# A block of mass 3.00 kg is placed against a horizontal spring of constant k = 875 N/m and pushed...

## Question:

A block of mass {eq}\rm 3.00 \ kg {/eq} is placed against a horizontal spring of constant {eq}\rm k = 875 \ N/m {/eq} and pushed so the spring compresses by {eq}\rm 0.070 \ m. {/eq}

(a) What is the elastic potential energy of the block spring system?

(b) If the block is now released and the surface is frictionless, calculate the block s speed after leaving the spring.

## Spring Elastic Potential Energy

The elastic potential energy on spring can be determined using a simple equation defined as {eq}E = \frac{1}{2}kx^2 {/eq}. Remind that the elastic potential energy is proportional to the compression distance and spring constant.

## Answer and Explanation:

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View this answer**Part (a):**We solve for the elastic potential energy using the equation

{eq}\displaystyle E = \frac{1}{2}kx^2 {/eq}

where,

- {eq}k = 875\ N/m {/eq}...

See full answer below.

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Chapter 17 / Lesson 11In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.