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Chapter 7: Problem 67

\(\mathrm{CALC}\) A \(3.00 \mathrm{~kg}\) 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 fish as it descends?

Short Answer

Expert verified
The speed of the fish after it has descended 0.0500 m is calculated in Step 5 and the maximum speed of the fish as it descends is calculated in Step 6 of the step-by-step solution. These are the final answers.

Step by step solution

01

Calculate initial potential energy

Calculate the initial potential energy of the spring when the fish is at rest. Since the spring is neither stretched nor compressed at this moment, the initial potential energy of the spring \(PE_{initial}\) is 0.
02

Formulate conservation of energy equation

According to the principle of conservation of mechanical energy, the initial mechanical energy should equal the final mechanical energy of the system. That is, \(KE_{initial} + PE_{initial} = KE_{final} + PE_{final}\). Since the fish was initially at rest, its initial kinetic energy is 0. So this equation simplifies to: \(0 + 0 = KE_{final} + PE_{final}\).
03

Calculate final potential energy

When the fish descends by 0.0500 m, the spring is compressed by this distance, so the final potential energy of the spring is given by \(PE_{final} = 1/2 * k * x^2 = 1/2 * 900 * 0.0500^2\).
04

Calculate final kinetic energy

By substituting the value of \(PE_{final}\) into the energy conservation equation from Step 2, the final kinetic energy can be calculated: \(KE_{final} = 0 - PE_{final}\).
05

Calculate speed after descending 0.0500 m

Since the kinetic energy is given by \(KE_{final} = 1/2 * m * v^2\), you can substitute \(KE_{final}\) and the given mass \(m = 3.00 kg\) to solve for \(v\) (speed of the fish after descending 0.0500m): \(v = \sqrt{{2 * KE_{final}} / m}\).
06

Calculate maximum speed of the fish

The maximum speed is attained when all the potential energy is transferred into kinetic energy (i.e., when \(PE_{final} = 0)\). By substituting into the energy conservation equation from Step 2, you get: \(0 + 0 = KE_{final} + 0\). This gives \(KE_{final} = 0\). By substituting this into the kinetic energy equation and solving for speed \(v\), you find the maximum speed of the fish: \(v_{max} = \sqrt{{2 * KE_{final}} / m}\).

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Key Concepts

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

Mechanical Energy
Mechanical energy is a key concept in physics, representing the sum of potential energy and kinetic energy in a system. It is conserved in a closed system where only conservative forces, like gravity, are acting. In our fish and spring scenario, the principle of conservation of mechanical energy is pivotal.
  • Mechanical energy in this exercise initially consists solely of gravitational potential energy when the fish is released.
  • As the fish descends, this energy is converted between potential and kinetic energy.
By understanding that mechanical energy is conserved, you can predict how the energies transform and calculate factors such as speed. This is represented by the formula:\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]which ensures that the total mechanical energy remains constant throughout the motion.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. In this exercise, it is given as 900 N/m, which indicates how much force is needed to stretch or compress the spring by one meter.
  • A higher spring constant means a stiffer spring that resists compression and extension.
  • Using the spring constant, you can determine the potential energy stored in the spring when it is either compressed or stretched.
The formula \( PE_{spring} = \frac{1}{2} k x^2 \) describes the potential energy stored in a spring, where \( x \) is the displacement from the spring's equilibrium position. Understanding \( k \) helps predict how changes in displacement affect the potential energy.
Potential Energy
Potential energy refers to the energy stored within a system due to the position of objects. In this problem, potential energy considers both gravitational and spring potential energy components.
  • Initially, the spring potential energy is zero because the spring is neither compressed nor stretched.
  • As the fish descends 0.0500 meters, the spring is compressed, converting gravitational potential energy into spring potential energy.
The gravitational potential energy \( PE_{gravity} = mgh \) reduces as the fish lowers, while the spring potential energy \( PE_{spring} = \frac{1}{2}kx^2 \) increases. By calculating these values at different points, you can track how energy transforms within the system, aiding in solving for other quantities like speed.
Kinetic Energy
Kinetic energy is the energy of an object due to its motion. It鈥檚 dynamic in the fish and spring scenario, as energy shifts between potential and kinetic forms.
  • Initially, the kinetic energy of the fish is zero because it starts from rest.
  • As it descends, the stored potential energy gets converted into kinetic energy, increasing the fish鈥檚 speed.
To find the kinetic energy of the fish, use the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the fish and \( v \) is its velocity. By knowing the transformations between potential and kinetic energy, you can determine the fish's speed at any given point during its descent.

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Most popular questions from this chapter

CALC 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?

\(\mathrm{A} 15.0 \mathrm{~kg}\) stone slides down \begin{tabular}{l} a snow- covered hill (Fig. \(\mathbf{P 7 . 4 5}\) ), \\ \hline \end{tabular} leaving point \(A\) at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). There is no friction on the hill between points \(A\) and \(B,\) but there is friction on the level ground at the bottom of the hill, between \(B\) and the wall. After entering the rough horizontal region, the stone travels \(100 \mathrm{~m}\) and then runs into a very long. light spring with force constant \(2.00 \mathrm{~N} / \mathrm{m}\). The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and \(0.80,\) respectively. (a) What is the speed of the stone when it reaches point \(B ?\) (b) How far will the stone compress the spring? (c) Will the stone move again after it has been stopped by the spring?

A small box with mass \(0.600 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline that slopes upward at \(37.0^{\circ}\) above the horizontal. The other end of the spring is attached to a wall. The coefficient of kinetic friction between the box and the surface of the incline is \(\mu_{k}=0.400\) The spring is released and the box travels up the incline, leaving the spring behind. What minimum elastic potential energy must be stored initially in the spring if the box is to travel \(2.00 \mathrm{~m}\) from its initial position to the top of the incline?

A proton with mass \(m\) moves in one dimension. The potential-energy function is \(U(x)=\left(\alpha / x^{2}\right)-(\beta / x),\) 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 $$ Graph \(U(x) .\) Calculate \(U\left(x_{0}\right)\) 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 \(\left(x=x_{0}\right.\) and \(\left.x=x_{1}\right),\) what are the maximum and minimum values of \(x\) reached during the motion? U(x)=\frac{\alpha}{x_{0}^{2}}\left[\left(\frac{x_{0}}{x}\right)^{2}-\frac{x_{0}}{x}\right] $$

A wooden block with mass \(1.50 \mathrm{~kg}\) 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 \(6.00 \mathrm{~m}\) up the incline from \(A\), the block is moving up the incline at \(7.00 \mathrm{~m} / \mathrm{s}\) 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.
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