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Chapter 4: Problem 21

An ideal spring with spring-constant \(k\) is hung from the ceiling and a block of mass \(m\) is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is (A) \(\frac{4 m g}{k}\) (B) \(\frac{2 m g}{k}\) (C) \(\frac{m g}{k}\) (D) \(\frac{m g}{2 k}\)

Short Answer

Expert verified
The maximum extension in the spring after the mass is released is (B) \(\frac{2mg}{k}\).

Step by step solution

01

Identify the forces acting on the block

In this problem, the forces acting on the block are gravitational force (weight) due to Earth's gravity and the restoring force due to the spring. The gravitational force (weight) can be written as: \( W = mg \) The restoring force due to the spring can be written using Hooke's law as: \( F_{spring} = -kx \) where x is the extension in the spring.
02

Apply conservation of mechanical energy

As there are no external forces acting on the system (the block and spring), the mechanical energy is conserved. The mechanical energy of the system can be represented as the sum of the potential energy due to the spring and the gravitational potential energy: \( E_{mechanical} = U_{spring} + U_{gravitational} \) Initially, when the spring is unstressed, all of the energy is in the form of gravitational potential energy: \( E_{initial} = mgh = mg(0) = 0 \) When the spring reaches its maximum extension x, the potential energy due to the spring and the gravitational force must be equal to the initial mechanical energy: \( E_{final} = \frac{1}{2}kx^2 + mgx \) As E_initial = E_final, we can say: \( \frac{1}{2}kx^2 + mgx = 0 \)
03

Solve for maximum extension x

To solve for the maximum extension x, we can factor the equation: \( x(\frac{1}{2}kx + mg) = 0 \) There are two possibilities: x = 0 or \((\frac{1}{2}kx + mg) = 0\). As we are interested in the maximum extension, x cannot be 0. Thus, we have: \( \frac{1}{2}kx + mg = 0 \) Rearranging for x, we get: \( x = \frac{-2mg}{k} \) But, as x must be a positive value and distance cannot be negative: \( x = \frac{2mg}{k} \) So, the correct option is (B) \(\frac{2 m g}{k}\).

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Most popular questions from this chapter

A uniform chain has a mass \(m\) and length \(L .\) It is placed on a frictionless table with length \(l_{0}\) hanging over the edge. The chain begins to slide down. The speed \(v\) with which the chain slides away from the edge is given by (A) \(v=\sqrt{\frac{g l_{0}}{L}\left(L+l_{0}\right)}\) (B) \(v=\sqrt{\frac{g l_{0}}{L}\left(L-l_{0}\right)}\) (C) \(v=\sqrt{\frac{g}{L}\left(L^{2}-l_{0}^{2}\right)}\) (D) \(v=\sqrt{2 g\left(L-l_{0}\right)}\)

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