# A 40.0 g object connected to a spring with a force constant of 40.0 N/m oscillates on a...

## Question:

A {eq}40.0 g {/eq} object connected to a spring with a force constant of {eq}40.0 \frac{N}{m} {/eq} oscillates on a horizontal, frictionless surface with an amplitude of {eq}7.00 cm {/eq}.

(a) Find the total energy of the system.

(b) Find the speed of the object when the position is {eq}1.15 cm {/eq}.

(c) Find the kinetic energy when the position is {eq}2.50 cm {/eq}.

(d) Find the potential energy when the position is {eq}2.50 cm {/eq}.

## Spring Potential Energy:

At any given moment, an object oscillating on a spring has a total energy consisting of kinetic energy {eq}K {/eq} and potential energy {eq}U {/eq}.

{eq}E = K + U {/eq}

What is the potential energy associated with a spring? It is equal to the work necessary to stretch the object back a given distance.

{eq}U = W {/eq}

In many cases, we can use the formula {eq}W = Fd {/eq}, but in this case the force is not constant. We need to use the integral definition of force.

{eq}U = \int Fdx {/eq}

The spring force is {eq}F = kx {/eq} ({eq}k {/eq} is the spring constant, {eq}x {/eq} is the displacement from equilibrium), so we can plug this in.

{eq}U = \int kx \ dx {/eq}

Finally, we can solve this integral using the power rule.

{eq}U = \frac{1}{2}kx^2 {/eq}

## Answer and Explanation: 1

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(a) The total energy of the system is equal to kinetic energy plus potential energy.

{eq}E = K + U\\ E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 {/eq}

W...

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#### Learn more about this topic: Hooke's Law & the Spring Constant: Definition & Equation

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Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.