A 3.00-kg fish is attached to the lower end of a vertical...

Name: Problem 67, Chapter 7: University Physics with Modern Physics
Uploaded: 2023-05-31T15:24:13-08:00
Duration: 4 min 42 s
Channel: Supratim Pal
Description: 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?

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?

Key Concepts

Conservation of Mechanical Energy

This principle states that, in the absence of non-conservative forces (like friction or air resistance), the total mechanical energy of a system remains constant. For a mass-spring system moving in a gravitational field, gravitational potential energy is converted into kinetic energy and elastic potential energy stored in the spring as the mass moves, allowing one to calculate quantities such as speed after a displacement using energy conservation equations.

Hooke's Law

Hooke's Law describes the behavior of springs, establishing that the force exerted by a spring is proportional to its displacement from its rest position. The corresponding potential energy stored in a spring is given by (1/2)kx², where k is the force constant and x is the displacement. This law is central to analyzing systems where elastic forces play a significant role in energy conversion and dynamics.

Static Equilibrium in Mass-Spring Systems

In a vertical mass-spring system, static equilibrium is reached when the downward gravitational force equals the upward spring force (mg = kx). This equilibrium condition is crucial because it defines the point at which the net force on the mass is zero. Understanding this balance is key for determining the point of maximum speed in oscillatory motion, as maximum kinetic energy is typically achieved near the equilibrium position.

Simple Harmonic Motion

Simple harmonic motion (SHM) describes the oscillatory movement of a mass about its equilibrium position in a system with a restoring force proportional to its displacement (as given by Hooke's Law). In SHM, the speed of the mass reaches a maximum value as it passes through the equilibrium position, where the conversion between potential and kinetic energy is most efficient. Recognizing SHM helps in analyzing the dynamic behavior of mass-spring systems including identifying peak speeds and energy distributions.

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:03 So here it's given that a 3 kg fish is attached to the lower end of a vertical spring that has got negligible mass right.

00:11 So the mass of the fish is given as 3 kgs right and it is attached to the lower end of a vertical spring having negligible mass and force constant k equal to 900 newton per meter right.

00:27 The spring initial is neither stretched nor compressed.

00:31 The fish is released from rest.

00:33 Now you have to calculate the speed after it descends 0 .05 meters from its initial position.

00:45 We can take x to be equal to 0 .05 meters right.

00:50 So now what you can do is you can conserve the energy to get the final speed of the fish right.

01:01 So what you have is let's say here we have this spring and here we have the fish right.

01:10 Now this fish has descended a distance of x right and as a result the spring is getting extended right sort of like this it will be.

01:23 That means the potential energy that was initially present in the fish that got converted to this spring potential energy plus the kinetic energy with which it is moving down right.

01:38 So that means you have got if you conserve the energy e1 let's say this is 0 .1 and this is 0 .2 right e1 will be equal to e2 right.

01:47 Energies will remain the same.

01:49 At e1 everything was stationary and the fish has got potential energy of mgx right.

01:56 Now e2 will be the spring potential energy due to elongation plus there will be the kinetic energy of the fish right half mv square...

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