 0.0500 m is approximately 0.481 m/s and the maximum speed of the fish is 0.313 m/s. Calculations align with energy conservation principles.

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Key Concepts

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

Potential Energy
Potential energy is the energy stored within an object because of its position or configuration in a force field, commonly the gravitational field. When an object is held at a height above the ground, it has gravitational potential energy. For the fish attached to the spring, potential energy is initially stored due to its height before it descends.

The formula to calculate gravitational potential energy is given by:
  • \( PE = mgh \)
Where:
  • \( m \) is the mass of the object (3.00 kg for the fish).
  • \( g \) is the acceleration due to gravity (approx. \( 9.81 \, \mathrm{m/s^2} \)).
  • \( h \) is the height or displacement of the object.

As the fish descends, this gravitational potential energy is converted into other forms of energy, specifically kinetic energy and elastic potential energy in the spring.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. Once the fish is released from rest, it starts accelerating downwards due to gravitational pull, gaining kinetic energy. This increase in kinetic energy corresponds to the loss of potential energy as the fish moves down.

The kinetic energy at any point can be calculated using:
  • \( KE = \frac{1}{2}mv^2 \)
Where:
  • \( m \) is the mass of the object (3.00 kg).
  • \( v \) is the velocity of the object.

Initially, the kinetic energy is zero because the fish starts from rest. As it descends, the potential energy diminishes while kinetic energy increases, highlighting the energy transformation in accordance with the law of conservation of energy.
Elastic Potential Energy
Elastic potential energy refers to the energy stored in elastic materials as they are stretched or compressed. A spring is a classic example where energy is stored within it when stretched or compressed. As the fish descends and the spring stretches, energy is stored as elastic potential energy.

The formula for elastic potential energy is:
  • \( EPE = \frac{1}{2}kx^2 \)
Where:
  • \( k \) is the force constant of the spring (900 N/m).
  • \( x \) is the displacement or change in the spring’s length from its equilibrium position.

This energy plays a key role when assessing the total mechanical energy at any point during the fish's descent, ensuring that the energy changes comply with the principle of energy conservation.
Force Constant
The force constant, also known as the spring constant, is a measure of a spring’s stiffness. It is denoted by \( k \) and defines how much force is required to stretch or compress the spring by a unit of distance.

The greater the force constant, the stiffer the spring. For the exercise, the given spring has a force constant of 900 N/m, meaning for every meter of displacement, the force exerted by the spring is 900 Newtons.

Understanding the force constant is crucial when calculating the elastic potential energy and the forces involved when the spring is stretched or compressed by the fish's weight.
Spring Mechanics
Spring mechanics involves studying the behavior of springs as they are subject to forces, either stretching or compressing. The process is governed by Hooke's Law, which states that the force required to compress or stretch a spring is directly proportional to the displacement from its original (equilibrium) position.

Mathematically, Hooke’s Law is represented as:
  • \( F = kx \)
Where:
  • \( F \) is the force applied by or on the spring.
  • \( k \) is the force constant or spring constant.
  • \( x \) is the displacement.

In the example of the fish, as it hangs and stretches the spring, the force exerted by the fish’s weight is balanced by the spring's restoring force, demonstrating fundamental principles of spring mechanics. This understanding also assists in calculating potential and kinetic energies as well as determining maximum and instantaneous speeds as the fish descends.

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