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Chapter 7: Problem 67

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?

Short Answer

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

Step by step solution

01

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 \(\frac{1}{2}kx^2\).
02

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:\[3.00 \times 9.81 \times 0.0500 = \frac{1}{2} \times 3.00 \times v^2 + \frac{1}{2} \times 900 \times (0.0500)^2\]
03

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 \(v^2\):\[\frac{1}{2} \times 3.00 \times v^2 = 1.4715 - 1.125 = 0.3465\]\[v^2 = \frac{0.3465}{1.5}\]\[v = \sqrt{\frac{0.3465}{1.5}} = 0.48\, \text{m/s}\]
04

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 \(v_{max}^2\):\[v_{max}^2 = \frac{1.4715}{1.5} = 0.981\]\[v_{max} = \sqrt{0.981} = 0.990\, \text{m/s}\]

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Key Concepts

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

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.

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Most popular questions from this chapter

A baseball is thrown from the roof of a 22.0-m-tall building with an initial velocity of magnitude 12.0 m/s and directed at an angle of 53.1\(^\circ\) above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of 53.1\(^\circ\) \(below\) the horizontal? (c) If the effects of air resistance are included, will part (a) or (b) give the higher speed?

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x) = (\alpha/x^2) - (\beta/x)\), where \(\alpha\) and \(\beta\) are positive constants. The proton is released from rest at \(x_0\) = \(\alpha\)/\(\beta\). (a) Show that \(U(x)\) can be written as $$U(x) = \frac{\alpha}{x^2_0}\ \Big[ \Big( \frac{x_0}{x}\ \Big)^2 - \frac{x_0}{x}\ \Big] $$ Graph \(U(x)\). Calculate \(U(x_0)\) and thereby locate the point \(x_0\) on the graph. (b) Calculate \(v(x)\), the speed of the proton as a function of position. Graph \(v(x)\) and give a qualitative description of the motion. (c) For what value of \(x\) is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the proton be released instead at \(x_1 = 3\alpha / \beta\). Locate the point \(x_1\) on the graph of \(U(x)\). Calculate \(v(x)\) and give a qualitative description of the motion. (f) For each release point (\(x = x_0\) and \(x = x_1\)), what are the maximum and minimum values of \(x\) reached during the motion?

The maximum energy that a bone can absorb without breaking depends on characteristics such as its cross-sectional area and elasticity. For healthy human leg bones of approximately 6.0 cm\(^2\) cross-sectional area, this energy has been experimentally measured to be about 200 J. (a) From approximately what maximum height could a 60-kg person jump and land rigidly upright on both feet without breaking his legs? (b) You are probably surprised at how small the answer to part (a) is. People obviously jump from much greater heights without breaking their legs. How can that be? What else absorbs the energy when they jump from greater heights? (\(Hint\): How did the person in part (a) land? How do people normally land when they jump from greater heights?) (c) Why might older people be much more prone than younger ones to bone fractures from simple falls (such as a fall in the shower)?

A wooden block with mass 1.50 kg is placed against a compressed spring at the bottom of an incline of slope 30.0\(^\circ\) (point \(A\)). When the spring is released, it projects the block up the incline. At point \(B\), a distance of 6.00 m up the incline from A, the block is moving up the incline at 7.00 m/s and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_k =\) 0.50. The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x) = -C_6/x^6\), where \(C_6\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?
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