Q11 **Problem Statement:**A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to:- ○ 0.96 J- ○ 1.7 J- ○ 0.69 J- ○ 1.3 J- ○ 1.2 J**Explanation:**The problem involves a block undergoing simple harmonic motion on a frictionless surface. It is attached to a spring, and we are interested in finding out the maximum elastic potential energy stored in the system.To calculate the maximum elastic potential energy (\(U\)), use the formula:\[U = \frac{1}{2} k x^2\]where:- \(k = 300 \, \text{N/m}\) is the force constant of the spring,- \(x = 0.080 \, \text{m}\) is the displacement from the equilibrium position.Plug in the values to find \(U\).

Given: Mass of the block (m) = 0.13 kg force constant (K)= 300 N/m displacement from equilibrium…

A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to: O 0.96 J O 1.7 J O 0.69 J 1.3 J O 1.2 J

A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to: O 0.96 J O 1.7 J O 0.69 J 1.3 J O 1.2 J

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Spring Potential Energy

The potential energy is the energy possessed by a body by virtue of its position or the energy held by the object due to its position relative to other objects, its force like stresses, charge, and other factors affecting the particles of the body. The potential energy is the measure of the net work done by or on the body. The spring potential energy is the potential energy stored due to the deformation of an elastic spring and this elasticity is due to the stretched spring. The elasticity is the ability of a solid-state matter to come back to its equilibrium position or original position after any kind of external force is acted on it. It is also known as Elastic potential energy. The potential energy here is caused due to the displacement of the spring from its equilibrium position to another position and this potential energy is equal to the net work done due to the stretching of the spring. It also resembles the same mechanics of the oscillation of a simple pendulum, where due to the external force, the object moves from its equilibrium position to its maximum ends and the periodic motion continues till it comes back to the equilibrium position to rest. 

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Q11

**Problem Statement:**

A 0.13 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 300 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The maximum elastic potential energy of the system is closest to:

- ○ 0.96 J
- ○ 1.7 J
- ○ 0.69 J
- ○ 1.3 J
- ○ 1.2 J

**Explanation:**

The problem involves a block undergoing simple harmonic motion on a frictionless surface. It is attached to a spring, and we are interested in finding out the maximum elastic potential energy stored in the system.

To calculate the maximum elastic potential energy (\(U\)), use the formula:
\[
U = \frac{1}{2} k x^2
\]
where:
- \(k = 300 \, \text{N/m}\) is the force constant of the spring,
- \(x = 0.080 \, \text{m}\) is the displacement from the equilibrium position.

Plug in the values to find \(U\).

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