# A body of mass 5 kg slides a distance of 6 m down a rough inclined plane (30 degrees). Then it...

## Question:

A body of mass 5 kg slides a distance of 6 m down a rough inclined plane (30 degrees). Then it moves on frictionless horizontal surface and compresses a spring. The coefficient of kinetic friction is 0.1 and the spring constant is 300 N/m. Find the maximum compression of the spring.

## Spring Energy:

Spring energy is a type of potential energy that exists in spring due to external mechanical work done on it. So we can say that the external work done is stored in the spring as its spring energy.

The spring energy is proportional to the square of the deformation length of the spring

## Answer and Explanation: 1

Given:

- The mass of the body is, {eq}m = 5\ \text{kg} {/eq}

- The sliding distance is, {eq}d = 6\ \text{m} {/eq}

- The coefficient of kinetic friction is {eq}\mu_k = 0.1 {/eq}

- The spring constant is {eq}k = 300 \text{N/m} {/eq}

- The inclination angle is, {eq}\theta = 30^\circ {/eq}

. Let the maximum compression of the spring is, {eq}x {/eq}

For the moving body, its initial gravitational potential energy at the top of the incline is equal to the sum of the work done by the frictional force and the final spring energy.

$$\begin{align} m\ g\ d\ \sin\ \theta & = \mu_k\ m\ g\ d \cos\ \theta + 0.5\ k\ x^2\\ (5\ \text{kg})\ (9.81\ \text{m/s}^2)\ ( 6\ \text{m})\ \sin\ 30^\circ &= 0.1\ (5\ \text{kg})\ (9.81\ \text{m/s}^2)\ ( 6\ \text{m})\ \cos\ 30^\circ + 0.5\ (300\ \text{N/m})\ x^2\\ \implies x& =\boxed{ 0.9\ \text{m}}\\ \end{align} $$

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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.