# A 2.9 kg block slides with a speed of 1.6 m/s on a frictionless horizontal surface until it...

## Question:

A 2.9 kg block slides with a speed of 1.6 m/s on a frictionless horizontal surface until it encounters a spring.

a.) If the block compresses the spring 4.8 cm before coming to rest, what is the force constant of the spring?

b.) What initial speed should the block have to compress the spring by 1.2 cm?

## Strain Energy:

When the spring is at the initial position, the position is termed as the unstretched position. If an object of certain mass strikes the spring, the spring gets compressed and the energy stored in the spring due to compression is called the strain energy.

#### Question (a)

We are given the following data:

• Mass of the block , {eq}m=2.9\ \text{kg} {/eq}
• Spring constant, {eq}k= {/eq}_____
• Compression of spring, {eq}x=4.8\ \text{cm} {/eq}
• Speed of the block,{eq}V=1.6\ \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}\times2.9\ \text{kg}\times(1.6\ \text{m/s})^{2}\\[0.3 cm] &=3.712\ \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] 3.712 \text{J}&=\dfrac{1}{2}kx^{2}\\[0.3 cm] 3.712 \text{J}&=\dfrac{1}{2}\times k\times (4.8\times10^{-2}\ \text{m})^{2}\\[0.3 cm] k&\approx \boxed{3222.2\ \text{N/m}} \end{align} {/eq}

#### Question (b)

If the spring is compressed by 1.2 cm, the strain energy stored in the spring is expressed by the following equation:

{eq}\text{Strain energy of spring}=\dfrac{1}{2}kx'^{2} {/eq}

Recall value of spring constant from part (a), we have:

{eq}\begin{align} E&=\dfrac{1}{2}kx'^{2}\\[0.3 cm] &=\dfrac{1}{2}\times(3222.2\ \text{N/m})\times(1.6\times10^{-2}\ \text{m})^{2}\\[0.3 cm] &=0.412\ \text{J} \end{align} {/eq}

By equating the strain energy stored in the spring to the kinetic energy of the block, we can calculate the initial velocity {eq}(V') {/eq} of the block:

Recall value of {eq}(m) {/eq} from part (a).

{eq}\begin{align} \text{K.E}&=E\\[0.3 cm] 0.412\ \text{J}&=\dfrac{1}{2}mV'^{2}\\[0.3 cm] 0.412\ \text{J}&=\dfrac{1}{2}\times2.9\ \text{kg}\times V'^{2}\\[0.3 cm] V'&=\boxed{0.533\ \text{m/s}} \end{align} {/eq} Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.4K

In 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.