# A 0.505kg block slides on a frictionless horizontal surface with a speed of 1.18m/s. The block...

## Question:

A 0.505kg block slides on a frictionless horizontal surface with a speed of 1.18m/s. The block encounters an un-stretched spring and compresses 23.2cm before coming to rest. What is the force constant in the spring?

## Force Constant:

The force constant is the property of the spring that defines the magnitude of external force required to compress the spring by one unit. The higher the value of k, the lower will be the force required to compress the spring. The force constant is measured in {eq}(\text{N/m}) {/eq} in the SI unit system.

## Answer and Explanation:

**We are given the following data:**

- Mass of the block , {eq}m=0.505\ \text{kg} {/eq}

- Spring constant, {eq}k= {/eq}_____

- Compression of spring, {eq}x=23.2\ \text{cm} {/eq}

- Speed of the block,{eq}V=1.18\ \text{m/s} {/eq}

**When the block is sliding on the frictionless surface, the kinetic energy stored in the block is expressed by the following equation:**

{eq}\begin{align} \text{K.E}&=\dfrac{1}{2}mV^{2}\\[0.3 cm] &=\dfrac{1}{2}\times0.505\ \text{kg}\times(1.18\ \text{m/s})^{2}\\[0.3 cm] &=0.297\ \text{J} \end{align} {/eq}

**When the block encounters an unstretched spring, the kinetic energy of the block is converted into the energy of the strain energy of the spring. We can calculate the force constant of spring by using the following equation:**

{eq}\begin{align} \text{K.E}&=\text{Strain energy of spring}\\[0.3 cm] 0.297 \text{J}&=\dfrac{1}{2}kx^{2}\\[0.3 cm] 0.297 \text{J}&=\dfrac{1}{2}\times k\times (23.2\times10^{-2}\ \text{m})^{2}\\[0.3 cm] k&\approx \boxed{11.1\ \text{N/m}} \end{align} {/eq}

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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.